# 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?

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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 $\triangle TSM$.

Updated on 10-Oct-2022 13:27:57