In ΔPQR and ΔMST, ∠P = 55°, ∠Q =25°, ∠M = 100° and ∠S = 25°. Is ΔQPR ~ ΔTSM? Why?
False
We know that, the sum of three angles of a triangle is 180°.
In ΔPQR, ∠P + ∠Q + ∠R = 180°
⇒ 55° + 25° + ∠R = 180°
⇒ ∠R = 180° – (55° + 25°)= 180° – 80° =100°
In ΔTSM, ∠T + ∠S + ∠M = 180°
⇒ ∠T + ∠25°+ 100° = 180°
⇒ ∠T = 180°-(25° +100°)
=180°-125°= 55°
In ΔPQR and A TSM, and
∠P = ∠T, ∠Q = ∠S,
and ∠R = ∠M
ΔPQR ~ ΔTSM [since, all corresponding angles are equal]
Hence, Δ QPR is not similar to ΔTSM, since correct correspondence is P ↔ T, Q < r→ S and R ↔M