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law of sines and cosines review continued find the area of the given tr…

Question

law of sines and cosines review continued
find the area of the given triangle.

  1. a = 45°, b = 30 ft, c = 16 ft

Explanation:

Step1: Recall the formula for the area of a triangle

The formula for the area of a triangle when two sides \(b\), \(c\) and the included angle \(A\) are known is \(S=\frac{1}{2}bc\sin A\)

Step2: Substitute the given values into the formula

Given \(b = 30\) ft, \(c=16\) ft, and \(A = 45^{\circ}\), \(\sin A=\sin45^{\circ}=\frac{\sqrt{2}}{2}\)

Substitute into the formula: \(S=\frac{1}{2}\times30\times16\times\sin45^{\circ}\)

First, calculate \(\frac{1}{2}\times30\times16 = 240\)

Then, \(S = 240\times\frac{\sqrt{2}}{2}\)

Step3: Simplify the expression

\(S=120\sqrt{2}\approx120\times1.414 = 169.68\) (square feet)

Answer:

The area of the triangle is approximately \(169.68\) square feet.