Question

Question 5. Two tangents inclined at an angle of 60° are drawn to a circle of radius 3 cm, then the length of each tangent is *

1 point

(3√3)/2 cm

6 cm

3 cm

3√3
​

Answer 1

✬ Tangent = 3√3 cm ✬

Step-by-step explanation:

Given:

• Two tangents are inclined at an angle of 60°.
• Radius of circle is 3 cm.

To Find:

• What is the length of each tangent ?

Solution: Let O be the centre of the circle and two tangents which are drawn from a point P outside the circle be PA & PB.

Some terms to remember :

• The tangent to any circle is perpendicular to the radius of the circle at point of contact .

• The lengths of the two tangents from an external point to a circle are equal.

Now we have

• AP = PB (tangents)
• AO = BO (radius)
• ∠OAP = OBP = 90°

Let's join the centre and external point i.e OP.

• From above , we conclude that ∆OAP ≅ ∆OBP by SSS criteria.

➮ ∠APO = ∠BPO {by CPCT}

➮ ∠APB = ∠APO + ∠BPO

➮ 60° = 2∠APO

➮ 30° = ∠APO = ∠BPO

Now in ∆OAP using tan30°

• AP {perpendicular}
• AO {base}
• ∠APO = 30°

★ tan θ = Perpendicular/base ★

$\implies{\rm }$ tan 30° = OA/AP

$\implies{\rm }$ tan 30° = 3/AP

$\implies{\rm }$ 1/√3 = 3/AP

$\implies{\rm }$ AP = 3√3

• AP = PB = 3√3

Hence, the length of each tangent is 3√3 cm.

(Option D is correct)

Answer 2

∠ APO = 60°/2 = 30°

OA (Radius) = 3cm

$\boxed{ \rm{tan30 \degree = \frac{Opposite \: side}{Adjacent \: side} = \frac{1}{ \sqrt{3} } }}$

$\dfrac{OA}{AP} = \dfrac{1}{ \sqrt{3} }$

$AP = OA \times \sqrt{3}$

$AP = 3 \times \sqrt{3}$

$\underline{\underline {AP = 3 \sqrt{3} \: cm}}$