Question

Please answer this question immediately
it is urgent

Answer 1

Required Answer:-

According to question,

Sin (A + B) = 1/2

From the above table,

⇒ A + B = 30° --------(1)

And, Cos ( A - B) = √3/2

Again from the above table,

⇒ A - B = 30° -------(2)

Adding (1) and (2),

⇒ A + B + A - B = 30° + 30°

⇒ 2A = 60°

⇒ A = 30°

Then, B = 60° - 30° = 30°

We just need the value of A,

That is, 30° (Answer A)

Note:-

• Hope you understand the procedure to solve these type of questions. I would suggest you to memorise the table, it saves time in exams$.$

Answer 2

$\green{ \boxed{\boxed{\begin{array}{cc} \maltese \bf \: given,\: \\ \\ \bf \: sin(A+B) = \frac{1}{2} \\ \bf \implies \: A+B = {sin}^{ - 1} ( \frac{1}{2}) \\ \bf \implies \: A+B = {30}^{ \circ} \: \: \: \: - - - - (1) \end{array}}}}$

$\blue{ \boxed{\boxed{\begin{array}{cc} \maltese \: \bf \: also \: \: given \\ \\ \bf \: cos(A-B) = \frac{ \sqrt{3} }{2} \\ \bf \implies \: A-B = {cos}^{ - 1}( \frac{ \sqrt{3} }{2} ) \\ \bf \implies \: A-B = {30}^{ \circ} \: \: \: \: - - - - (2) \end{array}}}}$

$\green{ \boxed{\boxed{\begin{array}{cc} \bf \: eqn.(1) + eqn.(2)\bf \implies \: \\ \\ A+B = {30}^{ \circ} \\ \underline{ A-B = {30}^{ \circ} } \\ 2A = {60}^{ \circ} \\ \therefore \: \pink{A = {30}^{ \circ} }\: \: \end{array}}}}$

→ More : About value of B

$\green{\qquad \boxed{\boxed{\begin{array}{cc} \bf \:A+B = {30}^{ \circ} \\ = > \bf B = {30}^{ \circ} - A \\ \bf = {30}^{ \circ} - {30}^{ \circ} \\ \bf = {0}^{ \circ} \\ \\ \therefore \: \red{B = {0}^{ \circ} }\: \: \end{array}}}}$