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

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
[tex]\frac{AC}{sin30}[/tex] = [tex]\frac{15}{sin40}[/tex] ( cross- multiply )
AC × sin40° = 15 × sin30° ( divide both sides by sin40° )
AC = [tex]\frac{15sin30}{sin40}[/tex] ≈ 11.668
Using the cosine ratio in right triangle ABC
cos70° = [tex]\frac{adjacent}{hypotenuse}[/tex] = [tex]\frac{BC}{AC}[/tex] = [tex]\frac{BC}{11.668}[/tex] ( multiply both sides by 11.668 )
11.668 × cos70° = BC , then
BC ≈ 4.0 ( to the nearest tenth )