From the given figure, find ∠P. [2 MARKS]
Concept: 1 Mark
Application: 1 Mark
In ΔABC and ΔPQR,
ABRQ=3.87.6=12
BCQP=612=12
and CAPR=12
⇒ABRQ=BCQP=CAPR
So, ΔABC∼ΔRQP [SSS similarity]
∴∠C=∠P [Corresponding angles of similar triangles]
But ∠C=180∘−∠A−∠B [Angle sum property]
∠C=180∘−80∘−60∘=40∘
So, ∠P=40∘