Question

6)
эр
tan (A+B) = 3
and
tan (A-B) =
و
o < ATBL 90'
and
I and B.
A7B; find
(3)​

Answer 1

Answer:

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Answer 2

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Correct Question

• If tan (A + B) = √3 and tan (A – B) = 1/√3 ; 0° < A + B ≤ 90°; A > B, find A and B?

Solution

• ★A = 45°, B = 15°

Explanation

★Given that

• tan (A + B) = √3

★But we know that tan 60° = √3

Thus,

• tan (A + B) = tan 60°

• ⇒A+B= 60°----------(1)

★&

$tan (A-B) = \: \frac{1}{ \sqrt{3} }$

★But we know that

• tan 30° = 1/√3

Thus

• tan ( AB) = tan 30°

• A- B = 30° ----------(2)

★Our equations are

• A + B = 60°---------(1)

• A-B = 30°--------------(2)

★Adding (1) and (2)

• A+B+A-B= 60° +30°

• 2A = 90°

• A = 90° / 2

• A = 45°

Putting A = 45° in (1)

★

• A+B= 60°

• 45° + B = 60°

• B = 60° - 45°

• B = 15°

★Hence A = 45°, B = 15°

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