Right angled Trigonometry

SOHCAHTOA

This part of trigonometry can be seen in the recurring relationships between the sides of right angles triangles. The relationship means that we can use two bits of information about a triangle to work out a third and so on until we know everything we need to know. As such it is a really useful tool both for further mathematics an in a nmber of very practical applications.

Learn about it

The following is a series of slides and videos that will help you understand, learn about and review this sub-topic.

Keep track of your progress and practice the exam questions on this Right angled trigonometry record sheet

Slides Gallery

Use these slides to review the material and key points covered in the videos.

Understanding trig ratios

This video explains the fundamental concept behind trigonometrical relationships.

Making Statements

Solving problems with trigonometry rests heavily on being able to take a given scenario and make correct statements of equality about it. These statements can then be used to find unknowns.

Rearranging equations

Often the thing that students find most difficult when solving trig problems is successfully rearranging the equations. This is a really important bit of algebraic manipulation that is useful in lots of places.

Solving for angles

Here we look at solving right angled trig problems where the unknown is an angle.

Solving for side lengths

Here we look at solving right angled trig problems where the unknown is an side length.

Angles of elevation and depression

These are important definitions to be aware of so you dont tae a wrong turn at the beginning of a question.

Revise

This section of the page can be used for quick review. The flashcards help you go over key points and the quiz lets you practice answering questions on this subtopic.

Flash Card

Review this condensed 'key point' flashcard to help you check and keep ideas fresh in your mind.

Practice Questions

Practice your understanding on these quiz questions. Check your answers when you are done and read the hints where you got stuck. If you find there are still some gaps in your understanding then go back to the videos and slides above. NOTE - Make sure that you have set your GDC to be in degrees. See the Your Graphical Display Calculator page for help with this.

Use your calculator to work out solve the following equations. Give your answers as decimals correct to 3sf.

a) x = sin 35

b) y = tan 75

c) z = cos 26

d) Sinp = 0.7

e) Cosq = 0.1

f) Tanr = 2.5

a) x =

b) y =

c) z =

d) p =

e) q =

f) r =

This is really just about correctly using the buttons on your calculator. For the last three parts when you have the sine of an angle, you need the inverse sine of that number to get the angle. For example,

part d) \(sin\quad p\quad =\quad 0.7\\ so\quad p\quad =\quad { sin }^{ -1 }(0.7)\quad =\quad 44.4\\ \\ \)

Use your calculator to work out solve the following equations. Give your answers as decimals correct to 3sf

a) \(sin\quad 30\quad =\quad \frac { x }{ 10 } \) , so x =

b) \(Cos\quad 40\quad =\quad \frac { 5 }{ x } \) , so x

c) \(Tan\quad x\quad =\quad \frac { 7 }{ 14 } \), so x =

a) \(sin\quad 30\quad =\quad \frac { x }{ 10 } \\ so\quad x\quad =\quad 10\quad \times \quad sin\quad 30\quad =\quad 5.00\)

b) \(Cos\quad 40\quad =\quad \frac { 5 }{ x } \\ so\quad x\quad =\quad \frac { 5 }{ cos\quad 40 } =\quad 6.53\)

c) \(Tan\quad x\quad =\quad \frac { 7 }{ 14 } \\ so\quad x\quad =\quad { tan }^{ -1 }\left( \frac { 7 }{ 14 } \right) =\quad 26.6\)

Consider the folloiwng diagram

Capital letters (in green) are angles. Lower case letters (in blue) are side lengths.

Next to each of the statements below, enter either 'T' for True of 'F' for False

\(Sin\quad A\quad =\quad \frac { f }{ b } \)
\(Cos\quad B\quad =\quad \frac { f }{ a } \)
\(Tan\quad P\quad =\quad \frac { f }{ d } \)
\(Cos\quad Q\quad =\quad \frac { e }{ a } \)
\(Tan\quad B\quad =\quad \frac { f }{ e } \)
\(Sin\quad Q\quad =\quad \frac { e }{ a } \)
\(Sin\quad A\quad =\quad Cos\quad P\)
\(Tan\quad A\quad =\quad Tan\quad B\)

This question is really about encouraging the making of statements that are true about a given situation. In practice you can then pick which ones will be mist hepful to you. Clearly it depends on remmebering the trig ratios - SOHCAHTOA and identifying the Hypotenuse as the longest side and the positions of the Opposite and adject sides relative to the angle in question.

Consider the triangle in the image below.

