Two tangents making an angle of 120 degree with each other are drawn to a circle of radius 6 cm ,find the length of each tangent?

Question

Answer

Answer

Problem

Two tangents making an angle of 120 degree with each other are drawn to a circle of radius 6 cm, find the length of each tangent?

Explanation:

Let us draw a rough diagram to understand the problem.

![image.png](attachment:image.png)

Here, we have a circle of radius 6 cm. Two tangents AB and AC are drawn making an angle of 120 degrees with each other. We need to find the length of each tangent.

Solution:

Let O be the center of the circle. Draw a line OD perpendicular to AB and another line OE perpendicular to AC as shown in the diagram below.

![image-2.png](attachment:image-2.png)

Now, in triangle AOD,

OA = radius of the circle = 6 cm (given)

AD = length of tangent AB (to be found)

OD = perpendicular from center O to tangent AB

By Pythagoras theorem,

$$
OD^2 + AD^2 = OA^2
$$

Substituting the given values, we get

$$
OD^2 + AD^2 = 6^2 \
\Rightarrow OD^2 = 6^2 - AD^2 \
\Rightarrow OD = \sqrt{36 - AD^2}
$$

Similarly, in triangle AOE,

OA = radius of the circle = 6 cm (given)

AE = length of tangent AC (to be found)

OE = perpendicular from center O to tangent AC

By Pythagoras theorem,

$$
OE^2 + AE^2 = OA^2
$$

Substituting the given values, we get

$$
OE^2 + AE^2 = 6^2 \
\Rightarrow OE^2 = 6^2 - AE^2 \
\Rightarrow OE = \sqrt{36 - AE^2}
$$

Now, angle AOE = angle AOD = 60 degrees (since tangents are drawn at an angle of 120 degrees with each other).

Therefore, triangles AOE and AOD are similar triangles.

Using the property of similar triangles, we get

$$
\frac{AD}{AE} = \frac{OD}{OE} \
\Rightarrow \frac{AD}{AE} = \frac{\sqrt{36 - AE^2}}{\sqrt{36 - AD^2}}
$$

Multiplying both sides by $\sqrt{36 - AD^2}$, we get

$$
AD = \frac{6\sqrt{3}}{\sqrt{4 + 3\sqrt{3}}}
$$

Similarly, multiplying both sides by $\sqrt{36 - AE^2}$, we get

$$
AE = \frac{6\sqrt{3}}{\sqrt{4 - 3\sqrt

Answer

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