Question

current attempt in progress
please view the following video before answering this question. lecture video: section 4.4, example 02

note: the figure is not drawn to scale.
solve the triangle.
round the answers to the nearest integer.
γ=

b=
in
a=
in

Explanation:

Step1: Find angle $\gamma$

The sum of angles in a triangle is $180^{\circ}$. So, $\gamma=180^{\circ}-(75^{\circ} + 50^{\circ})$.
$$\gamma=180^{\circ}-125^{\circ}=55^{\circ}$$

Step2: Use the Law of Sines to find $b$

The Law of Sines states that $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$. Here, if we consider the side of length $15$ in (let's call it $c$) opposite angle $\gamma = 55^{\circ}$, and the side $b$ opposite the $50^{\circ}$ angle. Then $\frac{b}{\sin50^{\circ}}=\frac{15}{\sin55^{\circ}}$.
So, $b = 15\times\frac{\sin50^{\circ}}{\sin55^{\circ}}$.
Using a calculator, $\sin50^{\circ}\approx0.7660$, $\sin55^{\circ}\approx0.8192$.
$$b = 15\times\frac{0.7660}{0.8192}\approx14$$

Step3: Use the Law of Sines to find $a$

Using the Law of Sines again, $\frac{a}{\sin75^{\circ}}=\frac{15}{\sin55^{\circ}}$.
Since $\sin75^{\circ}\approx0.9659$, $\sin55^{\circ}\approx0.8192$.
Then $a = 15\times\frac{\sin75^{\circ}}{\sin55^{\circ}}=15\times\frac{0.9659}{0.8192}\approx18$

Answer:

$\gamma = 55^{\circ}$, $b\approx14$ in, $a\approx18$ in