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

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.

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.

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

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

Calculate the area of this triangle: 7 cm 37° 11 cm Area = ½ x 7 x 11 x sin 37 = 23.17 cm²

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

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°

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 = 16.84 cm

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

Answers 10.69cm² 44.49m² 26.24mm² 202.11cm² 3.90cm 56°

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.

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

Find the length of the side marked a and give your answer correct to 3 s.f.   68° a   38° 9.4cm a = 6.2417… a = 6.24 cm (3 s.f.)

Find the size of the acute angle marked B and give your answer correct to 1 d.p.   72° 8.1cm   B sin b = 0.7860… 9.8cm    

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.

Answers 11.31cm 5.90m 9.36mm 13.89m 73° 37°

Answers 11.31cm 5.90m 9.36mm 13.89m 73° 37°

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

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

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

Find the length of the side marked a and give your answer correct to 3 s.f. a2 = 82 + 9.62 – 2 x 8 x 9.6 x Cos 40o a = (82 + 9.62 – 2 x 8 x 9.6 x Cos 40o) a = 6.20 cm (3 sf) 9.6 cm a 40o 8 cm

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

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 = 180 + 49.7 = 229.7°

Answers 8.03cm 7.91mm 5.94m 14.13cm 49° 54°