Question

If tangent QR,PR,PQ are drawn respectively at A,B,C to the circle circumscribing an acute angled triangle ABC so as to form another triangle PQR,then the angle RPQ is equal to?

Answer 1

Given: An acute angled triangle ABC , circumscribed by a circle and tangents QR, PR,PQ  are drawn at  A,B and C  such that they form a ΔPQR.

To Find: ∠RPQ

Construction: Join OP, OB and OC where O is the centre of circle Circumscribing triangle ABC.

Solution: →OB⊥PR and OC ⊥PQ→[The point of contact of radius and tangent to circle measures 90°.]

In Δ OBP and ΔOCP

→OB=OC (radii of same circle)

→ OP is common

→BP = CP→ (Lengths of tangents from external point to a circle are equal)

Δ OBP ≅ ΔOCP (SSS)

∠BPO=∠CPO (CPCT)

Let , ∠BPO=∠CPO=α

In Right ΔOBP

Tanα=$\frac{OB}{BP}$=$\frac{r}{BP}$→where ,r is radius of circle circumscribing ΔABC.Let BP=x

∠RPQ =Tan 2α= $\frac{2 tan\alpha}{1+tan^{2} \alpha}$

          = $\frac{\frac{2r}{x}}{1+\frac{r^{2}}{ x^{2}} }=\frac{2rx}{x^{2} +r^{2} }$