The altitude of a cone is `20 cm` and its semi-vertical angle is `30^o` . If the semi-vertical angle is increasing at the rate of `2^o//sec`  per second, then the radius of the base is increasing at the rate of

Options

A. (a) `30` `cm//sec`
B. (b) `160/3` `cm//sec`
C. (c) `10` `cm//sec`
D. (d) `160` `cm//sec`

Explanation

To solve the problem step by step, we will use the relationship between the radius of the base of the cone, the altitude, and the semi-vertical angle.

### Step 1: Understand the relationship
The altitude (H) of the cone is given as 20 cm, and the semi-vertical angle (α) is given as 30°. The radius (R) of the base of the cone can be expressed in terms of the altitude and the semi-vertical angle using the tangent function:
\[ R = H \cdot \tan(\alpha) \]

### Step 2: Substitute the known values
Substituting the known values into the equation:
\[ R = 20 \cdot \tan(30°) \]
Since \(\tan(30°) = \frac{1}{\sqrt{3}}\):
\[ R = 20 \cdot \frac{1}{\sqrt{3}} = \frac{20}{\sqrt{3}} \text{ cm} \]

### Step 3: Differentiate with respect to time
To find the rate at which the radius is increasing, we differentiate the equation \( R = H \cdot \tan(\alpha) \) with respect to time (t):
\[ \frac{dR}{dt} = H \cdot \sec^2(\alpha) \cdot \frac{d\alpha}{dt} \]
Here, \( \frac{d\alpha}{dt} \) is the rate of change of the angle, which is given as \( 2°/sec \).

### Step 4: Convert degrees to radians
Since we need to work in radians for differentiation, convert \( \frac{d\alpha}{dt} \) to radians:
\[ \frac{d\alpha}{dt} = 2° \cdot \frac{\pi}{180°} = \frac{\pi}{90} \text{ rad/sec} \]

### Step 5: Substitute known values into the differentiated equation
Now substitute \( H = 20 \) cm, \( \alpha = 30° \), and \( \frac{d\alpha}{dt} = \frac{\pi}{90} \):
\[ \frac{dR}{dt} = 20 \cdot \sec^2(30°) \cdot \frac{\pi}{90} \]
Since \( \sec(30°) = \frac{2}{\sqrt{3}} \), we have:
\[ \sec^2(30°) = \left(\frac{2}{\sqrt{3}}\right)^2 = \frac{4}{3} \]

### Step 6: Calculate \( \frac{dR}{dt} \)
Now substitute this value into the equation:
\[ \frac{dR}{dt} = 20 \cdot \frac{4}{3} \cdot \frac{\pi}{90} \]
\[ \frac{dR}{dt} = \frac{80\pi}{270} = \frac{8\pi}{27} \text{ cm/sec} \]

### Final Answer
Thus, the radius of the base of the cone is increasing at the rate of:
\[ \frac{8\pi}{27} \text{ cm/sec} \]