Question

1. draw a labelled diagram for each triangle. then calculate the required side length or angle measure.

a) in $\triangle sun$, $n = 58$ cm, $\angle n = 38^{\circ}$, and $\angle u = 72^{\circ}$. determine the length of side $u$.
b) in $\triangle pqr$, $\angle r = 73^{\circ}$, $\angle q = 32^{\circ}$, and $r = 23$ cm. determine the length of side $q$.
c) in $\triangle tam$, $t = 8$ cm, $m = 6$ cm, and $\angle t = 65^{\circ}$. determine the measure of $\angle m$.

Explanation:

Step1: Find the measure of angle \(S\)

In \(\triangle SUN\), the sum of angles in a triangle is \(180^{\circ}\). So, \(\angle S=180^{\circ}-\angle N - \angle U\).
\(\angle S = 180^{\circ}-38^{\circ}-72^{\circ}=70^{\circ}\)

Step2: Use the Law of Sines

The Law of Sines states that \(\frac{u}{\sin U}=\frac{n}{\sin N}\).
We know \(n = 58\mathrm{cm}\), \(\angle N=38^{\circ}\), \(\angle U = 72^{\circ}\).
From \(\frac{u}{\sin U}=\frac{n}{\sin N}\), we can solve for \(u\): \(u=\frac{n\sin U}{\sin N}\)
Substitute the values: \(u=\frac{58\times\sin72^{\circ}}{\sin38^{\circ}}\)
Since \(\sin72^{\circ}\approx0.9511\) and \(\sin38^{\circ}\approx0.6157\)
\(u=\frac{58\times0.9511}{0.6157}\)
\(u=\frac{55.1638}{0.6157}\approx89.6\mathrm{cm}\)

Answer:

The length of side \(u\) is approximately \(89.6\mathrm{cm}\)