13 4 problem solving with trigonometry answers

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13.4: Right Triangle Trigonometry

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Learning Objectives

• Use right triangles to evaluate trigonometric functions.
• Find function values for 30°(\(\dfrac{\pi}{6}\)),45°(\(\dfrac{\pi}{4}\)),and 60°(\(\dfrac{\pi}{3}\)).
• Use equal cofunctions of complementary angles.
• Use the definitions of trigonometric functions of any angle.
• Use right-triangle trigonometry to solve applied problems.

Mt. Everest, which straddles the border between China and Nepal, is the tallest mountain in the world. Measuring its height is no easy task and, in fact, the actual measurement has been a source of controversy for hundreds of years. The measurement process involves the use of triangles and a branch of mathematics known as trigonometry. In this section, we will define a new group of functions known as trigonometric functions, and find out how they can be used to measure heights, such as those of the tallest mountains.

We have previously defined the sine and cosine of an angle in terms of the coordinates of a point on the unit circle intersected by the terminal side of the angle:

\[ \begin{align*} \cos t &= x \\ \sin t &=y \end{align*} \]

In this section, we will see another way to define trigonometric functions using properties of right triangles .

Using Right Triangles to Evaluate Trigonometric Functions

In earlier sections, we used a unit circle to define the trigonometric functions . In this section, we will extend those definitions so that we can apply them to right triangles. The value of the sine or cosine function of \(t\) is its value at \(t\) radians. First, we need to create our right triangle. Figure \(\PageIndex{1}\) shows a point on a unit circle of radius 1. If we drop a vertical line segment from the point \((x,y)\) to the x -axis, we have a right triangle whose vertical side has length \(y\) and whose horizontal side has length \(x\). We can use this right triangle to redefine sine, cosine, and the other trigonometric functions as ratios of the sides of a right triangle.

\[ \cos t= \frac{x}{1}=x \]

Likewise, we know

\[ \sin t= \frac{y}{1}=y \]

These ratios still apply to the sides of a right triangle when no unit circle is involved and when the triangle is not in standard position and is not being graphed using \((x,y)\) coordinates. To be able to use these ratios freely, we will give the sides more general names: Instead of \(x\),we will call the side between the given angle and the right angle the adjacent side to angle \(t\). (Adjacent means “next to.”) Instead of \(y\),we will call the side most distant from the given angle the opposite side from angle \(t\). And instead of \(1\),we will call the side of a right triangle opposite the right angle the hypotenuse . These sides are labeled in Figure \(\PageIndex{2}\).

Understanding Right Triangle Relationships

Given a right triangle with an acute angle of \(t\),

\[\begin{align} \sin (t) &= \dfrac{\text{opposite}}{\text{hypotenuse}} \label{sindef}\\ \cos (t) &= \dfrac{\text{adjacent}}{\text{hypotenuse}} \label{cosdef}\\ \tan (t) &= \dfrac{\text{opposite}}{\text{adjacent}} \label{tandef}\end{align}\]

A common mnemonic for remembering these relationships is SohCahToa, formed from the first letters of “ S ine is o pposite over h ypotenuse, C osine is a djacent over h ypotenuse, T angent is o pposite over a djacent.”

how to: Given the side lengths of a right triangle and one of the acute angles, find the sine, cosine, and tangent of that angle

• Find the sine as the ratio of the opposite side to the hypotenuse.
• Find the cosine as the ratio of the adjacent side to the hypotenuse.
• Find the tangent is the ratio of the opposite side to the adjacent side.

Example \(\PageIndex{1}\): Evaluating a Trigonometric Function of a Right Triangle

Given the triangle shown in Figure \(\PageIndex{3}\), find the value of \(\cos α\).

The side adjacent to the angle is 15, and the hypotenuse of the triangle is 17, so via Equation \ref{cosdef}:

\[\begin{align*} \cos (α) &= \dfrac{\text{adjacent}}{\text{hypotenuse}} \\[4pt] &= \dfrac{15}{17} \end{align*}\]

Exercise \(\PageIndex{1}\)

Given the triangle shown in Figure \(\PageIndex{4}\), find the value of \(\sin t\).

\(\frac{7}{25}\)

Relating Angles and Their Functions

When working with right triangles, the same rules apply regardless of the orientation of the triangle. In fact, we can evaluate the six trigonometric functions of either of the two acute angles in the triangle in Figure \(\PageIndex{5}\). The side opposite one acute angle is the side adjacent to the other acute angle, and vice versa.

