1. GCSE Trigonometry of Right-Angled Triangles
Dr J Frost
Objectives:
Find unknown sides in right-angled triangles using a side and angle.
Find unknown angles in right-angled triangle using two sides.
Note: ‘Exact trigonometric ratios’ and 3D Trigonometry are covered in separate slides.
Last modified: 1st January 2019

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3. Frost Childhood Story Motivation 𝑦
I was in Year 9 and was trying to write a computer program that would draw an analogue clock (as you do).
I needed to work out the two coordinates to draw the minute hand between, and similarly for the hour hand, given the current hour, minute and desired length of each hand.
I couldn’t work out how to do it, until I (coincidentally) learnt trigonometry in maths the next day, and was able to finish my program that evening!
(𝑎,𝑏) 𝜃 𝑟 𝑥

4. Starter 13 𝑥 𝑦 3 4 5
Determine the length 𝑥.
Determine the length 𝑦. ? ?
5 2 + 𝑦 2 = 𝑦 2 =169 𝑦 2 =144 𝑦=12
= 𝑥 2 25= 𝑥 2 𝑥=5

5. But…
We used Pythagoras’ theorem if all 3 sides of a right-angled triangle were involved.
But what if we had two sides involved and an angle? (excluding the right-angle)
Unknown angle. 𝟑 𝟏𝟑 𝟒 𝒂 𝟏𝟒 𝒚 𝒙
Unknown side.
Unknown side.
𝟑𝟎° 𝟏𝟎

6. Names of sides relative to an angle
The hypotenuse is the longest side of a right-angled triangle, and is opposite the right-angle.
hypotenuse ?
opposite ?
The ‘opposite’ is the side opposite the angle of interest.
30°
adjacent ?
The ‘adjacent’ is the side adjacent (i.e. next to) the angle of interest.

7. Quickfire Side Naming 𝑥 𝑦 𝑧 1 2 𝑐 𝑎 𝑏 ? ? ? ? ? ? ? ? ? Hypotenuse
Opposite
Adjacent 𝑥 𝑦 𝑧 1 2 𝑐 𝑎 𝑏 𝑥 ? ? ?
60° 𝑧 𝑦 ? ? ? 1 𝑥 𝜽 2 ? ? ? 𝑐
20° 𝑎 𝑏

8. You can remember this using:
Sin/Cos/Tan
! sin, cos and tan are functions which take an angle and give us the ratio between pairs of sides in a right angle triangle.
Recall that ratio can just mean how many times bigger one quantity is than another.
sin 𝜃 = 𝑜 ℎ ? 𝜽 𝒐 𝒉 𝒂
cos 𝜃 = 𝑎 ℎ ?
Emphasise that sin for example is a ‘number machine’ which takes an angle as its input and outputs the ratio between the opposite and hypotenuse.
tan 𝜃 = 𝑜 𝑎 ?
You can remember this using:
“soh cah toa”

9. Find the value of 𝑥 (to 3sf)
Examples
Fro Tip: Put your h/a/o letters in circles – this avoids confusion with any variables you have used as side lengths (in this example, 𝑥)
Find the value of 𝑥 (to 3sf)
Step 1: Determine which sides are hyp/adj/opp.
𝟒𝟎° 𝟒 𝒙 ℎ
Step 2: Work out which trigonometric function we need.
We used 𝑎 and ℎ. Thinking about “soh cah toa”, we want the “cah”, i.e. cos:
cos 40° = 𝑥 4 𝑎
𝑥=4 cos 40° 𝑥=3.06 ×4 ×4
Fro Tip: The angle always follows the sin/cos/tan. I see a lot of students incorrectly write cos 𝑥 4 =40°
Step 3: Rearrange the equation to find the unknown.
𝑥 was being divided by 4, so we multiply both sides by 4 to cancel this out.

10. Find the value of 𝑥 (to 3sf)
Further Example
Find the value of 𝑥 (to 3sf)
𝐬𝐢𝐧 𝟐𝟎° = 𝒙 𝟕 𝒙=𝟕 𝐬𝐢𝐧 𝟐𝟎° 𝒙=𝟐.𝟑𝟗 ?
𝟐𝟎° 𝟕 𝒙 ℎ
Extremely Advanced Side Note: I often get asked how that calculator actually works out, for example, sin 20° . The process is complicated and would never ordinarily be done by a ‘human’. You won’t likely learn the technique for calculating it unless you do Further Maths A Level, using something called Maclaurin Series. But since you asked…
sin 𝑥 can be calculated using the infinitely long formula:
sin 𝑥 =𝑥− 𝑥 3 3! + 𝑥 5 5! − 𝑥 7 7! + 𝑥 9 9! −…
where 5! for example means 5×4×3×2×1. But the angle 𝑥 has to be in a unit called ‘radians’ (we can convert from degrees to radians by multiplying the angle by 𝜋 180 ). We can’t substitute into an infinitely long expression, but we can stop after the first several terms to get an accurate value, as the terms in the sum gradually approach 0. For example, sin 20 = …, and I got the same (up to 10 dp) by substituting into the above formula up to the 𝑥 9 term. 𝑜

