Question

1. in △pqr, m∠p = 53°, pq = 16, and qr = 10. find m∠q.

Explanation:

Step1: Apply the Law of Cosines

$PR^{2}=PQ^{2}+QR^{2}-2\cdot PQ\cdot QR\cdot\cos P$
Substitute $PQ = 16$, $QR = 10$, and $P=53^{\circ}$ (so $\cos P=\cos53^{\circ}\approx0.6018$).
$PR^{2}=16^{2}+10^{2}-2\times16\times10\times0.6018$
$PR^{2}=256 + 100-192.576$
$PR^{2}=163.424$
$PR=\sqrt{163.424}\approx 12.78$

Step2: Apply the Law of Sines

$\frac{\sin Q}{PR}=\frac{\sin P}{QR}$
$\sin Q=\frac{PR\cdot\sin P}{QR}$
Substitute $PR\approx12.78$, $\sin P=\sin53^{\circ}\approx0.7986$, and $QR = 10$
$\sin Q=\frac{12.78\times0.7986}{10}$
$\sin Q=\frac{10.205}{10}=1.0205$ (This is wrong. Let's use the Law of Cosines directly for $\angle Q$)
Using the Law of Cosines: $PR^{2}=PQ^{2}+QR^{2}-2\cdot PQ\cdot QR\cdot\cos Q$
$\cos Q=\frac{PQ^{2}+QR^{2}-PR^{2}}{2\cdot PQ\cdot QR}$
First, find $PR^{2}=16^{2}+10^{2}-2\times16\times10\times\cos53^{\circ}=256 + 100-192.576 = 163.424$
$\cos Q=\frac{16^{2}+10^{2}-163.424}{2\times16\times10}=\frac{256 + 100-163.424}{320}=\frac{192.576}{320}=0.6018$
$Q=\cos^{-1}(0.6018)\approx 53^{\circ}$

Answer:

$m\angle Q\approx53^{\circ}$