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

2 thoughts on “6)<br />эр<br />tan (A+B) = 3<br />and<br />tan (A-B) =<br />و<br />o < ATBL 90′<br />and<br />I and B.<br />A7B; find<br />(3)”

1. [tex]\huge\mathcal{\fcolorbox{cyan}{black}{\pink{Answer࿐}}}[/tex]

   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)

   ★&

   [tex]tan (A-B) = \: \frac{1}{ \sqrt{3} } [/tex]

   ★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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