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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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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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15 Trigonometry Questions And Practice Problems (KS3 & KS4): Harder GCSE Exam Style Questions Included

Beki christian.

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

Here we’ve provided 15 trigonometry questions to provide students with practice at the various sorts of trigonometry problems and GCSE exam style questions you can expect in KS3 and KS4 trigonometry.

You may also like:

Ks3 trigonometry questions – missing angles, ks4 trigonometry questions – sohcahtoa, ks4 trigonometry questions – exact values, ks4 trigonometry questions – 3d trigonometry, ks4 trigonometry questions – sine/cosine rule, ks4 trigonometry questions – area of a triangle, looking for more trigonometry questions and resources, looking for more ks3 and ks4 maths questions.

Free GCSE maths revision resources for schools As part of the Third Space Learning offer to schools, the personalised online GCSE maths tuition can be supplemented by hundreds of free GCSE maths revision resources from the secondary maths resources library including: – GCSE maths past papers – GCSE maths worksheets – GCSE maths questions – GCSE maths topic list

Trigonometry in the real world

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

Trigonometry in KS3 and KS4

In KS3 we learn about the trigonometric ratios sin, cos and tan and how we can use these to calculate sides and angles in right angled triangles. In KS4 trigonometry involves applying this to a variety of situations as well as learning the exact values of sin, cos and tan for certain angles.

In the higher GCSE syllabus we learn about the sine rule, the cosine rule, a new formula for the area of a triangle and we apply trigonometry to 3D shapes. In A Level maths trigonometry is developed further but that is not the focus of the trigonometry questions here.

Download this 15 Trigonometry Questions And Practice Problems (KS3 & KS4) Worksheet

Help your students prepare for their Maths GSCE with this free Trigonometry worksheet of 15 multiple choice questions and answers.

How to answer trigonometry questions

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

To answer the trigonometry question:

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 angled 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.

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

KS3 trigonometry questions

In KS3 trigonometry questions focus on understanding of sin, cos and tan (SOHCAHTOA) to calculate missing sides and angles in right triangles.

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KS3 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.

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.

Total height = 355.8+1=356.8m.

3. Triangle ABC is isosceles. Work out the height of triangle ABC.

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

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?

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?

Not enough information

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?

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

KS4 trigonometry questions

In KS4 maths, 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.

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

Trigonometry is covered by all exam boards, including Edexcel, AQA and OCR.

Read more: Question Level Analysis Of Edexcel Maths Past Papers (Foundation)

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

8. Kevin’s garden is in the shape of an isosceles trapezium (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?

To calculate the area of the trapezium, 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} .

We can then find the area of the trapezium:

Number of boxes: 88.215=5.88

Kevin will need 6 boxes.

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

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

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

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

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

We can then look at triangle BAC.

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.

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.

First we need to look at the right angled triangle.

Then we can look at the scalene triangle.

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

\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}

Third Space Learning’s free GCSE maths resource library contains detailed lessons with step-by-step instructions on how to solve ratio problems, as well as worksheets with trigonometry practice questions and more GCSE exam questions.

Take a look at the trigonometry lessons today – more are added every week.

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Trigonometric Word Problems

In these lessons, examples, and solutions we will learn the trigonometric functions (sine, cosine, tangent) and how to solve word problems using trigonometry.

Related Pages Trigonometry Word Problems Lessons On Trigonometry Inverse trigonometry Trigonometry Worksheets

The following diagram shows how SOHCAHTOA can help you remember how to use sine, cosine, or tangent to find missing angles or missing sides in a trigonometry problem. Scroll down the page for examples and solutions.

How To Solve Trigonometry Problems Or Questions?

Step 1: If no diagram is given, draw one yourself. Step 2: Mark the right angles in the diagram. Step 3: Show the sizes of the other angles and the lengths of any lines that are known. Step 4: Mark the angles or sides you have to calculate. Step 5: Consider whether you need to create right triangles by drawing extra lines. For example, divide an isosceles triangle into two congruent right triangles. Step 6: Decide whether you will need the Pythagorean theorem, sine, cosine or tangent. Step 7: Check that your answer is reasonable. The hypotenuse is the longest side in a right triangle.

How To Use Cosine To Calculate The Side Of A Right Triangle?

Solution: Use the Pythagorean theorem to evaluate the length of PR.