We will be asked to find all six trigonometric functions for a given angle in a triangle. Our strategy is to find the sine, cosine, and tangent of the angles first. Then, we can find the other trigonometric functions easily because we know that the reciprocal of sine is cosecant, the reciprocal of cosine is secant, and the reciprocal of tangent is cotangent.

how to: Given the side lengths of a right triangle, evaluate the six trigonometric functions of one of the acute angles

• If needed, draw the right triangle and label the angle provided.
• Identify the angle, the adjacent side, the side opposite the angle, and the hypotenuse of the right triangle.
• sine as the ratio of the opposite side to the hypotenuse
• cosine as the ratio of the adjacent side to the hypotenuse
• tangent as the ratio of the opposite side to the adjacent side
• secant as the ratio of the hypotenuse to the adjacent side
• cosecant as the ratio of the hypotenuse to the opposite side
• cotangent as the ratio of the adjacent side to the opposite side

Example \(\PageIndex{2}\): Evaluating Trigonometric Functions of Angles Not in Standard Position

Using the triangle shown in Figure \(\PageIndex{6}\), evaluate \( \sin α, \cos α, \tan α, \sec α, \csc α,\) and \( \cot α\).

\[ \begin{align*} \sin α &= \dfrac{\text{opposite } α}{\text{hypotenuse}} = \dfrac{4}{5} \\ \cos α &= \dfrac{\text{adjacent to }α}{\text{hypotenuse}}=\dfrac{3}{5} \\ \tan α &= \dfrac{\text{opposite }α}{\text{adjacent to }α}=\dfrac{4}{3} \\ \sec α &= \dfrac{\text{hypotenuse}}{\text{adjacent to }α}= \dfrac{5}{3} \\ \csc α &= \dfrac{\text{hypotenuse}}{\text{opposite }α}=\dfrac{5}{4} \\ \cot α &= \dfrac{\text{adjacent to }α}{\text{opposite }α}=\dfrac{3}{4} \end{align*}\]

Exercise \(\PageIndex{2}\)

Using the triangle shown in Figure \(\PageIndex{7}\), evaluate \( \sin t, \cos t,\tan t, \sec t, \csc t,\) and \(\cot t\).

\[\begin{align*} \sin t &= \frac{33}{65}, \cos t= \frac{56}{65},\tan t= \frac{33}{56}, \\ \\ \sec t &= \frac{65}{56},\csc t= \frac{65}{33},\cot t= \frac{56}{33} \end{align*}\]

Finding Trigonometric Functions of Special Angles Using Side Lengths

We have already discussed the trigonometric functions as they relate to the special angles on the unit circle. Now, we can use those relationships to evaluate triangles that contain those special angles. We do this because when we evaluate the special angles in trigonometric functions, they have relatively friendly values, values that contain either no or just one square root in the ratio. Therefore, these are the angles often used in math and science problems. We will use multiples of \(30°, 60°,\) and \(45°\), however, remember that when dealing with right triangles, we are limited to angles between \(0° \text{ and } 90°\).

Suppose we have a \(30°,60°,90°\) triangle, which can also be described as a \(\frac{π}{6}, \frac{π}{3},\frac{π}{2}\) triangle. The sides have lengths in the relation \(s,\sqrt{3}s,2s.\) The sides of a \(45°,45°,90° \)triangle, which can also be described as a \(\frac{π}{4},\frac{π}{4},\frac{π}{2}\) triangle, have lengths in the relation \(s,s,\sqrt{2}s.\) These relations are shown in Figure \(\PageIndex{8}\).

We can then use the ratios of the side lengths to evaluate trigonometric functions of special angles.

Given trigonometric functions of a special angle, evaluate using side lengths.

• Use the side lengths shown in Figure \(\PageIndex{8}\) for the special angle you wish to evaluate.
• Use the ratio of side lengths appropriate to the function you wish to evaluate.

Example \(\PageIndex{3}\): Evaluating Trigonometric Functions of Special Angles Using Side Lengths

Find the exact value of the trigonometric functions of \(\frac{π}{3}\), using side lengths.

\[\begin{align*} \sin (\dfrac{π}{3}) &= \dfrac{\text{opp}}{\text{hyp}}=\dfrac{\sqrt{3}s}{2s}=\dfrac{\sqrt{3}}{2} \\ \cos (\dfrac{π}{3}) &= \dfrac{\text{adj}}{\text{hyp}}=\dfrac{s}{2s}=\dfrac{1}{2} \\ \tan (\dfrac{π}{3}) &= \dfrac{\text{opp}}{\text{adj}} =\dfrac{\sqrt{3}s}{s}=\sqrt{3} \\ \sec (\dfrac{π}{3}) &= \dfrac{\text{hyp}}{\text{adj}} = \dfrac{2s}{s}=2 \\ \csc (\dfrac{π}{3}) &= \dfrac{\text{hyp}}{\text{opp}} =\dfrac{2s}{\sqrt{3}s}=\dfrac{2}{\sqrt{3}}=\dfrac{2\sqrt{3}}{3} \\ \cot (\dfrac{π}{3}) &= \dfrac{\text{adj}}{\text{opp}}=\dfrac{s}{\sqrt{3}s}=\dfrac{1}{\sqrt{3}}=\dfrac{\sqrt{3}}{3} \end{align*}\]

Exercise \(\PageIndex{3}\)

Find the exact value of the trigonometric functions of \(\frac{π}{4}\) using side lengths.

