Question

In triangle PQR, PR = 23mm, QR = 39mm, and m<R = 163 degrees. Find the area of the triangle to the nearest tenth.

Answer 1

Answer: 131.1287 square mm (approx)

Step-by-step explanation:

The area of a triangle,

$A=\frac{1}{2} \times s_1\times s_2\times sin \theta$

Where $s_1$ and $s_2$ are adjacent sides and $\theta$ is the include angle of these sides,

Here PR and QR are adjacent sides and ∠R is the included angle of these sides,

Thus, we can write,

$s_1 = PR= 23\text{ mm}$, $s_2=QR=39\text{ mm}$ and $\theta = 163^{\circ}$,

Thus, the area of the triangle PQR,

$A=\frac{1}{2} \times 23\times 39\times sin163^{\circ}$

$A=\frac{262.257419136}{2} = 131.128709568\approx 131.1287\text{ square mm}$