How To Use Tangent To Calculate The Side Of A Triangle?

Calculate the length of the side x, given that tan θ = 0.4

How To Use Sine To Calculate The Side Of A Triangle?

Calculate the length of the side x, given that sin θ = 0.6

How To Solve Word Problems Using Trigonometry?

The following video shows how to use the trigonometric ratio, tangent, to find the height of a balloon.

How To Solve Word Problems Using Sine?

This video shows how to use the trigonometric ratio, sine, to find the elevation gain of a hiker going up a slope.

Example: A hiker is hiking up a 12 degrees slope. If he hikes at a constant rate of 3 mph, how much altitude does he gain in 5 hours of hiking?

How To Use Cosine To Solve A Word Problem?

Example: A ramp is pulled out of the back of truck. There is a 38 degrees angle between the ramp and the pavement. If the distance from the end of the ramp to to the back of the truck is 10 feet. How long is the ramp? Step 1: Find the values of the givens. Step 2: Substitute the values into the cosine ratio. Step 3: Solve for the missing side. Step 4: Write the units

How To Solve Word Problems Using Tangent?

The following video shows how to use trigonometric ratio, tangent, to find the height of a building.

How To Solve Trigonometry Word Problems Using Tangent?

Example: Neil sees a rocket at an angle of elevation of 11 degrees. If Neil is located at 5 miles from the rocket launch pad, how high is the rocket?

How To Determining The Speed Of A Boat Using Trigonometry?

Example: A balloon is hovering 800 ft above a lake. The balloon is observed by the crew of a boat as they look upwards at an angle of 0f 20 degrees. 25 seconds later, the crew had to look at an angle of 65 degrees to see the balloon. How fast was the boat traveling?

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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

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Unit 4: Trigonometric equations and identities

About this unit, inverse trigonometric functions.

• Intro to arcsine (Opens a modal)
• Intro to arctangent (Opens a modal)
• Intro to arccosine (Opens a modal)
• Restricting domains of functions to make them invertible (Opens a modal)
• Domain & range of inverse tangent function (Opens a modal)
• Using inverse trig functions with a calculator (Opens a modal)
• Inverse trigonometric functions review (Opens a modal)
• Trigonometric equations and identities: FAQ (Opens a modal)
• Evaluate inverse trig functions Get 3 of 4 questions to level up!

Sinusoidal equations

• Solving sinusoidal equations of the form sin(x)=d (Opens a modal)
• Cosine equation algebraic solution set (Opens a modal)
• Cosine equation solution set in an interval (Opens a modal)
• Sine equation algebraic solution set (Opens a modal)
• Solving cos(θ)=1 and cos(θ)=-1 (Opens a modal)
• Solve sinusoidal equations (basic) Get 3 of 4 questions to level up!
• Solve sinusoidal equations Get 3 of 4 questions to level up!

Sinusoidal models

• Interpreting solutions of trigonometric equations (Opens a modal)
• Trig word problem: solving for temperature (Opens a modal)
• Trigonometric equations review (Opens a modal)
• Interpret solutions of trigonometric equations in context Get 3 of 4 questions to level up!
• Sinusoidal models word problems Get 3 of 4 questions to level up!

Angle addition identities

• Trig angle addition identities (Opens a modal)
• Using the cosine angle addition identity (Opens a modal)
• Using the cosine double-angle identity (Opens a modal)
• Proof of the sine angle addition identity (Opens a modal)
• Proof of the cosine angle addition identity (Opens a modal)
• Proof of the tangent angle sum and difference identities (Opens a modal)
• Using the trig angle addition identities Get 3 of 4 questions to level up!

Using trigonometric identities

• Finding trig values using angle addition identities (Opens a modal)
• Using the tangent angle addition identity (Opens a modal)
• Using trig angle addition identities: finding side lengths (Opens a modal)
• Using trig angle addition identities: manipulating expressions (Opens a modal)
• Using trigonometric identities (Opens a modal)
• Trig identity reference (Opens a modal)
• Find trig values using angle addition identities Get 3 of 4 questions to level up!

