Question 13 of 18intermediate🔧 ApplyShort Answer2 marks

If tan⁡(A+B)=3\tan(A + B) = \sqrt{3} and tan⁡(A−B)=13\tan(A - B) = \frac{1}{\sqrt{3}}; 0∘<A+B≤90∘0^\circ < A + B \leq 90^\circ; A>BA > B, find A and B.

Correct Answer

A = 45° and B = 15°

Exercise: EXERCISE 8.2 | Q: 3 | (Page Page 15)
For More Understanding

Explanation

This question tests the student's knowledge of standard trigonometric values from Table 8.1. Students need to recognize that tan 60° = 3\sqrt{3} and tan 30° = 1/3\sqrt{3}, then solve the resulting linear equations to find A and B.

Solution Steps

  1. Step 1: From Table 8.1, we know that tan 60° = 3\sqrt{3}. Since tan(A + B) = 3\sqrt{3}, therefore A + B = 60°.

  2. Step 2: From Table 8.1, we know that tan 30° = 1/3\sqrt{3}. Since tan(A - B) = 1/3\sqrt{3}, therefore A - B = 30°.

  3. Step 3: Solving equations A + B = 60° and A - B = 30° by adding them: 2A = 90°, so A = 45°.

  4. Step 4: Substituting A = 45° in A + B = 60°: 45° + B = 60°, so B = 15°.

  5. Step 5: Verification: A > B holds true (45° > 15°) and 0° < A + B ≤\leq 90° holds true (0° < 60° ≤\leq 90°).

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