\( \sin (\frac{π}{4})=\frac{\sqrt{2}}{2}, \cos (\frac{π}{4})=\frac{\sqrt{2}}{2}, \tan (\frac{π}{4})=1,\)

\( \sec (\frac{π}{4})=\sqrt{2}, \csc (\frac{π}{4})=\sqrt{2}, \cot (\frac{π}{4}) =1 \)

Using Equal Cofunction of Complements

If we look more closely at the relationship between the sine and cosine of the special angles relative to the unit circle, we will notice a pattern. In a right triangle with angles of \(\frac{π}{6}\) and \(\frac{π}{3}\), we see that the sine of \(\frac{π}{3}\), namely \(\frac{\sqrt{3}}{2}\), is also the cosine of \(\frac{π}{6}\), while the sine of \(\frac{π}{6}\), namely \(\frac{1}{2},\) is also the cosine of \(\frac{π}{3}\) (Figure \(\PageIndex{9}\)).

\[\begin{align*} \sin \frac{π}{3} &= \cos \frac{π}{6}=\frac{\sqrt{3}s}{2s}=\frac{\sqrt{3}}{2} \\ \sin \frac{π}{6} &= \cos \frac{π}{3}=\frac{s}{2s}=\frac{1}{2} \end{align*}\]

This result should not be surprising because, as we see from Figure \(\PageIndex{9}\), the side opposite the angle of \(\frac{π}{3}\) is also the side adjacent to \(\frac{π}{6}\), so \(\sin (\frac{π}{3})\) and \(\cos (\frac{π}{6})\) are exactly the same ratio of the same two sides, \(\sqrt{3} s\) and \(2s.\) Similarly, \( \cos (\frac{π}{3})\) and \( \sin (\frac{π}{6})\) are also the same ratio using the same two sides, \(s\) and \(2s\).

The interrelationship between the sines and cosines of \(\frac{π}{6}\) and \(\frac{π}{3}\) also holds for the two acute angles in any right triangle, since in every case, the ratio of the same two sides would constitute the sine of one angle and the cosine of the other. Since the three angles of a triangle add to π, π,and the right angle is \(\frac{π}{2}\), the remaining two angles must also add up to \(\frac{π}{2}\). That means that a right triangle can be formed with any two angles that add to \(\frac{π}{2}\)—in other words, any two complementary angles. So we may state a cofunction identity : If any two angles are complementary, the sine of one is the cosine of the other, and vice versa. This identity is illustrated in Figure \(\PageIndex{10}\).

Using this identity, we can state without calculating, for instance, that the sine of \(\frac{π}{12}\) equals the cosine of \(\frac{5π}{12}\), and that the sine of \(\frac{5π}{12}\) equals the cosine of \(\frac{π}{12}\). We can also state that if, for a certain angle \(t, \cos t= \frac{5}{13},\) then \( \sin (\frac{π}{2}−t)=\frac{5}{13}\) as well.

COFUNCTION IDENTITIES

The cofunction identities in radians are listed in Table \(\PageIndex{1}\).

how to: Given the sine and cosine of an angle, find the sine or cosine of its complement.

• To find the sine of the complementary angle, find the cosine of the original angle.
• To find the cosine of the complementary angle, find the sine of the original angle.

Example \(\PageIndex{4}\): Using Cofunction Identities

If \( \sin t = \frac{5}{12},\) find \(( \cos \frac{π}{2}−t)\).

According to the cofunction identities for sine and cosine,

\[ \sin t= \cos (\dfrac{π}{2}−t). \nonumber\]

\[ \cos (\dfrac{π}{2}−t)= \dfrac{5}{12}. \nonumber\]

Exercise \(\PageIndex{4}\)

If \(\csc (\frac{π}{6})=2,\) find \( \sec (\frac{π}{3}).\)

Using Trigonometric Functions

In previous examples, we evaluated the sine and cosine in triangles where we knew all three sides. But the real power of right-triangle trigonometry emerges when we look at triangles in which we know an angle but do not know all the sides.

how to: Given a right triangle, the length of one side, and the measure of one acute angle, find the remaining sides

• For each side, select the trigonometric function that has the unknown side as either the numerator or the denominator. The known side will in turn be the denominator or the numerator.
• Write an equation setting the function value of the known angle equal to the ratio of the corresponding sides.
• Using the value of the trigonometric function and the known side length, solve for the missing side length.

