15 Trigonometry Questions And Practice Problems To Do With High Schoolers

Trigonometry questions address the relationship between the angles of a triangle and the lengths of the sides. By using our knowledge of the rules of trigonometry and trigonometric functions, we can calculate missing angles or sides when we have been given some of the information.

Here we’ve provided 15 trigonometry questions that will help your students practice the various types of trigonometry questions they will encounter during high school.

Trigonometry in the real world

Trigonometry is used by architects, engineers, astronomers, crime scene investigators, flight engineers and many others.

Trigonometry Quiz

Need to identify the areas of strength and areas for focus in your high school classes? Use this trigonometry check for understanding quiz to understand how best to support your students with trigonometry. Includes topics such as right triangle trigonometry, law of sines, law of cosines and finding area of non-right triangles.

Trigonometry in high school

In trigonometry we learn about the sine function, tangent function, and cosine function. These trig functions are abbreviated as sin, cos, and tan. We can use these to calculate sides and angles in right angled triangles. Later, students will be applying this to a variety of situations as well as learning the exact values of sin, cos, and tan for certain angles.

Students learn about trigonometric ratios: the law of sines, law of cosines, a new formula for the area of a triangle and applying trigonometric theorems to 3D shapes.

Trigonometry for more senior high school students will introduce the reciprocal trig functions, cotangent, secant and cosecant, but you don’t have to worry about these right now!

The way to answer trigonometry questions depends on whether it is a right angled triangle or not.

How to answer trigonometry questions: right angled triangles

If your trigonometry question involves a right angled triangle, you can apply the following relationships, ie SOH, CAH, TOA

sin θ = opposite/hypotenuse

The acronym SOH CAH TOA is used so that you can remember which ratio to use.

1. Establish that it is a right angled triangle.
2. Label the opposite side (opposite the angle) the adjacent side (next to the angle) and the hypotenuse (longest side opposite the right angle).

3. Use the following triangles to help us decide which calculation to do:

How to answer trigonometry questions: non-right triangles

If the triangle is not a right angled triangle then we need to use the sine rule or the cosine rule.

There is also a formula we can use for the area of a triangle, which does not require us to know the base and height of the triangle.

Sine rule: \frac{a}{\sin(A)}=\frac{b}{\sin(B)}=\frac{c}{\sin(C)}

Cosine rule: a^{2}=b^{2}+c^{2}-2bc \cos(A)

Area of a triangle: Area = \frac{1}{2}ab \sin(C)

1. Establish that it is not a right angled triangle.
2. Label the sides of the triangle using lowercase a, b, c.
3. Label the angles of the triangle using upper case A, B and C.
4. Opposite sides and angles should use the same letter, for example, angle A is opposite to side a.

Trigonometry questions

In high school geometry, trigonometry questions focus on the understanding of sin, cos, and tan (SOHCAHTOA) to calculate missing sides and angles in right triangles.

Trigonometry questions: missing side

1. A zip wire runs between two posts, 25m apart. The zip wire is at an angle of 10^{\circ} to the horizontal. Calculate the length of the zip wire.

25.4m

144.0m

141.8m

24.6m

\begin{aligned} H&=\frac{A}{\cos(\theta)}\\\\ H&=\frac{25}{\cos(10)}\\\\ H&=25.4\mathrm{m} \end{aligned}

2. A surveyor wants to know the height of a skyscraper. He places his inclinometer on a tripod 1m from the ground. At a distance of 50m from the skyscraper, he records an angle of elevation of 82^{\circ} .

What is the height of the skyscraper? Give your answer to one decimal place.

355.8m

7.0m

356.8m

49.5m

\begin{aligned} O&=\tan(\theta) \times A\\\\ O&=\tan(82) \times 50\\\\ O&=355.8\mathrm{m} \end{aligned}

Total height = 355.8+1=356.8m.

