1. Right Angled Trigonometry
Tuesday, 27 November 2018

2. Pythagoras Theorem Hypotenuse NBc a b NB
To find the Hypotenuse we must square the other sides and then add together.
To find any other side we square the sides and subtract.
The side is then found by taking the square root.

3. 𝑃𝑄 2 = 21 2 + 15 2 𝑃𝑄 2 =666 𝑃𝑄 = 666 𝑃𝑄 =26m Example
Find the length of PQ correct to 2 significant figures
21 m
15 m P R Q
𝑃𝑄 2 =
𝑃𝑄 2 =666
𝑃𝑄 = 666
𝑃𝑄 =26m

4. Example
Find the value of y correct to 1 decimal place.
𝑦 2 = −
14.3 cm
3.7 cm y
𝑦 2 =190.8 𝑦=
𝑦=13.8cm

5. Example
A ladder 17m long reaches 15m up a vertical wall. If it rests on horizontal ground, how far is its foot from the foot of the wall?
𝑥 2 = 17 2 − 15 2
𝑥 2 =64
17m
𝑥= 64
15m
𝑥=8m 𝑥

6. Formulae to learn 𝑂𝑝𝑝 𝐴𝑑𝑗 tan 𝐴 = 𝑂𝑝𝑝 𝐻𝑦𝑝 sin 𝐴 = 𝐴𝑑𝑗 𝐻𝑦𝑝 cos 𝐴 =
Hyp
Opp
𝐴𝑑𝑗 𝐻𝑦𝑝
cos 𝐴 = A
Adj b
𝑎 2 + 𝑏 2 = 𝑐 2

7. 𝑎 𝑐 tan 𝐴 = 𝑎 𝑏 sin 𝐴 = 𝑐 𝑏 cos 𝐴 = 𝑎 𝑏 sin 𝐴 cos 𝐴 = × 𝑏 𝑐 = 𝑎 𝑏b a
𝑐 𝑏
cos 𝐴 = B A c
𝑎 𝑏
sin 𝐴 cos 𝐴 =
× 𝑏 𝑐
= 𝑎 𝑏
= 𝑎 𝑐
= tan 𝐴
𝑐 𝑏
tan 𝐴 = sin 𝐴 cos 𝐴

8. Trig Ratios for 45°
tan 45 = 1
1 2 1
sin 45 = 45
1 2
cos 45 = 1

9. tan 60 = 3 3 2 sin 60 = 1 2 3 cos 60 = tan 30 = 1 3 1 2 sin 30 = 3 2
Trig ratios for 30° and 60°
tan 60 = 3 30
3 2
sin 60 = 2
1 2 3
cos 60 = 60
tan 30 =
1 3 1
1 2
sin 30 =
3 2
cos 30 =

10. 𝐴𝑑𝑗 𝐻𝑦𝑝 cos 𝐴 = 𝑃𝑄 52 cos 32 = 52×cos 32 = 𝑃𝑄 𝑃𝑄= 44.10cm Example
Calculate the length of PQ in the triangle PQR
Opp
𝐴𝑑𝑗 𝐻𝑦𝑝
cos 𝐴 =
𝑃𝑄 52
cos 32 =
Hyp
Adj
52×cos 32 = 𝑃𝑄
𝑃𝑄=
44.10cm

11. 𝐴𝑑𝑗 𝐻𝑦𝑝 cos 𝐴 = 4.6 𝑋𝑌 cos 59 = XY×cos 59 = 4.6 4.6 cos 59 𝑋𝑌= 𝑋𝑌=
Example
Calculate the length XY in the triangle XYZ
𝐴𝑑𝑗 𝐻𝑦𝑝
Hyp
cos 𝐴 =
4.6 𝑋𝑌
cos 59 =
Opp
Adj
XY×cos 59 =
4.6
4.6 cos 59
𝑋𝑌=
𝑋𝑌=
8.93cm

12. 𝑂𝑝𝑝 𝐻𝑦𝑝 sin 𝐴 = 𝑥 5 sin 12 = 5×sin 12 = 𝑥 𝑥= 1.03…cm ∴𝐵𝐶= 2.08cm
Example
Calculate the length of BC in the triangle ABC.
𝑂𝑝𝑝 𝐻𝑦𝑝
Hyp
sin 𝐴 =
Opp 𝑥
𝑥 5 12
sin 12 =
Adj
5×sin 12 = 𝑥 𝑥=
1.03…cm
∴𝐵𝐶=
2.08cm

