`Delta`ABC is right angled at B. If `angle`A = `60^@`, then find the value of (secC + 2).

Options

A. `(2+2sqrt 3)//sqrt2`
B. 4/3
C. `(2+2 sqrt3)//sqrt3`
D. `4//sqrt3`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the triangle properties
Given that triangle ABC is a right-angled triangle with angle B = 90°. We also know that angle A = 60°.

### Step 2: Calculate angle C
In any triangle, the sum of the angles is 180°. Since angle B is 90° and angle A is 60°, we can find angle C using the formula:
\[ \text{Angle C} = 180° - \text{Angle A} - \text{Angle B} \]
Substituting the known values:
\[ \text{Angle C} = 180° - 60° - 90° = 30° \]

### Step 3: Find secant of angle C
The secant function is defined as the reciprocal of the cosine function:
\[ \sec C = \frac{1}{\cos C} \]
For angle C = 30°, we know:
\[ \cos 30° = \frac{\sqrt{3}}{2} \]
Thus,
\[ \sec 30° = \frac{1}{\cos 30°} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}} \]

### Step 4: Calculate sec C + 2
Now we need to find the value of \( \sec C + 2 \):
\[ \sec C + 2 = \frac{2}{\sqrt{3}} + 2 \]
To add these, we can express 2 with a common denominator:
\[ 2 = \frac{2\sqrt{3}}{\sqrt{3}} \]
Thus,
\[ \sec C + 2 = \frac{2}{\sqrt{3}} + \frac{2\sqrt{3}}{\sqrt{3}} = \frac{2 + 2\sqrt{3}}{\sqrt{3}} \]

### Final Answer
The value of \( \sec C + 2 \) is:
\[ \frac{2 + 2\sqrt{3}}{\sqrt{3}} \]

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