3. Triangle ABC is isosceles. Calculate the height of triangle ABC.

6cm

17.4cm

34.9cm

2.1cm

To solve this we split the triangle into two right angled triangles.

\begin{aligned} O&=\tan(\theta) \times A\\\\ O&=\tan(71) \times 6\\\\ O&=17.4\mathrm{cm} \end{aligned}

Trigonometry questions: missing angles

4. A builder is constructing a roof. The wood he is using for the sloped section of the roof is 4m long and the peak of the roof needs to be 2m high. What angle should the piece of wood make with the base of the roof?

26.6^{\circ}

60^{\circ}

0.008^{\circ}

30^{\circ}

\begin{aligned} \sin(\theta)&=\frac{O}{H}\\\\ \sin(\theta)&=\frac{2}{4}\\\\ \theta&=sin^{-1}(\frac{2}{4})\\\\ \theta&=30^{\circ} \end{aligned}

5. A ladder is leaning against a wall. The ladder is 1.8m long and the bottom of the ladder is 0.5m from the base of the wall. To be considered safe, a ladder must form an angle of between 70^{\circ} and 80^{\circ} with the floor. Is this ladder safe?

Yes

No

Not enough information

\begin{aligned} \cos(\theta)&=\frac{A}{H}\\\\ \cos(\theta)&=\frac{0.5}{1.8}\\\\ \theta&=cos^{-1}(\frac{0.5}{1.8})\\\\ \theta&=73.9^{\circ} \end{aligned}

Yes it is safe.

6. A helicopter flies 40km east followed by 105km south. On what bearing must the helicopter fly to return home directly?

201^{\circ}

159^{\circ}

339^{\circ}

\begin{aligned} \tan(\theta)&=\frac{O}{A}\\\\ \tan(\theta)&=\frac{40}{105}\\\\ \theta&=tan^{-1}(\frac{40}{105})\\\\ \theta&=21^{\circ} \end{aligned}

Since bearings are measured clockwise from North, we need to do 360-21=339^{\circ}.

In geometry, trigonometry questions ask students to solve a variety of problems including multi-step problems and real-life problems. We also need to be familiar with the exact values of the trigonometric functions at certain angles.

We look at applying trigonometry to 3D problems as well as using the sine rule, cosine rule, and area of a triangle.

Trigonometry questions: SOHCAHTOA

7.  Calculate the size of angle ABC. Give your answer to 3 significant figures.

24.6^{\circ}

38.4^{\circ}

18.8^{\circ}

21.8^{\circ}

\begin{aligned} A&=\frac{O}{\tan(\theta)}\\\\ A&=\frac{10}{\tan(28)}\\\\ A&=18.81\mathrm{m} \end{aligned}

\begin{aligned} \cos(\theta)&=\frac{A}{H}\\\\ \cos(\theta)&=\frac{18.81}{24}\\\\ \theta&=cos^{-1}(\frac{18.81}{24})\\\\ \theta&=38.4^{\circ} \end{aligned}

8. Kevin’s garden is in the shape of an isosceles trapezoid (the sloping sides are equal in length). Kevin wants to buy enough grass seed for his garden. Each box of grass seed covers 15m^2 . How many boxes of grass seed will Kevin need to buy?

6

4

5

10

To calculate the area of the trapezoid, we first need to find the height. Since it is an isosceles trapezium, it is symmetrical and we can create a right angled triangle with a base of \frac{10-5}{2} .

\begin{aligned} O&=\tan(\theta)\times A\\\\ O&=\tan(78)\times2.5\\\\ O&=11.76\mathrm{m} \end{aligned}

We can then find the area of the trapezoid:

\begin{aligned} \text{Area }&=\frac{1}{2}(a+b)h\\\\ \text{Area }&=\frac{1}{2}(5+10)\times 11.76\\\\ \text{Area }&=88.2\mathrm{m}^{2} \end{aligned}

Number of boxes: 88.215=5.88

Kevin will need 6 boxes.

Trigonometry questions: exact values

9.  Which of these values cannot be the value of \sin(\theta) ?

\frac{1}{2}

\frac{\sqrt{3}}{2}

\frac{5}{2}

\frac{-\sqrt{2}}{2}
\sin\theta \text{ cannot be } \frac{5}{2} \text{ because } \sin\theta \text{ is always between 1 and -1}

10. . Write 4sin(60) + 3tan(60) in the form a\sqrt{k}.

