Question

Please help me with this trigonometry question:

Answer 1

The distance from A to B is 736.2 to the nearest tenth foot

Explanation:

In ΔCAB

∵ m∠CAD = 30° ⇒ exterior angle of Δ at vertex A

∴ m∠CAD = m∠ACB + m∠ABC

∵ m∠ABC = 20°

∴ m∠ACB = 30° - 20° = 10°

We will use the sin rule to find the distance AB

∵
`(sin20)/(1450)=(sin10)/(AB)`

∴
`AB=(1450(sin10))/(sin20)=736.1842936` ≅ 736.2 to the nearest tenth foot