Question

in $delta vwx$, $x = 1.5$ cm, $mangle x = 26^{circ}$ and $mangle v = 71^{circ}$. find the length of $v$, to the nearest 10th of a centimeter.

Explanation:

Step1: Find angle W

The sum of angles in a triangle is 180°. So, $m\angle W=180^{\circ}-(m\angle V + m\angle X)$.
$m\angle W=180^{\circ}-(71^{\circ}+ 26^{\circ})=83^{\circ}$.

Step2: Use the Law of Sines

The Law of Sines states that $\frac{v}{\sin V}=\frac{x}{\sin X}$.
We know $x = 1.5$ cm, $m\angle X = 26^{\circ}$, and $m\angle V=71^{\circ}$.
So, $v=\frac{x\sin V}{\sin X}$.
Substitute the values: $v=\frac{1.5\times\sin(71^{\circ})}{\sin(26^{\circ})}$.
Since $\sin(71^{\circ})\approx0.9455$ and $\sin(26^{\circ})\approx0.4384$.
$v=\frac{1.5\times0.9455}{0.4384}=\frac{1.41825}{0.4384}\approx3.2$.

Answer:

$3.2$ cm