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In triangles PQR and MST, \angleP = 55°, \angleQ = 25°, \angleM = 100° and \angleS = 25°. Is \DeltaQPR \sim \DeltaTSM? Why?

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Answer: [False]

\Delta PQR, \angle P + \angle Q + \angle R = 180^{o}\text{ ({Interior angle of triangle}) }

55+25+\angle R=180

\angle R=180-80

\angle R=100^{o}

\text{In }  \Delta TSM, \angle T + \angle S + \angle M = 180^{o}\text{[Interior angle of the triangle] }

\angle T + 25^{o}+100^{o}=180^{o}

\angle T =180^{o}-125

\angle T =55^{o}

\text{In }  \Delta PQR \text{and } \Delta TSM

\angle P=\angle T

\angle Q=\angle S

\angle R=\angle M

Also are know that if all corresponding angles of two triangles are equal then triangles are similar.

\therefore \Delta PQR\sim \Delta TSM

Hence, the given statement is false because \angleQPR is not similar to \angleTSM.

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