Question

If 0
° ≤ , ≤ 45° such that sin
( + ) = 1 and cos( − ) =
√3
/2
, value of A
and B respectively is:

a) 30°
, 60°

b) 60°
, 30°

c) 45°
, 45°

d) 65°
, 65°​

Answer 1

i ) Given Sin (A+B) = 1

$\implies sin(A+B) = sin 90\degree$

$\implies A + B = 90\degree \: --(1)$

$ii ) and\: Cos (A-B) = \frac{\sqrt{3}}{2}$

$\implies Cos (A-B) = cos 30\degree$

$\implies A-B = cos 30\degree \: --(2)$

/* Add Equations (1) and (2), we get */

$2A = 120\degree$

$\implies A = \frac{120}{2}$

$\implies A = 60\degree$

$Put \: Value \: A \ : in \: equation \:(1) ,we \:get$

$60\degree + B = 90\degree$

$\implies B = 90\degree - 60\degree$

$\implies B = 30\degree$

Therefore.,

$\red{ Value \:of \: A } \green { = 60\degree}$

$\red{ Value \:of \: B } \green { = 30\degree}$

•••♪

Answer 2

A = 60° and B = 30°

Solution :-

• Sin (A + B ) = 1 -------( Given )

=> Sin (A + B ) = Sin90° [ °.° 1 =Sin90°]

=> A + B = 90° ----------(1)

• Cos ( A - B ) = √3/2 ------( Given )

=> Cos ( A - B ) = Cos30° [ °.° √3/2 = Cos30° ]

=> A - B = 30° ----------(2)

Adding eqn. (1) & (2) ,

=> A + B + A - B = 90° + 30°

=> 2A = 120°

=> A = 60°

Put A=60° in eqn. (2)

=> 60° - B = 30°

=> -B = 30° - 60°

=> -B = -30°

=> B = 30°