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
∵ [tex]\frac{sin20}{1450}=\frac{sin10}{AB}[/tex]
∴ [tex]AB=\frac{1450(sin10)}{sin20}=736.1842936[/tex] ≅ 736.2 to the nearest tenth foot