Example \(\PageIndex{5}\): Finding Missing Side Lengths Using Trigonometric Ratios

Find the unknown sides of the triangle in Figure \(\PageIndex{11}\).

We know the angle and the opposite side, so we can use the tangent to find the adjacent side.

\[ \tan (30°)= \dfrac{7}{a} \nonumber\]

We rearrange to solve for \(a\).

\[\begin{align} a &=\dfrac{7}{ \tan (30°)} \\ & =12.1 \end{align} \nonumber\]

We can use the sine to find the hypotenuse.

\[ \sin (30°)= \dfrac{7}{c} \nonumber\]

Again, we rearrange to solve for \(c\).

\[\begin{align*} c &= \dfrac{7}{\sin (30°)} =14 \end{align*}\]

Exercise \(\PageIndex{5}\):

A right triangle has one angle of \(\frac{π}{3}\) and a hypotenuse of 20. Find the unknown sides and angle of the triangle.

\(\mathrm{adjacent=10; opposite=10 \sqrt{3}; }\) missing angle is \(\frac{π}{6}\)

Using Right Triangle Trigonometry to Solve Applied Problems

Right-triangle trigonometry has many practical applications. For example, the ability to compute the lengths of sides of a triangle makes it possible to find the height of a tall object without climbing to the top or having to extend a tape measure along its height. We do so by measuring a distance from the base of the object to a point on the ground some distance away, where we can look up to the top of the tall object at an angle. The angle of elevation of an object above an observer relative to the observer is the angle between the horizontal and the line from the object to the observer's eye. The right triangle this position creates has sides that represent the unknown height, the measured distance from the base, and the angled line of sight from the ground to the top of the object. Knowing the measured distance to the base of the object and the angle of the line of sight, we can use trigonometric functions to calculate the unknown height. Similarly, we can form a triangle from the top of a tall object by looking downward. The angle of depression of an object below an observer relative to the observer is the angle between the horizontal and the line from the object to the observer's eye. See Figure \(\PageIndex{12}\).

how to: Given a tall object, measure its height indirectly

• Make a sketch of the problem situation to keep track of known and unknown information.
• Lay out a measured distance from the base of the object to a point where the top of the object is clearly visible.
• At the other end of the measured distance, look up to the top of the object. Measure the angle the line of sight makes with the horizontal.
• Write an equation relating the unknown height, the measured distance, and the tangent of the angle of the line of sight.
• Solve the equation for the unknown height.

Example \(\PageIndex{6}\): Measuring a Distance Indirectly

To find the height of a tree, a person walks to a point 30 feet from the base of the tree. She measures an angle of 57° 57° between a line of sight to the top of the tree and the ground, as shown in Figure \(\PageIndex{13}\). Find the height of the tree.

We know that the angle of elevation is \(57°\) and the adjacent side is 30 ft long. The opposite side is the unknown height.

The trigonometric function relating the side opposite to an angle and the side adjacent to the angle is the tangent. So we will state our information in terms of the tangent of \(57°\), letting \(h\) be the unknown height.

\[\begin{array}{cl} \tan θ = \dfrac{\text{opposite}}{\text{adjacent}} & \text{} \\ \tan (57°) = \dfrac{h}{30} & \text{Solve for }h. \\ h=30 \tan (57°) & \text{Multiply.} \\ h≈46.2 & \text{Use a calculator.} \end{array} \]

The tree is approximately 46 feet tall.

Exercise \(\PageIndex{6}\):

How long a ladder is needed to reach a windowsill 50 feet above the ground if the ladder rests against the building making an angle of \(\frac{5π}{12}\) with the ground? Round to the nearest foot.

About 52 ft

Access these online resources for additional instruction and practice with right triangle trigonometry.

• Finding Trig Functions on Calculator
• Finding Trig Functions Using a Right Triangle
• Relate Trig Functions to Sides of a Right Triangle
• Determine Six Trig Functions from a Triangle
• Determine Length of Right Triangle Side

Visit this website for additional practice questions from Learningpod.

Key Equations

Cofunction Identities

Key Concepts

• We can define trigonometric functions as ratios of the side lengths of a right triangle. See Example .
• The same side lengths can be used to evaluate the trigonometric functions of either acute angle in a right triangle. See Example .
• We can evaluate the trigonometric functions of special angles, knowing the side lengths of the triangles in which they occur. See Example .
• Any two complementary angles could be the two acute angles of a right triangle.
• If two angles are complementary, the cofunction identities state that the sine of one equals the cosine of the other and vice versa. See Example .
• We can use trigonometric functions of an angle to find unknown side lengths.
• Select the trigonometric function representing the ratio of the unknown side to the known side. See Example .
• Right-triangle trigonometry permits the measurement of inaccessible heights and distances.
• The unknown height or distance can be found by creating a right triangle in which the unknown height or distance is one of the sides, and another side and angle are known. See Example .

