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 :

(a) length of ST

(b) value of ∠OPQ, i.e. θ

(c) ∠QUR, in nearest degree

(a) We know that,

The radius of a circle and tangent are perpendicular at the point of contact.

∴ ∠PQO = 90°

In right-angled triangle PQO,

⇒ PO² = PQ² + OQ²

⇒ 25² = PQ² + 7²

⇒ 625 = PQ² + 49

⇒ PQ² = 625 - 49

⇒ PQ² = 576

⇒ PQ = $\sqrt{576}$ = 24 cm.

In △ POQ,

⇒ tan θ = $\dfrac{OQ}{PQ}$

⇒ tan θ = $\dfrac{7}{24}$ ……………(1)

From figure,

⇒ ∠PAS = 90°

⇒ PA = OP - OA = 25 - 7 = 18 cm.

⇒ tan θ = $\dfrac{AS}{PA}$

⇒ tan θ = $\dfrac{AS}{18}$ ………..(2)

From equation (1) and (2), we get :

$\Rightarrow \dfrac{7}{24} = \dfrac{AS}{18} \\[1em] \Rightarrow AS = \dfrac{7}{24} \times 18 \\[1em] \Rightarrow AS = \dfrac{7 \times 3}{4} = \dfrac{21}{4} \text{ cm}.$

From figure,

⇒ ST = 2 × AS = $2 \times \dfrac{21}{4} = \dfrac{21}{2}$ = 10.5 cm

Hence, ST = 10.5 cm.

(b) From equation (1),

⇒ tan θ = $\dfrac{7}{24}$

⇒ tan θ = 0.292

⇒ tan θ = tan 16° 16'

⇒ θ = 16° 16'.

Hence, θ = 16° 16'.

(c) In △ POQ,

⇒ ∠POQ + ∠PQO + ∠QPO = 180°

⇒ ∠POQ + 90° + 16° 16' = 180°

⇒ ∠POQ + 106° 16' = 180°

⇒ ∠POQ = 180° - 106° 16' = 73° 44' = 74°.

We know that,

Tangent from an external point to the circle are equal in length.

∴ PQ = PR.

Also,

OR = OQ (Both equal to radius of circle)

In △ POQ and △ POR,

⇒ PQ = PR (Proved above)

⇒ OQ = OR (Radius of same circle)

⇒ PO = PO (Common side)

∴ △ POQ ≅ △ POR (By S.S.S. axiom)

We know that,

Corresponding parts of congruent triangles are equal.

∴ ∠POR = ∠POQ = 74°

From figure,

⇒ ∠QOR = ∠POR + ∠POQ = 74° + 74° = 148°.

We know that,

The angle subtended by an arc of a circle at its center is twice the angle it subtends anywhere on the circle's circumference.

⇒ ∠QOR = 2∠QUR

⇒ ∠QUR = $\dfrac{∠QOR}{2} = \dfrac{148°}{2}$ = 74°.

Hence, ∠QUR = 74°.