13. 𝑂𝑝𝑝 𝐴𝑑𝑗 Tan 𝐴 = 𝑃𝑅 1000 Tan 20 = 1000×Tan 20 = 𝑃𝑅 𝑃𝑅= 363.97m Example
A surveyor, P, is 1000m away on horizontal ground from the foot of a radio mast, QR. From P the angle of elevation of the top, R, of the mast is 20. Find the height of the mast. Assume the surveyor’s height to be negligible. S P R
1000 20
Hyp
opp
Adj
With a written problem the first step is to draw a sketch.
𝑂𝑝𝑝 𝐴𝑑𝑗
Tan 𝐴 =
𝑃𝑅 1000
Tan 20 =
1000×Tan 20 = 𝑃𝑅
𝑃𝑅=
363.97m

14. 𝑂𝑝𝑝 𝐴𝑑𝑗 tan 𝐴 = 100 𝑑 tan 32 = d×tan 32 = 100 100 tan 32 𝑑= 𝑑= 160.03m
Example
From the top of a vertical cliff 100m high the angle of depression of a boat out at sea is 32. How far is the boat from the foot of the cliff? 32
100 d
Hyp
Opp
𝑂𝑝𝑝 𝐴𝑑𝑗
tan 𝐴 =
Adj
100 𝑑
tan 32 =
d×tan 32 =
100
100 tan 32 𝑑= 𝑑=
160.03m

15. 𝑂𝑝𝑝 𝐻𝑦𝑝 sin 𝐴 = 4 6 sin 𝐴 = 𝐴= sin −1 0.66 𝐴= 41.8° Example
Calculate the size of angle A in the triangle ABC
Opp
𝑂𝑝𝑝 𝐻𝑦𝑝
sin 𝐴 =
Adj
4 6
sin 𝐴 =
Hyp 𝐴=
sin − 𝐴=
41.8°

16. 𝑂𝑝𝑝 𝐴𝑑𝑗 tan 𝐴 = 1.2 4.7 Tan 𝑋 = 𝑋= tan −1 0.255 𝑋= 14.32° 𝑂𝑝𝑝 𝐴𝑑𝑗
Example
Find the size of angle x in each of the following
(i)
(ii)
𝑂𝑝𝑝 𝐴𝑑𝑗
tan 𝐴 =
Tan 𝑋 = 𝑋=
tan − 𝑋=
14.32°
𝑂𝑝𝑝 𝐴𝑑𝑗
tan 𝐴 =
10 3.7
tan 𝑋 = 𝑋=
tan − 𝑋=
69.70°

17. (iii)
𝐴𝑑𝑗 𝐻𝑦𝑝
cos 𝐴 =
7 12
Cos 𝑋 = 𝑋=
cos − 𝑋=
54.31°

18. → 𝐴𝑑𝑗 𝐻𝑦𝑝 cos 𝐴 = 3 8 Cos 𝑄 = 𝑄= cos −1 0.375 𝑄= 68.0° Example
Calculate the size of angle PQR in the triangle below. P
𝐴𝑑𝑗 𝐻𝑦𝑝
cos 𝐴 = 8 →
3 8
Cos 𝑄 = 𝑄=
cos − Q 3 𝑄=
68.0°

19. Example
The figure is a pyramid on a square base ABCD. The edges of the base are 30cm long and the height, EH, of the pyramid is 42cm. Find
a) the length of AC
b) the angle EAH

20. 𝑂𝑝𝑝 𝐴𝑑𝑗 tan 𝐴 = 42 21.21 tan 𝐴 = 𝐴= tan −1 0.1.97 𝐴= 63.2°
a) the length of AC
Using pythagoras theorem with the base
42cm A D
𝐴𝐶=
30cm
30cm
𝐴𝐶=42.42
30cm B C
30cm
b) the angle EAH
𝑂𝑝𝑝 𝐴𝑑𝑗 E
tan 𝐴 =
Draw the right angled triangle for EAH
42cm
tan 𝐴 = 𝐴=
tan − A 𝐴=
63.2° H
21.21cm

24. 5. In triangle DEF, Angle F = 90, Angle D = 21 and DF = 3. 2cm
5. In triangle DEF, Angle F = 90, Angle D = 21 and DF = 3.2cm. Find the length of EF

25. 6. For the diagram drawn below find the lengths of x and y
15cm
32 x
20 y

27. 8. Find the height of these stairs correct to one decimal place
35
3.6m h