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Trig Calculator

The sine and cosine trigonometric functions, tan, cot, sec, and csc, calculated from trig identities., trigonometry in a right triangle, other trigonometric calculators.

Welcome to this trigonometric calculator, a trig tool created to:

• Calculate any trigonometric function by inputting the angle at which you want to evaluate it; and
• Solve for the sides or angles of right triangles by using trigonometry.

Keep reading this article to learn more about trigonometric functions and the trig identities that relate them.

Trig functions are functions that take an angle as the argument. We define these functions by using the angle of a right triangle that is inserted in a unitary circle . Then, we relate that angle to the sides of such a triangle.

As the right triangle is circumscribed in a unit circle, the length of its hypotenuse equals the circle's radius (which equals one unit).

Sine and cosine are the fundamental trigonometric functions arising from the previous diagram:

• The sine of theta ( sin θ ) is the hypotenuse's vertical projection (green line); and
• The cosine of theta ( cos θ ) is the hypotenuse's horizontal projection (blue line).

We can rotate the radial line through the four quadrants and obtain the values of the trig functions from 0 to 360 degrees , as in the diagram below:

• For example, for an angle that leads to the second quadrant (90-180°), the cosine will be negative as the horizontal projection of the hypotenuse will point to the left. That's the case of 135°, whose cosine (horizontal projection, or x coordinate) equals -√2/2.
• For the third quadrant (180-270°), the cosine and the sine of the angles lying on it will be negative.
• For the fourth quadrant angles (270-360)°, the cosines will be positive and the sines negative.

Beyond 360 degrees

The previous behavior repeats cyclically, so trigonometric functions are not limited to 360°. We can keep rotating counterclockwise, and once we reach 360 degrees, the sine and cosine functions start to repeat the same behavior. As a consequence, we can relate the functions at different angles with the following trig identities for any n integer:

• sin(θ + 2πn) = sin(θ) ;
• cos(θ + 2πn) = cos(θ); and

For example a trig function at 90° (π/2) will be mathematically the same as at 450° (5π/2), as 5π/2 = π/2 + 2π.

Negative angles

Negative angles imply the same way to calculate sine and cosine (vertical and horizontal projections, respectively), with the difference that angular rotation occurs in the clockwise direction. For example, a trigonometric function at 270° is the same as at -90°, as their radial lines are the same (you can check it with this calculator)

Once you know the value of sine and cosine, you can use the following trigonometric identities to obtain the values of the other four functions:

Tangent is the sine-to-cosine ratio

tan(α) = sin(α)/cos(α)

Cosecant is the reciprocal of the sine

csc(α) = 1/sin(α)

Secant is the reciprocal of the cosine

sec(α) = 1/cos(α)

Cotangent is the reciprocal of the tangent

cot(α) = 1/tan(α)

From the previous analysis, we can obtain some valuable formulas that relate the angle of a right triangle to its sides. Of course, this doesn't limit to unit circles, so we can use it for hypotenuses of any length.

We relate the angle of the right triangle to its sides in the following way:

sin(α) = opposite/hypotenuse

cos(α) = adjacent/hypotenuse

tan(α) = opposite/adjacent

Remember that cotangent, secant, and cosecant are the inverse of the previous functions:

csc(α) = 1/sin(α) = hypotenuse/opposite

sec(α) = 1/cos(α) = hypotenuse/adjacent

cot(α) = 1/tan(α) = adjacent/opposite

You also can apply the previous formulas for the other acute angle ( β ), but consider that the legs of the triangle will switch: the adjacent will now be the hypotenuse and vice versa.

If you like this calculator, you may find these other tools interesting:

• Trigonometry calculator ;
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• Tangent ratio calculator ; and
• Tangent angle calculator .

How do I solve a 45-90-45 degree triangle by trig formulas?

If one leg of a 45 45 90 triangle is equal to a , then:

• The second leg also equals a ;
• The hypotenuse equals a √2 (from the hypotenuse formula c = √( a ² + a ²) = a √2 );
• The area is A = a ²/2 ; and
• The perimeter equals a (2 + √2) (the sum of the two sides plus the hypotenuse).

What are the values of the 6 trig functions at 90 degrees?

