Question

If sin(A+B) =0 and sin(A-B)=1/2, then find A and B.
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Answer 1

Required Answer:-

Given:

• sin(A + B) = 0
• sin(A - B) = 1/2

To find:

• ∠A and ∠B

Solution:

Given that,

➡ sin(A + B) = 0

➡ sin(A - B) = 1/2

From Trigonometry Ratio Table,

➡ sin(A + B) = 0

➡ sin(A + B) = sin(0°)

➡ A + B = 0° .......(i)

Also,

➡ sin(A - B) = 1/2

➡ sin(A - B) = sin(30°)

➡ A - B = 30° .......(ii)

Adding equations (i) and (ii), we get,

➡ 2A = 30°

➡ ∠A = 15°

From (i), we get,

∠B = -∠A

= -15°

So,

➡ ∠A = 15°

➡ ∠B = -15°

Answer:

• ∠A = 15°
• ∠B = -15°

Trigonometry Ratio Table:

$\sf Trigonometry\: Value \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle x & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin(x) & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos(x)& 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan(x) & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec(x) & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec(x)& 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot(x)& \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0 \end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}$