In Fig. 3, from an external point P, two tangents PT and PS are drawn to a circle with centre O and radius r. If $O P = 2 r$ , show that $∠ O T S = ∠ O S T = 30^{\circ}$ .

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Solution

We have,

$O P = 2 r$

Let $∠ T O P = θ$

In $Δ O T P, c o s θ = \frac{O T}{O P} = \frac{r}{2 r} = \frac{1}{2}$

$∵ θ = 60^{\circ}$

Hence, $∠ T O S = 2 θ = 2 \times 60^{\circ} = 120^{\circ}$

In $Δ T O S$

$∠ T O S + ∠ O T S + ∠ O S T = 180^{\circ}$

$120^{\circ} + 2 ∠ O T S = 180^{\circ} (∵ ∠ O T S = ∠ O S T)$

$2 ∠ O T S = 180^{\circ} - 120^{\circ}$

$∠ O T S = 30^{\circ}$

Hence, $∠ O T S = ∠ O S T = 30^{\circ}$

Hence Proved.

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In Fig. 3, from an external point P, two tangents PT and PS are drawn to a circle with centre O and radius r. If OP=2r, show that ∠OTS=∠OST=30∘.

We have, OP=2r Let ∠TOP=θ In ΔOTP, cos θ=OTOP=r2r=12 ∵ θ=60∘ Hence, ∠TOS=2θ=2×60∘=120∘ In ΔTOS ∠TOS+∠OTS+∠OST=180∘ 120∘+2∠OTS=180∘ (∵ ∠OTS=∠OST) 2∠OTS=180∘−120∘ ∠OTS=30∘ Hence, ∠OTS=∠OST=30∘ Hence Proved.

In Fig. 3, from an external point P, two tangents PT and PS are drawn to a circle with centre O and radius r. If $O P = 2 r$ , show that $∠ O T S = ∠ O S T = 30^{\circ}$ .