The values of the 6 trig functions for 90 degrees (π/2) are the following ones:

• sin(90°) = 1;
• cos(90°) = 0;
• tan(90°)= undefined;
• cot(90°) = 0;
• sec(90°) = undefined; and
• csc(90°) = 1.

Polish notation

Triangle length.

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Trigonometry Worksheets

Free worksheets with answer keys.

Enjoy these free sheets. Each one has model problems worked out step by step, practice problems, as well as challenge questions at the sheets end. Plus each one comes with an answer key.

(This sheet is a summative worksheet that focuses on deciding when to use the law of sines or cosines as well as on using both formulas to solve for a single triangle's side or angle)

• Law of Sines
• Ambiguous Case of the Law of Sines
• Law Of Cosines
• Sine, Cosine, Tangent, to Find Side Length
• Sine, Cosine, Tangent Chart
• Inverse Trig Functions
• Real World Applications of SOHCATOA
• Mixed Review
• Vector Worksheet
• Unit Circle Worksheet
• Graphing Sine and Cosine Worksheet

Ultimate Math Solver (Free) Free Algebra Solver ... type anything in there!

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Trigonometry Questions

Trigonometry questions given here involve finding the missing sides of a triangle with the help of trigonometric ratios and proving trigonometry identities. We know that trigonometry is one of the most important chapters of Class 10 Maths. Hence, solving these questions will help you to improve your problem-solving skills.

What is Trigonometry?

The word ‘trigonometry’ is derived from the Greek words ‘tri’ (meaning three), ‘gon’ (meaning sides) and ‘metron’ (meaning measure). Trigonometry is the study of relationships between the sides and angles of a triangle.

The basic trigonometric ratios are defined as follows.

sine of ∠A = sin A = Side opposite to ∠A/ Hypotenuse

cosine of ∠A = cos A = Side adjacent to ∠A/ Hypotenuse

tangent of ∠A = tan A = (Side opposite to ∠A)/ (Side adjacent to ∠A)

cosecant of ∠A = cosec A = 1/sin A = Hypotenuse/ Side opposite to ∠A

secant of ∠A = sec A = 1/cos A = Hypotenuse/ Side adjacent to ∠A

cotangent of ∠A = cot A = 1/tan A = (Side adjacent to ∠A)/ (Side opposite to ∠A)

Also, tan A = sin A/cos A

cot A = cos A/sin A

Also, read: Trigonometry

Trigonometry Questions and Answers

1. From the given figure, find tan P – cot R.

From the given,

In the right triangle PQR, Q is right angle.

By Pythagoras theorem,

PR 2 = PQ 2 + QR 2

QR 2 = (13) 2 – (12) 2

= 169 – 144

tan P = QR/PQ = 5/12

cot R = QR/PQ = 5/12

So, tan P – cot R = (5/12) – (5/12) = 0

2. Prove that (sin 4 θ – cos 4 θ +1) cosec 2 θ = 2

L.H.S. = (sin 4 θ – cos 4 θ +1) cosec 2 θ

= [(sin 2 θ – cos 2 θ) (sin 2 θ + cos 2 θ) + 1] cosec 2 θ

Using the identity sin 2 A + cos 2 A = 1,

= (sin 2 θ – cos 2 θ + 1) cosec 2 θ

= [sin 2 θ – (1 – sin 2 θ) + 1] cosec 2 θ

= 2 sin 2 θ cosec 2 θ

= 2 sin 2 θ (1/sin 2 θ)

3. Prove that (√3 + 1) (3 – cot 30°) = tan 3 60° – 2 sin 60°.

LHS = (√3 + 1)(3 – cot 30°)

= (√3 + 1)(3 – √3)

= 3√3 – √3.√3 + 3 – √3

= 2√3 – 3 + 3

RHS = tan 3 60° – 2 sin 60°

= (√3) 3 – 2(√3/2)

= 3√3 – √3

Therefore, (√3 + 1) (3 – cot 30°) = tan 3 60° – 2 sin 60°.

Hence proved.

4. If tan(A + B) = √3 and tan(A – B) = 1/√3 ; 0° < A + B ≤ 90°; A > B, find A and B.

tan(A + B) = √3

tan(A + B) = tan 60°

A + B = 60°….(i)

tan(A – B) = 1/√3

tan(A – B) = tan 30°

A – B = 30°….(ii)

Adding (i) and (ii),

A + B + A – B = 60° + 30°

Substituting A = 45° in (i),

45° + B = 60°

B = 60° – 45° = 15°

Therefore, A = 45° and B = 15°.

