Question

5) In the figure, triangle ABC is a right triangle at B. If CD = 15, find BC to the nearest tenth.

Answer 1

Answer:

BC ≈ 4.0

Step-by-step explanation:

∠ DCA = 180° - 70° = 110° ( adjacent angles )

∠ DAC = 180° - (30 + 110)° ← sum of angles in triangle

∠ DAC = 180° - 140° = 40°

Using the Sine rule in Δ ACD to find common side AC

$\frac{AC}{sin30}$ = $\frac{15}{sin40}$ ( cross- multiply )

AC × sin40° = 15 × sin30° ( divide both sides by sin40° )

AC = $\frac{15sin30}{sin40}$ ≈ 11.668

Using the cosine ratio in right triangle ABC

cos70° = $\frac{adjacent}{hypotenuse}$ = $\frac{BC}{AC}$ = $\frac{BC}{11.668}$ ( multiply both sides by 11.668 )

11.668 × cos70° = BC , then

BC ≈ 4.0 ( to the nearest tenth )