Question

23. If sin (A + B) = 1 and cos (A - B) = 1, 0° S(A + B) < 90° and A > B, then
find A and B.​

Answer 1

Answer:

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°.

Answer 2

Answer :-

• Value of A, B are 45°,45°

Given :-

sin(A+B) = 1

cos(A-B) = 1

To find :-

Value of A, B

Solution :-

As they given sin(A+B) = 1 Let R.H.S also we shall convert in terms of "sin" only As similarly cos (A-B) = 1 We shall convert in forms of "cos" by using the trigonometric table

• sin 90°= 1
• cos 0° = 1

So,

sin(A+B) = sin 90°

Removing "sin" on both sides that gives ,

A+B = 90°{eq-1}

cos(A-B) = cos 0°

Removing "cos" on both sides that gives ,

A-B = 0° {eq-2}

We shall add the both equations which gives any variable value

A+B +(A-B) = 0°+90°

A+B+A-B = 90°

2A= 90°

A = 90/2

A = 45°

So, the value of A  is 45°

Now , we shall subtract the both equations which gives the other variable value

A+B -(A-B) = 90°-0°

A+B-A+B = 90°

2B = 90°

B = 90/2

B =45°

So, the value of B is 45°

Know more Trigonometric values

• sin0° = 0
• sin30° = 1/2
• sin45° = 1/√2
• sin60° = √3/2
• sin90° = 1

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• cos0° = 1
• cos30° = √3/2
• cos45° = 1/√2
• cos60° = 1/2
• cos90° = 0

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Note :- If you  remember sin values  and cos values Then other Trigonometric values you can calculate easily .