5. If sin 3A = cos (A – 26°), where 3A is an acute angle, find the value of A.

sin 3A = cos(A – 26°); 3A is an acute angle

cos(90° – 3A) = cos(A – 26°) {since cos(90° – A) = sin A}

⇒ 90° – 3A = A – 26

⇒ 3A + A = 90° + 26°

⇒ 4A = 116°

⇒ A = 116°/4

6. If A, B and C are interior angles of a triangle ABC, show that sin (B + C/2) = cos A/2.

We know that, for a given triangle, the sum of all the interior angles of a triangle is equal to 180°

A + B + C = 180° ….(1)

B + C = 180° – A

Dividing both sides of this equation by 2, we get;

⇒ (B + C)/2 = (180° – A)/2

⇒ (B + C)/2 = 90° – A/2

Take sin on both sides,

sin (B + C)/2 = sin (90° – A/2)

⇒ sin (B + C)/2 = cos A/2 {since sin(90° – x) = cos x}

7. If tan θ + sec θ = l, prove that sec θ = (l 2 + 1)/2l.

tan θ + sec θ = l….(i)

We know that,

sec 2 θ – tan 2 θ = 1

(sec θ – tan θ)(sec θ + tan θ) = 1

(sec θ – tan θ) l = 1 {from (i)}

sec θ – tan θ = 1/l….(ii)

tan θ + sec θ + sec θ – tan θ = l + (1/l)

2 sec θ = (l 2 + 1)l

sec θ = (l 2 + 1)/2l

8. Prove that (cos A – sin A + 1)/ (cos A + sin A – 1) = cosec A + cot A, using the identity cosec 2 A = 1 + cot 2 A.

LHS = (cos A – sin A + 1)/ (cos A + sin A – 1)

Dividing the numerator and denominator by sin A, we get;

= (cot A – 1 + cosec A)/(cot A + 1 – cosec A)

Using the identity cosec 2 A = 1 + cot 2 A ⇒ cosec 2 A – cot 2 A = 1,

= [cot A – (cosec 2 A – cot 2 A) + cosec A]/ (cot A + 1 – cosec A)

= [(cosec A + cot A) – (cosec A – cot A)(cosec A + cot A)] / (cot A + 1 – cosec A)

= cosec A + cot A

9. Prove that: (cosec A – sin A)(sec A – cos A) = 1/(tan A + cot A)

[Hint: Simplify LHS and RHS separately]

LHS = (cosec A – sin A)(sec A – cos A)

= (cos 2 A/sin A) (sin 2 A/cos A)

= cos A sin A….(i)

RHS = 1/(tan A + cot A)

= (sin A cos A)/ (sin 2 A + cos 2 A)

= (sin A cos A)/1

= sin A cos A….(ii)

From (i) and (ii),

i.e. (cosec A – sin A)(sec A – cos A) = 1/(tan A + cot A)

10. If a sin θ + b cos θ = c, prove that a cosθ – b sinθ = √(a 2 + b 2 – c 2 ).

a sin θ + b cos θ = c

Squaring on both sides,

(a sin θ + b cos θ) 2 = c 2

a 2 sin 2 θ + b 2 cos 2 θ + 2ab sin θ cos θ = c 2

a 2 (1 – cos 2 θ) + b 2 (1 – sin 2 θ) + 2ab sin θ cos θ = c 2

a 2 – a 2 cos 2 θ + b 2 – b 2 sin 2 θ + 2ab sin θ cos θ = c 2

a 2 + b 2 – c 2 = a 2 cos 2 θ + b 2 sin 2 θ – 2ab sin θ cos θ

a 2 + b 2 – c 2 = (a cos θ – b sin θ ) 2

⇒ a cos θ – b sin θ = √(a 2 + b 2 – c 2 )

Video Lesson on Trigonometry

Practice Questions on Trigonometry

Solve the following trigonometry problems.

• Prove that (sin α + cos α) (tan α + cot α) = sec α + cosec α.
• If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
• If sin θ + cos θ = √3, prove that tan θ + cot θ = 1.
• Evaluate: 2 tan 2 45° + cos 2 30° – sin 2 60°
• Express cot 85° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.

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Trigonometry : Solving Word Problems with Trigonometry

Study concepts, example questions & explanations for trigonometry, all trigonometry resources, example questions, example question #1 : solving word problems with trigonometry.

You can draw the following right triangle using the information given by the question:

Since you want to find the height of the platform, you will need to use tangent.

You can draw the following right triangle from the information given by the question.

In order to find the height of the flagpole, you will need to use tangent.

You can draw the following right triangle from the information given in the question:

In order to find out how far up the ladder goes, you will need to use sine.

In right triangle ABC, where angle A measures 90 degrees, side AB measures 15 and side AC measures 36, what is the length of side BC?

This triangle cannot exist.

Example Question #5 : Solving Word Problems With Trigonometry

A support wire is anchored 10 meters up from the base of a flagpole, and the wire makes a 25 o angle with the ground. How long is the wire, w? Round your answer to two decimal places.

