If sin(A + B) = cos(A - B) = 1/√2, then find A and B, where 0° < A + B < 90° and A > B
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Step-by-step explanation:
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°.
VONGOLADECIMO:
This is the wrong answer bro.
no it's right
just solve and see
In my book, the answer is mentioned. It's A = 45° and B = 0°
The question is sin(A+B) = cos(A-B) = 1/√2
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