Question

2
Q2 If sin(A + B) = 1 and cos(A - B) = 1.0° <A+B < 90°,A> B then find A and B.
2 + 3 = 2 and 49
Q3 Solve​

Answer 1

$\\ \red{\rm\bigstar\large\underline{\underline{Solution :- }}}$

Sin ( A + B ) = 1 or sin ( A + B ) = sin90°

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: - \: \: \: [As \: \sin 90\degree = 1 ] - - - - (1)$

Again , $\sf \: \cos(A + B) = 1$

$\\ \: \: \: \: \sf \: \: \: \: \: \: \: \: \: \: \: \: \: = \cos0 \degree$

$\\ \: \: \: \: \sf \: \: \: \: \: \: \: \: \: \: (A - B) = 0 - - - (2)$

Adding (1) and (2) , we get

$\\ \: \: \: \: \: \: \: \: \: \sf \: \: \: \: 2A = 90 \degree \: or \: A \: = 45 \degree$

Putting A = 45° in (1) we get

$\\ \: \: \: \: \: \: \: \: \: \sf \: \: \: \: 45 \degree + B = 90 \degree \: or \: B \: = 45 \degree$

Hence,

$\\ \: \: \: \: \: \: \: \: \: \: \underline{\boxed{\frak \orange{ A = 45 \degree and \: B = 45 \degree}}}$

Answer 2

$\\⠀⠀\bigstar \:⠀⠀⠀\underbrace\bold{\large\underline{Concept: - }}$

If sin(A + B) = 1 and cos(A - B) = 1.0° <A+B < 90°,A> B then find A and B.

2 + 3 = 2 and 49

$\\⠀⠀\bigstar \:⠀⠀⠀\underbrace\bold{\large\underline{Solution: - }}$

sin (A + B)= 1 or sin (A + B) = sin 90° [As sin 90° = 1]

• A + B = 90° …(1)

Again, Cos(A-B) = 1 = cos 0° A – B = 0 …(2)

Adding (1) and (2),

we get 2A = 90°

or A = 45°

Putting A = 45° in (1) we get

45° + B = 90°

or B = 45°

• Hence, A = 45° and B = 45°.