Question

If tan(A-B)=\(\frac{1}{\sqrt3}\) and sin A= \(\frac{1}{\sqrt2}\),  then the value of B is 

• A. 45°
• B. 15°
• C. 30°
• D. 60°

Answer 1

The Correct Option is B

\[ \tan(A - B) = \frac{1}{\sqrt{3}} \quad \text{and} \quad \sin A = \frac{1}{\sqrt{2}}. \]

Step 1: Determine the value of \( A \).

From \( \sin A = \frac{1}{\sqrt{2}} \), we recognize this as a standard trigonometric value. The angle \( A \) that satisfies this is:

\[ A = 45^\circ \quad (\text{since } \sin 45^\circ = \frac{1}{\sqrt{2}}). \]

Step 2: Use the given \( \tan(A - B) = \frac{1}{\sqrt{3}} \).

Substitute \( A = 45^\circ \):

\[ \tan(45^\circ - B) = \frac{1}{\sqrt{3}}. \]

Recognize that \( \frac{1}{\sqrt{3}} \) is the tangent of \( 30^\circ \):

\[ \tan(45^\circ - B) = \tan 30^\circ. \]

Step 3: Solve for \( B \).

Since \( \tan(45^\circ - B) = \tan 30^\circ \), we equate the angles:

\[ 45^\circ - B = 30^\circ. \]

Solve for \( B \):

\[ B = 45^\circ - 30^\circ = 15^\circ. \]

Final Answer: The value of \( B \) is \( \mathbf{15^\circ} \), which corresponds to option \( \mathbf{(2)} \).