Question

$\text{Observe the given triangles and find the}\text{value of}\angle P.$

Answer 1

$\text{In}\Delta ABC\text{and}\Delta RQP,$

$\frac{AB}{RQ}=\frac{3.8}{7.6}=\frac{1}{2}$

$\frac{BC}{QP}=\frac{6}{12}=\frac{1}{2}$ (1 mark)

$\text{and}\frac{CA}{PR}=\frac{3\sqrt{3}}{6\sqrt{3}}=\frac{1}{2}$

$\Rightarrow \frac{AB}{RQ}=\frac{BC}{QP}=\frac{CA}{PR}$

$\text{So,}\Delta ABC\sim \Delta RQP\text{(By SSS similarity criterion)}$ (1 mark)

$\therefore \angle C=\angle P\text{(Corresponding angles of similar}\text{triangles are equal)}$

$\text{But,}\angle C={180}^{\circ }-\angle A-\angle B\text{(By angle sum property)}$ (1 mark)

$\angle C={180}^{\circ }-{80}^{\circ }-{60}^{\circ }={40}^{\circ }$

$\text{So,}\angle P={40}^{\circ }$ (1 mark)