In triangles \( \mathrm{PQR} \) and \( \mathrm{MST}, \angle \mathrm{P}=55^{\circ}, \angle \mathrm{Q}=25^{\circ}, \angle \mathrm{M}=100^{\circ} \) and \( \angle \mathrm{S}=25^{\circ} \). Is \( \triangle \mathrm{QPR} \sim \triangle \mathrm{TSM} \) ? Why?
Given:
In triangles \( \mathrm{PQR} \) and \( \mathrm{MST}, \angle \mathrm{P}=55^{\circ}, \angle \mathrm{Q}=25^{\circ}, \angle \mathrm{M}=100^{\circ} \) and \( \angle \mathrm{S}=25^{\circ} \).
To do:
We have to find whether \( \triangle \mathrm{QPR} \sim \triangle \mathrm{TSM} \).
Solution:
We know that,
The sum of the angles of a triangle is $180^o$.
In $\triangle QPR$
$\angle P + \angle Q + \angle R = 180^o$
$55^o + 25^o + \angle R = 180^o$
$\angle R = 180^o - (55^o + 25^o)$
$= 180^o - 80^o$
$=100^o$
In $\triangle TSM$,
$\angle T + \angle S + \angle M = 180^o$
$\angle T + \angle 25^o+ 100^o = 180^o$
$\angle T = 180^o-125^o$
$= 55^o$
In $\triangle PQR$ and $\triangle TSM$,
$\angle P = \angle T, \angle Q = \angle S$ and $\angle R = \angle M$
Therefore,
$\triangle PQR \sim \triangle TSM$
Here,
The correct correspondence is $P ↔ T, Q ↔ S$ and $R ↔M$
Hence, $\triangle QPR$ is not similar to $\triangle TSM$.
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