23.81 meters

28.31 meters

21.83 meters

To make sense of the problem, start by drawing a diagram. Label the angle of elevation as 25 o , the height between the ground and where the wire hits the flagpole as 10 meters, and our unknown, the length of the wire, as w. 

Now, we just need to solve for w using the information given in the diagram. We need to ask ourselves which parts of a triangle 10 and w are relative to our known angle of 25 o . 10 is opposite this angle, and w is the hypotenuse. Now, ask yourself which trig function(s) relate opposite and hypotenuse. There are two correct options: sine and cosecant. Using sine is probably the most common, but both options are detailed below.

We know that sine of a given angle is equal to the opposite divided by the hypotenuse, and cosecant of an angle is equal to the hypotenuse divided by the opposite (just the reciprocal of the sine function). Therefore:

To solve this problem instead using the cosecant function, we would get:

The reason that we got 23.7 here and 23.81 above is due to differences in rounding in the middle of the problem. 

Example Question #6 : Solving Word Problems With Trigonometry

When the sun is 22 o above the horizon, how long is the shadow cast by a building that is 60 meters high?

To solve this problem, first set up a diagram that shows all of the info given in the problem. 

Next, we need to interpret which side length corresponds to the shadow of the building, which is what the problem is asking us to find. Is it the hypotenuse, or the base of the triangle? Think about when you look at a shadow. When you see a shadow, you are seeing it on something else, like the ground, the sidewalk, or another object. We see the shadow on the ground, which corresponds to the base of our triangle, so that is what we'll be solving for. We'll call this base b.

Therefore the shadow cast by the building is 150 meters long.

If you got one of the incorrect answers, you may have used sine or cosine instead of tangent, or you may have used the tangent function but inverted the fraction (adjacent over opposite instead of opposite over adjacent.)

Example Question #7 : Solving Word Problems With Trigonometry

From the top of a lighthouse that sits 105 meters above the sea, the angle of depression of a boat is 19 o . How far from the boat is the top of the lighthouse?

423.18 meters

318.18 meters

36.15 meters

110.53 meters

To solve this problem, we need to create a diagram, but in order to create that diagram, we need to understand the vocabulary that is being used in this question. The following diagram clarifies the difference between an angle of depression (an angle that looks downward; relevant to our problem) and the angle of elevation (an angle that looks upward; relevant to other problems, but not this specific one.) Imagine that the top of the blue altitude line is the top of the lighthouse, the green line labelled GroundHorizon is sea level, and point B is where the boat is.

Merging together the given info and this diagram, we know that the angle of depression is 19 o  and and the altitude (blue line) is 105 meters. While the blue line is drawn on the left hand side in the diagram, we can assume is it is the same as the right hand side. Next, we need to think of the trig function that relates the given angle, the given side, and the side we want to solve for. The altitude or blue line is opposite the known angle, and we want to find the distance between the boat (point B) and the top of the lighthouse. That means that we want to determine the length of the hypotenuse, or red line labelled SlantRange. The sine function relates opposite and hypotenuse, so we'll use that here. We get:

Example Question #8 : Solving Word Problems With Trigonometry

Angelina just got a new car, and she wants to ride it to the top of a mountain and visit a lookout point. If she drives 4000 meters along a road that is inclined 22 o to the horizontal, how high above her starting point is she when she arrives at the lookout?

9.37 meters

1480 meters

3708.74 meters

10677.87 meters

1616.1 meters

As with other trig problems, begin with a sketch of a diagram of the given and sought after information.

Angelina and her car start at the bottom left of the diagram. The road she is driving on is the hypotenuse of our triangle, and the angle of the road relative to flat ground is 22 o . Because we want to find the change in height (also called elevation), we want to determine the difference between her ending and starting heights, which is labelled x in the diagram. Next, consider which trig function relates together an angle and the sides opposite and hypotenuse relative to it; the correct one is sine. Then, set up:

Therefore the change in height between Angelina's starting and ending points is 1480 meters. 

Example Question #9 : Solving Word Problems With Trigonometry

Two buildings with flat roofs are 50 feet apart. The shorter building is 40 feet tall. From the roof of the shorter building, the angle of elevation to the edge of the taller building is 48 o . How high is the taller building?

To solve this problem, let's start by drawing a diagram of the two buildings, the distance in between them, and the angle between the tops of the two buildings. Then, label in the given lengths and angle. 

Example Question #10 : Solving Word Problems With Trigonometry

Two buildings with flat roofs are 80 feet apart. The shorter building is 55 feet tall. From the roof of the shorter building, the angle of elevation to the edge of the taller building is 32 o . How high is the taller building?

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