Question

Are these two given triangles similar?


Answer as yes or no.

Answer 1

The given two triangles are,


$\text{Angle sum property of a triangle:}$
Sum of all three interior angles of a triangle is always ${180}^o$.

In $△ABC$,
$m\angle A+m\angle B+m\angle C={180}^o$
$\Rightarrow m\angle A+{50}^o+{60}^o={180}^o$
$\Rightarrow m\angle A+{110}^o={180}^o$
Subtracting ${110}^o$ from both sides,
$\Rightarrow m\angle A+{110}^o-{110}^o={180}^o-{110}^o$
$\Rightarrow m\angle A={70}^o$
Hence, the measure of the angles of $△ABC$ are:
$\bullet m\angle A={70}^o$
$\bullet m\angle B={50}^o$
$\bullet m\angle C={60}^o$

In $△PQR$,
$m\angle P+m\angle Q+m\angle R={180}^o$
$\Rightarrow {70}^o+{50}^o+m\angle R={180}^o$
$\Rightarrow {120}^o+m\angle R={180}^o$
Subtracting ${120}^o$ from both sides,
$\Rightarrow {120}^o+m\angle R-{120}^o={180}^o-{120}^o$
$\Rightarrow m\angle R={60}^o$
Hence, the measure of the angles of $△PQR$ are,
$\bullet m\angle P={70}^o$
$\bullet m\angle Q={50}^o$
$\bullet m\angle R={60}^o$

Clearly, the corresponding angles of $△ABC$ and $△PQR$ are equal in measure.

$\therefore △ABC\sim △PQR$