If sin-A - B - 1 and sin -A -B - 1-2- such that 0 - A - B - 90- and A - B- then find A and B-

We have Sin (A +B) = 1 = Sin 90° A + B = 90° ...........(i) and Sin (A B) = 1/2 = sin 30° A B = 30° ......(ii) Adding (I) + (ii), We get 2A = 120° A = 60° S ...

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Standard IX

Mathematics

Trigonometric Ratios

If sin⁡A + B ...

Question

If sin⁡(A + B) = 1 and sin ⁡(A -B) = $\frac{1}{2}$ , such that 0 ≤ A + B ≤ 90° and A > B, then find A and B.

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Solution

We have Sin (A +B) = 1 = Sin 90°
A + B = 90° ...........(i)
and Sin (A - B) = $\frac{1}{2}$ = sin 30°
A - B = 30° ......(ii)
Adding (I) + (ii),
We get 2A = 120°
A = 60°
Substituting value of A in either (i) or (ii), we get B = 30°

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If sin⁡(A + B) = 1 and sin ⁡(A -B) = 12, such that 0 ≤ A + B ≤ 90° and A > B, then find A and B.

We have Sin (A +B) = 1 = Sin 90° A + B = 90° ...........(i) and Sin (A - B) = 12 = sin 30° A - B = 30° ......(ii) Adding (I) + (ii), We get 2A = 120° A = 60° Substituting value of A in either (i) or (ii), we get B = 30°

If sin⁡(A + B) = 1 and sin ⁡(A -B) = $\frac{1}{2}$ , such that 0 ≤ A + B ≤ 90° and A > B, then find A and B.

We have Sin (A +B) = 1 = Sin 90°
A + B = 90° ...........(i)
and Sin (A - B) = $\frac{1}{2}$ = sin 30°
A - B = 30° ......(ii)
Adding (I) + (ii),
We get 2A = 120°
A = 60°
Substituting value of A in either (i) or (ii), we get B = 30°