Question

7 use the law of sines to find ac. round to the nearest tenth. law of sines: \\(\frac{\sin a}{a} = \frac{\sin b}{b}\\) \\(\boldsymbol{\text{\textcircled{a}}}\\) 7.1 \\(\boldsymbol{\text{\textcircled{b}}}\\) 31.9 \\(\boldsymbol{\text{\textcircled{c}}}\\) 8.2 \\(\boldsymbol{\text{\textcircled{d}}}\\) 7.0

Explanation:

Step1: Identify angles and sides

In $\triangle ABC$, $\angle A=62^\circ$, $\angle B=35^\circ$, $BC=11$ ft. $AC$ is opposite $\angle B$, $BC$ opposite $\angle A$.

Step2: Apply Law of Sines

$\frac{AC}{\sin B} = \frac{BC}{\sin A}$
$AC = \frac{BC \cdot \sin B}{\sin A}$

Step3: Calculate values

$\sin 35^\circ \approx 0.5736$, $\sin 62^\circ \approx 0.8829$
$AC = \frac{11 \cdot 0.5736}{0.8829} \approx \frac{6.3096}{0.8829} \approx 7.1$

Answer:

A. 7.1