Question

From the given figure, find $\angle P$. [2 MARKS]

Answer 1

Concept: 1 Mark
Application: 1 Mark

In $\Delta ABC$ and $\Delta PQR$,

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

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

and $\frac{CA}{PR}=\frac{1}{2}$

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

So, $\Delta ABC\sim \Delta RQP$ [SSS similarity]

$\therefore \angle C=\angle P$ [Corresponding angles of similar triangles]

But $\angle C={180}^{\circ }-\angle A-\angle B$ [Angle sum property]

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

So, $\angle P={40}^{\circ }$