Question

In $\Delta PQR,r=9$ inches, $p=6.6$ inches, and $\angle Q=6°$. Find the area of $\Delta PQR$, to the nearest one decimal of a square inch.

Answer 1

Find the area of the triangle.

Given that: In $\Delta PQR,r=9$ inches, $p=6.6$ inches, and $\angle Q=6°$.

$A=\frac{1}{2}rp\sin \left(\angle Q\right)$

Substitute $r=9,p=6.6and\angle Q=6^{\circ }$:

$$\begin{gathered} \Rightarrow A=\frac{1}{2}·9·6.6·\sin \left(6^{\circ }\right) \\ \Rightarrow A=3.1i{n}^2 \end{gathered}$$

Hence, the area of the triangle is $3.1i{n}^2.$