If two tangents inclined at an angle `60^@` are drawn to a circle of radius `3` `cm`, then find the length of each tangent.

Options

A. `(3)/(2)sqrt3cm`
B. 6 cm
C. 3 cm
D. `3sqrt3cm`

Explanation

To find the length of each tangent drawn from a point outside a circle to the circle, we can follow these steps:

### Step 1: Understand the Problem
We have a circle with a radius of 3 cm and two tangents drawn from a point P outside the circle, forming an angle of 60° between them.

### Step 2: Identify Key Points
Let:
- O be the center of the circle.
- A and C be the points where the tangents touch the circle.
- OP be the line segment from the center O to the point P.

### Step 3: Use Angle Bisector Theorem
Since the angle between the two tangents (∠APC) is 60°, the angle bisector OP divides this angle into two equal angles of 30° each:
- ∠APO = 30° (angle between OP and PA)
- ∠CPO = 30° (angle between OP and PC)

### Step 4: Apply Right Triangle Properties
Since the tangent at point A (PA) is perpendicular to the radius OA, we have:
- ∠PAO = 90°.

Now, triangle OAP is a right triangle where:
- OA = radius = 3 cm (perpendicular)
- PA = tangent length (we need to find this).

### Step 5: Use Trigonometric Ratios
In triangle OAP:
- We can use the tangent function:
\[ \tan(\angle APO) = \frac{OA}{PA} \]
Here, ∠APO = 30°.

Substituting the known values:
\[ \tan(30°) = \frac{OA}{PA} = \frac{3}{PA} \]

### Step 6: Solve for PA
We know that:
\[ \tan(30°) = \frac{1}{\sqrt{3}} \]
So, we can set up the equation:
\[ \frac{1}{\sqrt{3}} = \frac{3}{PA} \]

Cross-multiplying gives:
\[ PA = 3\sqrt{3} \]

### Step 7: Conclusion
Since both tangents PA and PC are equal in length, the length of each tangent is:
\[ PA = PC = 3\sqrt{3} \text{ cm} \]

### Final Answer
The length of each tangent is \( 3\sqrt{3} \text{ cm} \).

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