Use trigonometry to work out the lengths of sides a and b in the diagram. Give your answer correct to 3 sf.

a =

b =

a, \(Sin40\quad =\quad \frac { a }{ 10 } \\ so\quad a\quad =\quad 10\times sin40\\ a\quad =\quad 6.43\quad (3sf)\)

b, \(Cos40\quad =\quad \frac { b }{ 10 } \\ so\quad b\quad =\quad 10\times cos40\\ a\quad =\quad 7.66\quad (3sf)\)

Consider the triangle in the image below.

Use trigonometry to work out the lengths of sides a and b in the diagram. Give your answer correct to 3 sf.

a =

b =

a, \(sin28\quad =\quad \frac { 15 }{ a } \\ so\quad a\quad =\quad \frac { 15 }{ sin28 } \\ a\quad =\quad 32.0\quad (3sf)\)

b, \(tan28\quad =\quad \frac { b }{ 15 } \\ so\quad b\quad =\quad 15\times tan28\\ a\quad =\quad 7.98\quad (3sf)\)

Consider the triangle in the image below.

Use trigonometry to work out the lengths of angles A and B in the diagram. Give your answer correct to 3 sf.

A =

B =

A, \(cos\quad A\quad =\quad \frac { 41 }{ 56 } \\ so,\quad A\quad =\quad { cos }^{ -1 }\left( \frac { 41 }{ 56 } \right) \\ A\quad =\quad 42.9\quad (3sf)\)

B, \(sin\quad B\quad =\quad \frac { 41 }{ 56 } \\ so,\quad B\quad =\quad { sin }^{ -1 }\left( \frac { 41 }{ 56 } \right) \\ A\quad =\quad 47.1\quad (3sf)\)

NOTE, A + B = 90 as they should!

The diagram below shows a ladder put against a house for cleaning the gutters. The yop of the ladder is 5m above the ground and the angle the ladder makes with the horizontal must be 65°.

a) How far away (x) from the wall must the base of the ladder be put to make the angle 65°

b) How long does the ladder need to be? (Use trigonometry and unrounded numbers to answer this question)

c) If the length of the ladder is fixed at 6m and the height against the house remains at 5m, what would the angle between the ladder and the gorund become?

d) What is the precentage error in the new angle from the required 65°? (Use the rounded answer to part c)

a) x = m

b) length of ladder = m

c) New angle betwen the ladder and the ground = °

d) Percentage error = %

a) \(tan\quad 65\quad =\quad \frac { 5 }{ x } \\ so,\quad x\quad =\quad \frac { 5 }{ tan\quad 65 } \\ A\quad =\quad 2.33\quad (3sf)\)

b) \(sin\quad 65\quad =\quad \frac { 5 }{ l } \\ so,\quad l\quad =\quad \frac { 5 }{ sin\quad 65 } \\ A\quad =\quad 5.52\quad (3sf)\\ \\ OR\quad { l }^{ 2 }\quad =\quad { 5 }^{ 2 }+{ x }^{ 2 }\\ Using\quad the\quad unrounded\quad answer\quad for\quad x\quad gives\quad the\quad same\quad result\quad for\quad l\)

NOTE, you could have used cosine with x and 65°. Again if you used the unrounded anser for x you would have got the same result.

c) \(sin\quad A\quad =\quad \frac { 5 }{ 6 } \\ so,\quad A\quad =\quad { sin }^{ -1 }\left( \frac { 5 }{ 6 } \right) \\ A\quad =\quad 56.4\quad (3sf)\)

d) = \(\frac { 65\quad -\quad 56.4 }{ 65 } \times \quad 100\quad =\quad 13.2\)

The diagram below shows part of a metal structure and lists some its geometrical properties.

Find the folllowing giving all answers correct to 3sf

a) The length AC = m

b) \(\hat { CAD } \) = °

c) \(\hat { mAD } \) =

In changes to the design it is decided that \(\hat { mAD } \) needs to be exactly halved

d) What is the new length AM? m

a) \({ AC }^{ 2 }\quad =\quad { 4 }^{ 2 }+{ 5 }^{ 2 }\\ AC\quad =\sqrt { 25 } =5.00\quad (3sf)\)

b) \(tan\hat { CAD } =\frac { 4 }{ 3 } \\ so,\quad \hat { CAD } \quad =\quad { tan }^{ -1 }\left( \frac { 4 }{ 3 } \right) \\ \hat { CAD } \quad =\quad 53.1°\)

c) \(tan\hat { mAD } =\frac { 2 }{ 3 } \\ so,\quad \hat { mAD } \quad =\quad { tan }^{ -1 }\left( \frac { 2 }{ 3 } \right) \\ \hat { mAD } \quad =\quad 45.0°\)

d) \(cos22.5\quad =\quad \frac { 3 }{ mA } \\ so,\quad mA\quad =\quad \frac { 3 }{ cos\quad 22.5 } \\ mA\quad =\quad 3.25m\quad (3sf)\)

USe the diagram below for the following questions.

