Question

in $delta uvw$, $w = 1.4$ cm, $mangle w = 63^{circ}$ and $mangle u = 29^{circ}$. find the length of $v$, to the nearest 10th of a centimeter.

Explanation:

Step1: Find angle V

The sum of angles in a triangle is 180°. So, $m\angle V=180^{\circ}-m\angle U - m\angle W$.
$m\angle V = 180^{\circ}-29^{\circ}-63^{\circ}=88^{\circ}$

Step2: Use the Law of Sines

The Law of Sines states that $\frac{v}{\sin V}=\frac{w}{\sin W}$. We want to find $v$, so $v=\frac{w\sin V}{\sin W}$.
Substitute $w = 1.4$ cm, $V = 88^{\circ}$, and $W=63^{\circ}$ into the formula.
$v=\frac{1.4\times\sin88^{\circ}}{\sin63^{\circ}}$
We know that $\sin88^{\circ}\approx0.9994$ and $\sin63^{\circ}\approx0.8910$.
$v=\frac{1.4\times0.9994}{0.8910}\approx1.6$

Answer:

$1.6$ cm