solving trigonometry questions involving length and angles

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Trigonometry problems and questions with solutions - grade 10.

Grade 10 trigonometry problems and questions with answers and solutions are presented.

The area of a right triangle is 50. One of its angles is 45°. Find the lengths of the sides and hypotenuse of the triangle.
In a right triangle ABC, tan(A) = 3/4. Find sin(A) and cos(A).
In a right triangle ABC with angle A equal to 90°, find angle B and C so that sin(B) = cos(B).
A rectangle has dimensions 10 cm by 5 cm. Determine the measures of the angles at the point where the diagonals intersect.
The lengths of side AB and side BC of a scalene triangle ABC are 12 cm and 8 cm respectively. The size of angle C is 59°. Find the length of side AC.
From the top of a 200 meters high building, the angle of depression to the bottom of a second building is 20 degrees. From the same point, the angle of elevation to the top of the second building is 10 degrees. Calculate the height of the second building.
Karla is riding vertically in a hot air balloon, directly over a point P on the ground. Karla spots a parked car on the ground at an angle of depression of 30°. The balloon rises 50 meters. Now the angle of depression to the car is 35 degrees. How far is the car from point P?
If the shadow of a building increases by 10 meters when the angle of elevation of the sun rays decreases from 70° to 60°, what is the height of the building?

Solutions to the Above Problems

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High school geometry

Unit 5: lesson 6, solving for a side in right triangles with trigonometry, let's look at an example..

Let's focus on angle B \goldD B B start color #e07d10, B, end color #e07d10 since that is the angle that is explicitly given in the diagram.

A right triangle A B C. Angle A C B is a right angle. Angle A B C is fifty degrees and is highlighted. Side A C is unknown. Side A B is six units.

Note that we are given the length of the hypotenuse \purpleC{\text{hypotenuse}} hypotenuse start color #aa87ff, start text, h, y, p, o, t, e, n, u, s, e, end text, end color #aa87ff , and we are asked to find the length of the side opposite \blueD{\text{opposite}} opposite start color #11accd, start text, o, p, p, o, s, i, t, e, end text, end color #11accd angle B \goldD B B start color #e07d10, B, end color #e07d10 . The trigonometric ratio that contains both of those sides is the sine . [I'd like to review the trig ratios.]

sin ⁡ ( B ) = opposite hypotenuse Define sine. sin ⁡ ( 5 0 ∘ ) = A C 6 Substitute. 6 sin ⁡ ( 5 0 ∘ ) = A C Multiply both sides by 6. 4.60 ≈ A C Evaluate with a calculator. \begin{aligned}\sin( \goldD{ B}) &= \dfrac{ \blueD{\text{ opposite}} \text{ } }{\purpleC{\text{ hypotenuse} }} ~~~~~~~~\small{\gray{\text{Define sine.}}}\\\\ \sin (\goldD{50^\circ})&= \dfrac{\blueD{AC}}{\purpleC6}~~~~~~~~~~~~~~~~~~~~~~~\small{\gray{\text{Substitute.}}} \\\\\\\\ 6\sin ({50^\circ})&= {{AC}} ~~~~~~~~~~~~~~~~~~~~~~~~~\small{\gray{\text{Multiply both sides by }6.}}\\\\\\\\ 4.60&\approx AC~~~~~~~~~~~~~~~~~~~~~~~~~\small{\gray{\text{Evaluate with a calculator.}}} \end{aligned} sin ( B ) sin ( 5 0 ∘ ) 6 sin ( 5 0 ∘ ) 4 . 6 0 = hypotenuse opposite Define sine. = 6 A C Substitute. = A C Multiply both sides by 6 . ≈ A C Evaluate with a calculator.

Now let's try some practice problems.

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Circles and Trigonometry

For simplicity, let's fix the hypotenuse of the triangle at a given length: we'll call it r for now. We'll call one of the acute angles θ (the other acute angle is 90° – θ .)

As we vary the angle θ, the opposite and adjacent sides of the triangle also vary.

At this point, you might note that as we "rotate" the hypotenuse of length r, we end up with a circle. The circle has a radius r, and its center point is the vertex corresponding to the angle θ.

If we overlay a set of coordinate axes on this circle, with the origin on the center, we see that the adjacent side of the right triangle is a distance along the x -axis and the opposite side is a distance parallel to the y -axis. Furthermore, the vertex of the triangle on the circle must have coordinates ( x 1 , y 1 ), where x 1 is the length of the adjacent side and y 1 is the length of the opposite side.

By the Pythagorean theorem, we know the following relationship among r, x 1 , and y 1 .

If we "normalize" the hypotenuse to a length of 1 (that is, set r equal to 1), then we call the resulting circle a unit circle. In this case, the circle intercepts the axes at y = ±1 and x = ±1.

