1. Starter Calculate the area of this triangle. Hint: Area = ½ x b x h
height = 4 x tan 64
height = 8.2 cm
height
area = ½ x 4 x 8.2
64°
area = 16.4 cm²
4cm

2. Calculate the area of this triangle:
7 cm
37°
11 cm
Why is it difficult to find the area of this triangle?
Because we don’t know the height.

3. The vertices of a triangle are labelled with capital letters
The vertices of a triangle are labelled with capital letters. The triangle shown is triangle ABC. c a b C A B
Area of a triangle = ½ base x height
Derive a formula to find the area of this triangle.

4. B c a h C A b Area of a triangle = ½ base x height
sin = opp  sin C = h  h = a sin C
hyp a
Area of a triangle = ½ b x a sin C
= ½ a b sin C
½ a b sin C = ½ b c sin A = ½ a c sin B

5. ½ a b sin C = ½ b c sin A = ½ a c sin B

6. Calculate the area of this triangle:
7 cm
37°
11 cm
Area = ½ x 7 x 11 x sin 37
= cm²

7. Example 1: Find the area of this triangle. Give your answer to 3sf
5.8 cm
43°
12.3 cm
Area = ½ x 12.3 x 5.8 x sin 43
= 24.3 cm²

8. Sometimes questions give us the area but ask us to find either angles or sides.
Example 2: In triangle ABC, a = 3.6 cm, b = 4.9 cm and the area is 5 cm². Find the size of angle BCA.
Hint: if it’s tricky, draw a piccy!
3.6 cm
4.9 cm
5 = ½ x 3.6 x 4.9 x sin C
0.5668… = sin C
C = 34.5°

9. Example 3: In triangle ABC, b = 13 cm, Angle CAB = 66° and the area is 100 cm². Find the length of side ‘c’. A B C
13 cm
66°
Area = 100cm²
100 = ½ x 13 x c x sin 66
c = cm

10. Extension
Use the measurements given to find the area of this quadrilateral shaped field. Give your answer to 3sf.
50m
45m
62º
70º
60m
25m

11. Answers
10.69cm² m²
26.24mm² cm²
3.90cm 56°

12. Starter
The 12-sided window is made up of squares, equilateral triangles and a regular hexagon.
The perimeter of the window is 15.6m.
Calculate the area of the window.

13. c a b C A B sin C = sin A = sin B c a b ...is the same as... c = a = b
We can use this formula to find missing sides or angles of non-right-angled triangles

14. Find the length of the side marked a and give your answer correct to 3 s.f.
68° a
38°
9.4cm
a = …
a = 6.24 cm (3 s.f.)

15. Find the size of the acute angle marked B and give your answer correct to 1 d.p.
72°
8.1cm B
sin b = …
9.8cm

16. Extension task
Quadrilateral ABCD is made up of two triangles ABD and BCD.
Use the sine rule to work out the area of triangle BCD.

17. Answers
11.31cm 5.90m
9.36mm m
73° 37°

18. Answers
11.31cm 5.90m
9.36mm m
73° 37°

19. The angle of elevation of the top of a building measured from point A is 25o. At point D which is 15m closer to the building, the angle of elevation is 35o Calculate the height of the building. T B A D
35o
25o
10o
36.5
145o
15 m
Angle TDA =
180 – 35 = 145o
Angle DTA =
180 – 170 = 10o

20. Starter Rearrange a² = b² + c² - 2bc cosA
to make cosA the subject of the formula

21. c a b C A B a² = b² + c² - 2bc cosA ...is the same as...
cosA = b² + c² - a²
2bc
We can use this formula to find missing sides or angles of non-right-angled triangles

22. Find the length of the side marked a and give your answer correct to 3 s.f.
a2 = – 2 x 8 x 9.6 x Cos 40o
a = ( – 2 x 8 x 9.6 x Cos 40o)
a = 6.20 cm (3 sf)
9.6 cm a
40o
8 cm

23. Find the size of the acute angle marked A and give your answer correct to 1 d.p.
cosA = 6² + 4.9² - 3.2²
2 x 6 x 4.9
6 cm
A = Cos-1 (6² + 4.9² - 3.2²)
(2 x 6 x 4.9)
3.2 cm
A = 32.2° (1 dp) A
4.9 cm

24. Extension task P
670 miles W
530 miles Q
520 miles
An AWACS aircraft takes off from RAF Waddington (W) on a navigation exercise. It flies 530 miles North to a point (P) as shown, It then turns left and flies to a point (Q), 670 miles away. Finally it flies back to base, a distance of 520 miles.
Find the bearing of Q from point P.
cosP = 670² + 530² - 520²
2 x 670 x 530
P = 49.7° (1 dp)
Bearing = = 229.7°

25. Answers
8.03cm 7.91mm
5.94m cm
49° 54°