Question

let abc be a triangle such that angle a is opposite side a, angle b is opposite side b, and angle c is opposite side c. if ( b = 120.87^{circ} ), ( a = 15 ), and ( b = 20 ), solve for angle a. round your answer to the nearest hundredth of a degree.( mangle a=) 40.17

Explanation:

Step1: Apply the Law of Sines

The Law of Sines states that \(\frac{a}{\sin A}=\frac{b}{\sin B}\).
Substitute the given values: \(\frac{15}{\sin A}=\frac{20}{\sin(120.87^{\circ})}\).

Step2: Solve for \(\sin A\)

Cross - multiply: \(20\sin A = 15\sin(120.87^{\circ})\).
Then \(\sin A=\frac{15\sin(120.87^{\circ})}{20}\).
Calculate \(\sin(120.87^{\circ})\approx0.859\).
So \(\sin A=\frac{15\times0.859}{20}=\frac{12.885}{20}=0.64425\).

Step3: Find angle \(A\)

Take the inverse sine: \(A=\sin^{-1}(0.64425)\).
Using a calculator, \(A\approx40.17^{\circ}\).

Answer:

\(40.17\)