11. Find the value of 𝑦 (to 3sf)
Test Your Understanding So Far
Find the value of 𝑦 (to 3sf)
𝟐𝟓° 𝒚 𝟓 ?
𝐭𝐚𝐧 𝟐𝟓° = 𝒚 𝟓 𝒚=𝟓 𝐭𝐚𝐧 𝟐𝟓° 𝒚=𝟐.𝟑𝟑 𝑜 𝑎

12. Harder Examples
60 ° 𝒙 12
If the variable is in the denominator, you can apply something called the ‘swapsie trick’.
Notice that we can rearrange:
8 4 =2 → 8 2 =4
i.e. we can ‘swap’ the thing we’re dividing by and the result of the division. In this trig question, we could swap the 𝑥 and the sin 60 ?
𝒔𝒊𝒏 𝟔𝟎 = 𝟏𝟐 𝒙 𝒙= 𝟏𝟐 𝒔𝒊𝒏 𝟔𝟎 =𝟏𝟑.𝟖𝟔
30° 4 𝒙 ?
𝒕𝒂𝒏 𝟑𝟎 = 𝟒 𝒙 𝒙= 𝟒 𝒕𝒂𝒏 𝟑𝟎 =𝟔.𝟗𝟑

13. Test Your Understanding1 2
8.63 ?
Working:
sin 32°= 𝐵𝐶 47 𝐵𝐶=24.906
tan 51° = 𝐵𝐷 𝐵𝐷= tan 51° =20.2
20.2 ?

14. Exercise 1 ? ? ? ? ? ? ? (questions on provided sheet) 1
Find 𝑥, giving your answers to 3 significant figures.
𝑥=11.0 ?
𝟕𝟎° 15 𝒙
𝟒𝟎° 22 𝒙 c
𝑥=16.9 ?
𝟕𝟎° 4 𝒙 a b
𝑥=14.1 ?
𝟕𝟎° 𝒙 𝟒 d
𝟖𝟎° 20 𝒙
𝑥=20.3 ? f e
𝟓𝟓° 10 𝒙
𝑥=7.00 ?
𝑥=11.7 ? g
𝟔𝟏° 3 𝒙
𝑥=6.19 ?
Q2-7 on next slide…

15. Exercise 1 ? ? ? ? ? ? (questions on provided sheet) 5 2 3 6 4
In the shape drawn below, 𝐴𝐵=𝐵𝐶=𝐶𝐷. Work out the area of 𝐴𝐵𝐶𝐷, giving your answer to 1 dp.
89.2 cm2 2
I put a ladder 1.5m away from a tree. The ladder is inclined at 70° above the horizontal. What is the height of the tree? 𝟒.𝟏𝟐𝒎
Ship B is 100m east of Ship A, and the bearing of Ship B from Ship A is 30°. How far due North is the ship? 𝟏𝟎𝟎÷ 𝐭𝐚𝐧 𝟑𝟎 =𝟏𝟕𝟑.𝟐𝒎 5 ? ? 3
[OCR GCSE(9-1) SAM 6H Q16aii]
Simon cuts the corners off a square piece of card to leave the regular octagon shown below.  O is the centre of the octagon. A and B are vertices of the octagon. OA = OB = 5 cm. Angle AOB = 45°.
Work out the area of the original square piece of card cm2 6 ?
[Edexcel GCSE June2012-2H Q18] 
The diagram shows a quadrilateral ABCD. AB = 16 cm. AD = 12 cm. Angle BCD = 40°. Angle ADB = angle CBD = 90°.
Calculate the length of CD.
16.5 cm 4 ? ? N
[IMC] The semicircle and isosceles triangle have equal areas. Find tan 𝑥 .
𝝅 𝟐 ?

16. Frost Childhood Story So what is 𝑎,𝑏 then? 𝑟 y 𝒂 (𝑎,𝑏) 𝒃 𝜃 x
sin 𝜃 = 𝑎 𝑟 → 𝑎=𝑟 sin 𝜃
cos 𝜃 = 𝑏 𝑟 → 𝑏=𝑟 cos 𝜃
Therefore I drew the hand between 0,0 and 𝑟 sin 𝜃 , 𝑟 cos 𝜃

17. 4 30 ° 𝒙 STARTER RECAP When would we use trigonometry?
When we have a right-angled triangle.
When we’re involving two side lengths and an angle. ?
30 ° 4 𝒙
𝑥= 4 tan 30 =4 3 𝑜𝑟 6.93 ?