The angle of elevation from Point A to some looking out of the window at point B is 60°

a) What is the height, h, of point B above the ground?

A person looking at point A from point C is 35m above the ground.

b) What is the angle of depression, D, they are looking at?

a) Height h =

b) Angle of depression, a =

a) \(tan60\quad =\quad \frac { h }{ 12 } \\ so,\quad h\quad =\quad 12\times tan60\\ h\quad =\quad 20.9m\quad (3sf)\)

b) \(tanD\quad =\quad \frac { 35 }{ 12 } \\ so,\quad D\quad =\quad { tan }^{ -1 }\left( \frac { 35 }{ 12 } \right) \\ h\quad =\quad 71.1°\quad (3sf)\)

The diagram below shows ta narrow chanel of water between the cliff coasts of two islands. Some distances are shown.

The diagram also shows.....

The top of a lighthouse at point A
The view from the base of the light house at point B
A boat at point C
The view from the top of the cliff at point D

The angle of elevation of point D from point C is 65°. The angle of depression of point C from point D is 40°.

a) Find the width of the water chanel by finding the respective distances x and y to 3sf. (Use rounded answers to get the total width)

The angle of elevation from point C to point A is 75° .

b) What is the height of the lighthouse above the the ground? (Use rounded ansers to any distances you have calculated)

c) What is the angle, p, of depression from the top of the light house to point D? (Use rounded ansers to any distances you have calculated)

a) x = m to (3sf), y = m (3sf),

Total distance = m

b) h = m

a) \(tan40\quad =\quad \frac { 10 }{ x } \\ x\quad =\quad \frac { 10 }{ tan40 } \\ x\quad =\quad 11.9\quad (3sf)\)

\(tan65\quad =\quad \frac { 18 }{ y } \\ y\quad =\quad \frac { 18 }{ tan65 } \\ y\quad =\quad 8.39\quad (3sf)\)

Total distance = 8.39 + 11.9 = 20.29 , 20.3 (3sf)

b) \(tan75\quad =\quad \frac { h+10 }{ 11.9 } \\ h\quad +\quad 10\quad =\quad 11.9\times tan75\\ h\quad =\quad 11.9\times tan75\quad -\quad 10\\ h\quad =\quad 34.4\quad m\quad (3sf)\)

c) \(tanp\quad =\quad \frac { 10\quad +\quad 34.4\quad -\quad 18 }{ 11.9 } =\frac { 26.4 }{ 20.3 } \\ p\quad =\quad { tan }^{ -1 }\left( \frac { 26.4 }{ 20.3 } \right) \\ p\quad =\quad 52.4\quad (3sf)\)

Exam Style Questions

The following questions are based on IB exam style questions from past exams. You should print these off (from the document at the top) and try to do these questions under exam conditions. Then you can check your work with the video solution. NOTE - Make sure that you have set your GDC to be in degrees. See the Your Graphical Display Calculator page for help with this.

Question 1

Question 2

Question 3

.....

IBDP Maths Studies

Start here

Your Graphical Display Calculator

Number & Algebra

Mathematical Models

Statistics

Geometry & Trigonometry

Logic, sets & Probability

Calculus

Exams & Assessment

Right angled Trigonometry

SOHCAHTOA

Learn about it

Slides Gallery

Understanding trig ratios

Making Statements

Rearranging equations

Solving for angles

Solving for side lengths

Angles of elevation and depression

Revise

Flash Card

Practice Questions

Exam Style Questions

Question 1

Question 2

Question 3

IBDP Maths Studies

Geometry and Trigonometry Home Start here Your Graphical Display Calculator Number and Algebra Mathematical Models Statistics Logic, sets and Probability Calculus Exams & Assessment

Right angled Trigonometry

SOHCAHTOA

Learn about it

Slides Gallery

Understanding trig ratios

Making Statements

Rearranging equations

Solving for angles

Solving for side lengths

Angles of elevation and depression

Revise

Flash Card

Practice Questions

Exam Style Questions

Question 1

Question 2

Question 3

Geometry and Trigonometry
Home
Start here
Your Graphical Display Calculator
Number and Algebra
Mathematical Models
Statistics
Logic, sets and Probability
Calculus
Exams & Assessment