5\sqrt{3}

2+3\sqrt{3}

\frac{3}{2} \sqrt{3}

\frac{1}{2}\sqrt{12} +3
\sin(60)=\frac{\sqrt{3}}{2}, \tan(60)=\sqrt{3} \begin{aligned} 4\sin(60)+3\tan(60) &= 4\times \frac{\sqrt{3}}{2}+3\sqrt{3}\\\\ &=2\sqrt{3}+3\sqrt{3}\\\\ &=5\sqrt{3} \end{aligned}

Trigonometry questions: 3D trigonometry

11. Work out angle a, between the line AG and the plane ADHE.

14.3^{\circ}

15.6^{\circ}

15.9^{\circ}

90^{\circ}

We need to begin by finding the length AH by looking at the triangle AEH and using pythagorean theorem.

\begin{aligned} &AH^2=14^2+3^2 \\\\ &AH^2=205 \\\\ &AH=14.32cm \end{aligned}

We can then find angle a by looking at the triangle AGH.

\begin{aligned} \tan(\theta)&=\frac{O}{A}\\\\ \tan(\theta)&=\frac{4}{14.32}\\\\ \theta&=tan^{-1}(\frac{4}{14.32})\\\\ \theta&=15.6^{\circ} \end{aligned}

12.  Work out the length of BC.

30.0mm

19.0mm

23.3mm

29.6mm

First we need to find the length DC by looking at triangle CDE.

\begin{aligned} H&=\frac{O}{\sin(\theta)}\\\\ H&=\frac{15}{\sin(52)}\\\\ H&=19.04\mathrm{cm} \end{aligned}

We can then look at triangle BAC.

\begin{aligned} H&=\frac{o}{\sin(\theta)}\\\\ H&=\frac{19.04}{\sin(40)}\\\\ H&=29.6\mathrm{mm} \end{aligned}

Trigonometry questions: sine/cosine rule

13. Ship A sails 40km due West and ship B sails 65km on a bearing of 050^{\circ} . Find the distance between the two ships.

76.3km

99.0km

52.5km

84.9km

The angle between their two paths is 90+50=140^{\circ} .

\begin{aligned}
a^{2}&=b^{2}+c^{2}-2bc \cos(A)\\\\
a^{2}&=40^{2}+65^{2}-2\times 40 \times 65 \cos(140)\\\\
a^{2}&=5825-5200 \cos(140)\\\\
a^{2}&=9808.43\\\\
a&=99.0\mathrm{km}
\end{aligned}

14.   Find the size of angle B.

78^{\circ}

31.8^{\circ}

9.3^{\circ}

28.8^{\circ}

First we need to look at the right angled triangle.

\begin{aligned} &O=tan()A\\\\ &O=tan(22)23\\\\ &O=9.29cm \end{aligned}

Then we can look at the scalene triangle.

\begin{aligned} \frac{\sin(A)}{A}&=\frac{\sin(B)}{B}\\\\ \frac{\sin(51)}{15}&=\frac{\sin(B)}{9.29}\\\\ 9.29\times \frac{\sin(51)}{15} &= \sin(B)\\\\ 0.481&=\sin(B)\\\\ \sin^{-1}(0.481)&=B\\\\ 28.8^{\circ}&=B \end{aligned}

Trigonometry questions – area of a triangle

15. The area of the triangle is 16cm^2 . Find the length of the side x .

5.0cm

10.0cm

5.3cm

4cm

\begin{aligned}
\text{Area }&=\frac{1}{2}ab \sin(C)\\\\
16&=\frac{1}{2} \times x \times 2x \times \sin(40)\\\\
16&=x^{2} \sin(40)\\\\
\frac{1}{\sin(40)}&=x^{2}\\\\
24.89&=x^{2}\\\\
5.0&=x
\end{aligned}

Looking for more high school trigonometry math questions and word problems?

Try these:

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