Question

If two tangents are inclined at 60˚ are drawn to a circle of radius 3cm then
find length of each tangent.

Answer 1

Answer:

3√3 cm.

Step-by-step explanation:

As both the tangents are drawn to the circle from a single point, the tangents are equal in length.

Let's say that the origin point of tangents be X, the tangents meet the circle at A and B, and the centre of the circle be O.

We know tangent of circle is perpendicular to the radius drawn to the meeting point of tangent nd circle.

So, OA ⊥ AX and OB ⊥ BX.

Therefore, Δ OAX and Δ OBX are right angled at A and B.

Now,

OA = OB = radius of circle, ∠OAX = ∠OBX = 90°, AX = BX (equal tangents)  

Therefore, Δ OAX ≅ Δ OBX (by SAS congruency)

So, ∠OXA = ∠OXB (by c.p.c.t)

∠AXB = ∠OXA + ∠OXB = 2∠OXA = 2∠OXB = 60°

So, ∠OXA = ∠OXB = 30°

In right triangle OAX, ∠OXA = 30°.

So, OA/AX = tan30° = 1/√3

AX = OA√3 = 3√3 cm.

Answer 2

Answer:

Answer:

3√3 cm.

Step-by-step explanation:

As both the tangents are drawn to the circle from a single point, the tangents are equal in length.

Let's say that the origin point of tangents be X, the tangents meet the circle at A and B, and the centre of the circle be O.

We know tangent of circle is perpendicular to the radius drawn to the meeting point of tangent nd circle.

So, OA ⊥ AX and OB ⊥ BX.

Therefore, Δ OAX and Δ OBX are right angled at A and B.

Now,

OA = OB = radius of circle, ∠OAX = ∠OBX = 90°, AX = BX (equal tangents)  

Therefore, Δ OAX ≅ Δ OBX (by SAS congruency)

So, ∠OXA = ∠OXB (by c.p.c.t)

∠AXB = ∠OXA + ∠OXB = 2∠OXA = 2∠OXB = 60°

So, ∠OXA = ∠OXB = 30°

In right triangle OAX, ∠OXA = 30°.

So, OA/AX = tan30° = 1/√3

AX = OA√3 = 3√3 cm.

HOPE IT IS HELPFUL...

Step-by-step explanation: