Question

Please help me with this trigonometry question:

Answer 1

Answer:

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

Step-by-step 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

∵ $\frac{sin20}{1450}=\frac{sin10}{AB}$

∴ $AB=\frac{1450(sin10)}{sin20}=736.1842936$ ≅ 736.2 to the nearest tenth foot