So what does this all do for us? Why should we care? Well, first, notice that we've joined our concepts of coordinate graphs, circles, and right triangles together--they are interrelated. Thus, logically, we expect trigonometry to have a role in our understanding of circles as well as right triangles. But before we delve further into this relationship, we must first define some properties of the angle θ in this context.

Angle Measurements

Thus far, we have relied exclusively on degrees as the unit of angle measure. Another unit with which you should be familiar is the radian. First, note that the circumference of a circle is 2 πr, where r is the radius. If we divide out the radius from the circumference, we are left simply 2 π: this is in some sense how "far" the circle goes around its center, regardless of the radius. In other words, 2 π is the angle swept by the radius of the circle.

solving trigonometry questions involving length and angles

As you probably recognize, then, 2 π is also equivalent to 360°. The unit of angle in this case is the radian. One radian is defined as the angle formed such that the portion of the circle (or arc length ) swept by that angle is equal to the radius of the circle.

Another way to look at it is this: if we took a segment of length r (the radius) and molded it onto the circle, the angle formed by the radii connecting the center of the circle to the endpoints of the arc would have a measure of 1 radian. As it turns out, one radian is equal to approximately 57.3°.

By convention, the angle θ is measured from the x- axis in the counterclockwise direction. If the clockwise direction is used, the angle is simply assigned a negative value. Thus, for instance, π radians (180°) and –π radians (–180°) correspond to the same point on the circle.

For our purposes, we will primarily deal with angles between –2 π and 2 π, inclusive (that is, –2 π ≤ θ ≤ 2 π ). Angles larger in magnitude than 2 π are defined, but they overlap with lower-value angles, so we won't consider them in much detail.

Practice Problem: Convert each of the following angle measures to radians. Be sure to use exact expressions for your answers.

a. 60° b. 270° c. -90° d. 120°

Solution: First, note that 360° is equal to 2 π radians. We can use this ratio to convert from degrees to radians. Below is the calculation for part a; the other parts follow the same pattern. Note that the negative sign (part c) translates directly from degrees to radians.

Practice Problem: Convert each of the following angle measures from radians to degrees.

Solution: Here, simply use the same process as the previous practice problem, but use the reciprocal ratio of radians and degrees. Again, only the calculation for part a is shown below; the rest are similar.

a. -22.5° b. 360° c. 36°

Application to Trigonometric Functions

Now, we can finally look at what circles have to do with trigonometry. Let's look again at our unit circle.

Each point P on the unit circle, designated by the coordinates ( x 1 , y 1 ), can be expressed in terms of the angle θ as well. Note the following, where we apply what we learned about right-triangle trig:

In other words, point Phas coordinates (cos θ, sin θ ), where θ is the angle formed between the x- -axis and the radius to P. What we've also done is define two functions:

sin θ = distance of point P from the x -axis

cos θ = distance of point P from the y -axis

Practice Problem: A tire has a radius (outer radius) of 10 inches. A small mark is painted at the very top of the tire, and then the tire is rolled forward slightly so that the mark rotates through an angle of pi/4 radians. How far above the ground is the mark at this point?

Solution: This problem requires that we apply much of what we've learned. First, draw a "tire" (circle) with a radius of 10 inches, and draw a point at the top of the tire to indicate the mark.

We can use the basic facts of angles to redraw this situation in a more familiar form.

Now, we want to find the height of the mark above the ground. Using the definitions in the figure below, this height is 10 + y.

Our job is to then calculate y. We can use trigonometry as follows.

Thus, the mark is about 17.07 inches above the ground.

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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.

$solving trigonometry questions involving length and angles$

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.

$solving trigonometry questions involving length and angles$

Example Question #2 : Solving Word Problems With Trigonometry

$solving trigonometry questions involving length and angles$

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.

$solving trigonometry questions involving length and angles$

Example Question #3 : Solving Word Problems With Trigonometry

$solving trigonometry questions involving length and angles$

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.

$solving trigonometry questions involving length and angles$

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?

$solving trigonometry questions involving length and angles$

This triangle cannot exist.

$solving trigonometry questions involving length and angles$

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.

21.83 meters

28.31 meters

$solving trigonometry questions involving length and angles$

23.81 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:

$solving trigonometry questions involving length and angles$

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

$solving trigonometry questions involving length and angles$

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

$solving trigonometry questions involving length and angles$

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.

$solving trigonometry questions involving length and angles$

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?

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:

$solving trigonometry questions involving length and angles$

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?

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:

$solving trigonometry questions involving length and angles$

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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You can draw the following right triangle from the information given by the question. 2. In order to find the height of the flagpole, you will need to use
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