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1 answer:
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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
∵
∴ ≅ 736.2 to the nearest tenth foot