Question

4 fill in the blank 1 point a path in a park forms a triangle as shown. what is the measure of angle a? round to the nearest tenth of a degree. m∠a≈ type your answer...

Explanation:

Step1: Apply the Law of Sines

The Law of Sines states that \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\). In \(\triangle ABC\), let \(a = 135\), \(b = 72\), and \(C=119^{\circ}\). Then \(\frac{135}{\sin A}=\frac{72}{\sin B}\), and also \(A + B+119^{\circ}=180^{\circ}\), so \(B = 61^{\circ}-A\).

By the Law of Sines: \(\frac{135}{\sin A}=\frac{72}{\sin(61^{\circ}-A)}\)

Using the formula \(\sin(A - B)=\sin A\cos B-\cos A\sin B\), \(\sin(61^{\circ}-A)=\sin61^{\circ}\cos A-\cos61^{\circ}\sin A\approx0.875\cos A - 0.485\sin A\)

The equation \(\frac{135}{\sin A}=\frac{72}{0.875\cos A - 0.485\sin A}\)

Cross - multiply: \(135\times(0.875\cos A - 0.485\sin A)=72\sin A\)

\(118.125\cos A-65.475\sin A = 72\sin A\)

\(118.125\cos A=(72 + 65.475)\sin A\)

\(118.125\cos A=137.475\sin A\)

\(\tan A=\frac{118.125}{137.475}\approx0.859\)

Step2: Find the angle

\(A=\arctan(0.859)\approx40.7^{\circ}\)

Another way:
By the Law of Sines \(\frac{BC}{\sin A}=\frac{AB}{\sin C}\)

First, find \(AB\) using the Law of Cosines \(AB^{2}=72^{2}+135^{2}-2\times72\times135\times\cos119^{\circ}\)

\(AB^{2}=5184 + 18225-19440\times(- 0.485)\)

\(AB^{2}=23409+9438.4\)

\(AB^{2}=32847.4\)

\(AB\approx181.2\)

Then by the Law of Sines \(\frac{135}{\sin A}=\frac{181.2}{\sin119^{\circ}}\)

\(\sin A=\frac{135\times\sin119^{\circ}}{181.2}\)

\(\sin A=\frac{135\times0.875}{181.2}\)

\(\sin A=\frac{118.125}{181.2}\approx0.652\)

\(A=\arcsin(0.652)\approx40.7^{\circ}\)

Answer:

\(40.7\)