18. 5 3 𝒂 But what if the angle is unknown? sin 𝑎 = 3 5
We can use the same process… ℎ
Step 1: Determine which sides are hyp/adj/opp. 𝑜
Step 2: Work out which trigonometric function we need.
sin 𝑎 = 3 5
Remember that the angle always goes after the sin/cos/tan.
(si𝑛 −1 )
(si𝑛 −1 )
𝑎= sin − =36.9°
Step 3: Rearrange the equation to find the unknown.
We want to get the angle 𝑎 on its own. But how do we get rid of the 𝑠𝑖𝑛 on front of it?
We know we get rid of something in an equation by doing the opposite.
The opposite of sin is ‘inverse sin’, written si𝑛 −1 . Use the shift button to get it on your calculator.

19. What is the missing angle?𝟓 𝒂 𝟒
cos −
cos −
sin −
sin −

20. What is the missing angle?𝒂 𝟏 𝟐
cos −
sin −1 2
tan −1 2
tan −

21. What is the missing angle?𝟓 𝟑 𝒂
cos −
sin −
tan −
sin −

22. What is the missing angle?𝟑 𝒂 𝟐
cos −
sin −
sin −
tan −

23. Test Your Understanding
[Edexcel GCSE Nov2007-4I Q25, Nov2007-6H Q14]
PQR is a right-angled triangle.
PR = 12 cm. QR = 4.5 cm. Angle 𝑃𝑅𝑄=90°.
Work out the value of 𝑥. Give your answer correct to one decimal place. ?
tan − =20.6 cm
[Edexcel IGCSE Jan2014(R)-3H Q17] The diagram shows triangle ABC. D is the point on AB, such that CD is perpendicular to AB. AC = 8.3 cm. AD = 4.7 cm. BD = 7.5 cm.
Calculate the size of angle ABC. Give your answer correct to 1 decimal place.
𝑪𝑫= 𝟖.𝟑 𝟐 − 𝟒.𝟕 𝟐 =𝟔.𝟖𝟒𝟏𝟎𝟓 ∠𝑨𝑩𝑪= 𝐭𝐚𝐧 −𝟏 𝟔.𝟖𝟒𝟏𝟎𝟓 𝟕.𝟓 =𝟒𝟐.𝟒° ?

24. Exercise 2 ? ? ? ? ? ? ? Find 𝜃, giving your answer to 3sf. 7 2 3 1 𝜃
(questions on provided sheet)
Find 𝜃, giving your answer to 3sf.
𝜃=30.0° ? 1 a
𝜃=31.0° ? c 7 2 b 3 1 𝜃 𝜃 𝜃 4 5
𝜃=55.2° ? 2
[Edexcel GCSE June2014-2H Q15b] The diagram shows the positions of three turbines A, B and C.
Calculate the bearing of C from A.
Give your answer correct to the nearest degree. 3 d 4 5
𝜃=37.3° ? 𝜃 7 10
𝜃=51.3° ? ?
217°
60° 𝜃 e 7 11 𝜃
𝜃=50.5° ?

25. Exercise 2
(questions on provided sheet)
[Edexcel IGCSE May2015-3H Q15]
𝐴𝐵𝐶𝐷 is a trapezium. 𝐴𝐵=25 cm. 𝐵𝐶=24 cm. 𝐶𝐷=10 cm. Angle 𝐴𝐵𝐶= angle 𝐵𝐶𝐷°
Calculate the size of angle 𝐶𝐷𝐴. Give your answer correct to 3 significant figures.
Solution: 𝟏𝟐𝟐° 5 4 13 𝜃 4 ?
[Edexcel GCSE(9-1) Nov F Q22, Nov H Q7] 𝐴𝐵𝐶𝐷 is a trapezium. Work out the size of angle 𝐶𝐷𝐴. Give your answer correct to 1dp. 6 3
𝜽= 𝐜𝐨𝐬 −𝟏 𝟓 𝟏𝟑 =𝟔𝟕.𝟒° ? 1 1 N 1 ?
Solution: 𝟑𝟐.𝟑°
The angles 𝜃 1 , 𝜃 2 , … form a sequence. Give the formula for the 𝑛th term of the sequence.
𝜽 𝒏 =𝐭𝐚𝐧 −𝟏 𝟏 𝒏
𝜃 3 1
𝜃 2 ?
𝜃 1 1

26. Related Slides In other slides on www.drfrostmaths.com we’ll cover:
Exact Trigonometric Ratios e.g. What is the exact value of sin 60° ? (without using a calculator)
3D Trigonometry & Pythagoras