Challenging trigonometry problems

• Trig challenge problem: area of a triangle (Opens a modal)
• Trig challenge problem: area of a hexagon (Opens a modal)
• Trig challenge problem: cosine of angle-sum (Opens a modal)
• Trig challenge problem: arithmetic progression (Opens a modal)
• Trig challenge problem: maximum value (Opens a modal)
• Trig challenge problem: multiple constraints (Opens a modal)
• Trig challenge problem: system of equations (Opens a modal)

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

• x = 10 / tan(51°) = 8.1 (2 significant digits) H = 10 / sin(51°) = 13 (2 significant digits)
• Area = (1/2)(2x)(x) = 400 Solve for x: x = 20 , 2x = 40 Pythagora's theorem: (2x) 2 + (x) 2 = H 2 H = x √(5) = 20 √(5)
• BH perpendicular to AC means that triangles ABH and HBC are right triangles. Hence tan(39°) = 11 / AH or AH = 11 / tan(39°) HC = 19 - AH = 19 - 11 / tan(39°) Pythagora's theorem applied to right triangle HBC: 11 2 + HC 2 = x 2 solve for x and substitute HC: x = √ [ 11 2 + (19 - 11 / tan(39°) ) 2 ] = 12.3 (rounded to 3 significant digits)
• Since angle A is right, both triangles ABC and ABD are right and therefore we can apply Pythagora's theorem. 14 2 = 10 2 + AD 2 , 16 2 = 10 2 + AC 2 Also x = AC - AD = √( 16 2 - 10 2 ) - √( 14 2 - 10 2 ) = 2.69 (rounded to 3 significant digits)
• Use right triangle ABC to write: tan(31°) = 6 / BC , solve: BC = 6 / tan(31°) Use Pythagora's theorem in the right triangle BCD to write: 9 2 + BC 2 = BD 2 Solve above for BD and substitute BC: BD = √ [ 9 + ( 6 / tan(31°) ) 2 ] = 13.4 (rounded to 3 significant digits)
• The triangle is right and the size one of its angles is 45°; the third angle has a size 45° and therefore the triangle is right and isosceles. Let x be the length of one of the sides and H be the length of the hypotenuse. Area = (1/2)x 2 = 50 , solve for x: x = 10 We now use Pythagora to find H: x 2 + x 2 = H 2 Solve for H: H = 10 √(2)
• Let a be the length of the side opposite angle A, b the length of the side adjacent to angle A and h be the length of the hypotenuse. tan(A) = opposite side / adjacent side = a/b = 3/4 We can say that: a = 3k and b = 4k , where k is a coefficient of proportionality. Let us find h. Pythagora's theorem: h 2 = (3k) 2 + (5k) 2 Solve for h: h = 5k sin(A) = a / h = 3k / 5k = 3/5 and cos(A) = 4k / 5k = 4/5
• Let b be the length of the side opposite angle B and c the length of the side opposite angle C and h the length of the hypotenuse. sin(B) = b/h and cos(B) = c/h sin(B) = cos(B) means b/h = c/h which gives c = b The two sides are equal in length means that the triangle is isosceles and angles B and C are equal in size of 45°.
• Let x be the length of side AC. Use the cosine law 12 2 = 8 2 + x 2 - 2 · 8 · x · cos(59°) Solve the quadratic equation for x: x = 14.0 and x = - 5.7 x cannot be negative and therefore the solution is x = 14.0 (rounded to one decimal place).

More References and links on Trigonometry

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Trigonometry mixed homework including problem solving

Subject: Mathematics

Age range: 14-16

Resource type: Worksheet/Activity

Last updated

26 April 2018

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Trigonometry questions designed to test students ability to apply their knowledge of basic trigonometry using the sine, cosine and tangent ratios. Includes problem solving questions. Solutions provided!

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Word Problem: Rachel has 17 apples. She gives some to Sarah. Sarah now has 8 apples. How many apples did Rachel give her?

Simplified Equation: 17 - x = 8

Word Problem: Rhonda has 12 marbles more than Douglas. Douglas has 6 marbles more than Bertha. Rhonda has twice as many marbles as Bertha has. How many marbles does Douglas have?

Variables: Rhonda's marbles is represented by (r), Douglas' marbles is represented by (d) and Bertha's marbles is represented by (b)

Simplified Equation: {r = d + 12, d = b + 6, r = 2 �� b}

Word Problem: if there are 40 cookies all together and Angela takes 10 and Brett takes 5 how many are left?

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