Question

Construct $∆PQR$ if $PQ=5\mathrm{cm},\mathrm{m}\angle PQR={105}^{°}\mathrm{and}\mathrm{m}\angle QRP={40}^{°}$. (Hint: Recall the angle-sum property of a triangle).

Answer 1

Steps 1. Draw a line segment $PQ=5\mathrm{cm}$

Step 2. At point $Q$ drawn an angle $\angle PQR=105°$

Step 3. Calculate angle $\angle RPQ$ by using angle sum property.

Apply the sum of the angles of a triangle is $180°$ to find measure of $\angle RPQ$:

$\begin{array}{rcl}\angle PQR+\angle QRP+\angle RPQ& =& {180}^{°}\\ & \Rightarrow & 105+40+\angle RPQ=180\\ & \Rightarrow & \angle RPQ=180-145\\ & \Rightarrow & \angle RPQ={35}^{°}\end{array}$

Thus, the measure of $\angle RPQ$ is ${35}^{°}$.

Step 4. At point $P$, drawn an angle of $35°$ and join to point $R$.

Hence, the required triangle is,