Question

Two tangents RQ and RP are drawn from an external point R to the circle with Centre O. if angle PRQ = 120 degree then prove that OR =PR+RQ.

FIND FIGURE 3 BELOW..

Answer 1

construction: join OP and OQ

proof:

In ∆OPR and ∆OQR

OP = OQ ... (radii of same circle)

RP = RQ

... (lengths of two tangents from an external point are equal)

OR = OR ... (common side)

∆OPR congruent to ∆OQR ... (by SSS test)

angle ORP = angle ORQ = ½ angle PRQ ... (cpct)

angle ORP = angle ORQ = ½(120)° = 60°

angle POR = angle QOR ... (cpct)

angle POR = angle QOR = 90° - 60°

angle POR = angle QOR = 30°

sin(angle POR) = sin(angle QOR) = sin30°

$\frac{PR}{OR} = \frac{RQ}{OR} = \frac{1}{2} \\</p><p>\frac{PR}{OR} = \frac{1}{2} \: \: \: => PR = \frac{1}{2} OR$ ... (i)

$</p><p>\frac{RQ}{OR} = \frac{1}{2} \\</p><p>RQ = \frac{1}{2} OR$ ... (ii)

adding (i) and (ii), we get

PR + RQ = ½OR + ½OR

PR + RQ = OR

... Hence Proved!