Questions: Find the smallest angle of triangle PQR.

Solution Steps

We are given two sides and the included angle of triangle PQR. We can use the Law of Cosines to find the third side QR. The Law of Cosines states that for any triangle with sides a, b, and c, and angle C opposite side c, $c^2 = a^2 + b^2 - 2ab\cos(C)$.

In our case, let $a = RP = 24$ m, $b = PQ = 25$ m, and $C = \angle P = 115^\circ$. We want to find $c = QR$. $c^2 = 24^2 + 25^2 - 2(24)(25)\cos(115^\circ)$ $c^2 = 576 + 625 - 1200\cos(115^\circ)$ $c^2 = 1201 - 1200(-0.4226)$ $c^2 = 1201 + 507.12$ $c^2 = 1708.12$ $c = \sqrt{1708.12} \approx 41.33$ m

The smallest angle is opposite the smallest side. The smallest side is PR = 24 m, so we want to find $\angle Q$. Using the Law of Sines:

$\frac{\sin Q}{PR} = \frac{\sin P}{QR}$ $\frac{\sin Q}{24} = \frac{\sin 115^\circ}{41.33}$ $\sin Q = \frac{24\sin 115^\circ}{41.33}$ $\sin Q = \frac{24(0.9063)}{41.33}$ $\sin Q = \frac{21.7512}{41.33} \approx 0.5263$ $Q = \arcsin(0.5263) \approx 31.75^\circ$

The sum of angles in a triangle is $180^\circ$. We have $\angle P = 115^\circ$ and $\angle Q \approx 31.75^\circ$. So, $\angle R = 180^\circ - 115^\circ - 31.75^\circ$ $\angle R \approx 33.25^\circ$

Final Answer