Question

In the adjoining diagram PQ, PR and ST are the tangents to the circle with centre O and radius 7 cm. Given OP = 25 cm.

Find:

1. length of ST
2. value of ∠OPQ, i.e., θ
3. ∠QUR, in nearest degree (use mathematical tables)

Answer 1

Given: PQ, PR and ST are tangents to the circle with centre O and radius 7 cm; OP = 25 cm.

Step-wise calculation:

1. Length of each tangent PQ and PR:

In right triangle OPQ,

OQ = 7

OP = 25

And OQ ⟂ PQ (radius to tangent)

By Pythagoras: `PQ = sqrt(OP^2 - OQ^2)`

= `sqrt(25^2 - 7^2)`

= `sqrt(625 - 49)`

= `sqrt(576)`

= 24 cm

2. Length ST:

Put O at (0, 0), so the top tangent is y = 7 and P at (0, 25).

Equation of a tangent through P: y = mx + 25. 

Tangency condition gives `m^2 = 576/49` so, `m = ±24/7`.

Intersections with y = 7:

`7 = (±24/7) x + 25` 

⇒ `x = ±21/4`

= ± 5.25

Thus, `S = (-21/4, 7)`

`T = (21/4, 7)`

And `ST = 21/4 - (-21/4)`

= `21/2`

= 10.5 cm

3. Angle θ = ∠OPQ:

In triangle OPQ right at Q, 

`sin θ = (OQ)/(OP)`

= `7/25`

`cos θ = (PQ)/(OP)`

= `24/25`

`tan θ = 7/24`

Hence, `θ = arctan(7/24) ≈ 16.26^circ` (θ ≈ 16.3° to 1 d.p.).

4. Angle ∠QUR:

QOR is the central angle subtended by chord QR.

Using perpendicularity, QOR = 180° – 2θ.

Angle at the circumference on the opposite arc at U is half the central angle:

`∠QUR = 1/2 xx QOR`

= 90° – θ

Numerically ∠QUR = 90° – 16.26° ≈ 73.74°, which to the nearest degree is 74°.