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Home » In ΔPQR and ΔMST, ∠P = 55°, ∠Q =25°, ∠M = 100° and ∠S = 25°. Is ΔQPR ~ ΔTSM? Why?
In ΔPQR and ΔMST, ∠P = 55°, ∠Q =25°, ∠M = 100° and ∠S = 25°. Is ΔQPR ~ ΔTSM? Why?
CBSE | Class 10 | Exercise 6.2 | Maths | NCERT Exemplar | Triangles
In ΔPQR and ΔMST, ∠P = 55°, ∠Q =25°, ∠M = 100° and ∠S = 25°. Is ΔQPR ~ ΔTSM? Why?

Solution:

We all know that,

When the three angles of a triangle are added then their sum equals to 180°.

Then, from triangle PQR,

\angle P\text{ }+~\angle Q\text{ }+~\angle R\text{ }=\text{ }180{}^\circ

55{}^\circ \text{ }+\text{ }25{}^\circ \text{ }+~\angle R\text{ }=\text{ }180{}^\circ

As a result, we get,

\angle R\text{ }=\text{ }180{}^\circ \text{ }\text{ }\left( 55{}^\circ \text{ }+\text{ }25{}^\circ \right)\text{ }=\text{ }180{}^\circ -\text{ }80{}^\circ \text{ }=\text{ }100{}^\circ

Similarly, from the triangle TSM,

\angle T\text{ }+~\angle S\text{ }+~\angle M\text{ }=\text{ }180{}^\circ

\angle T\text{ }+~\angle 25{}^\circ +\text{ }100{}^\circ =\text{ }180{}^\circ

As a result, we get,

\angle T\text{ }=\text{ }180{}^\circ -\text{ }(\angle 25{}^\circ +\text{ }100{}^\circ )

\angle T\text{ }=\text{ }180{}^\circ 125{}^\circ =\text{ }55{}^\circ

In triangles PQR and TSM,

We have,

\angle P\text{ }=~\angle T,

\angle Q\text{ }=~\angle S

\angle R\text{ }=~\angle M

As a result, ∆PQR ∼ ∆TSM

As, all corresponding angles are equal,

Therefore, ∆QPR is similar to ∆TSM.

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