Question

If sin(A - B) = 1/2 , cos(A + B) = 1/2, 0°< A + B ≤ 90°, A > B. Find A and B

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Answer 1

Answer:

A= 45° and B= 15°

Step-by-step explanation:

sin(A-B) = 1/2 = sin(A-B) = sin30°

= A-B=30------(i)

cos(A+B) = 1/2 = cos(A+B) = cos60°

= A+B = 60 ---------(ii)

By adding eq (i) and (ii),

A-B+A+B = 30+60

2A = 90

A= 45°

By putting the value of A in eq (ii),

A+B =60

45+B = 60

B= 15° .

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:0 < A \leqslant 90\degree$

$\rm :\longmapsto\:0 < B \leqslant 90\degree$

$\rm :\longmapsto\:A > B$

Further, given that

$\rm :\longmapsto\:sin(A - B) = \dfrac{1}{2}$

$\rm :\longmapsto\:sin(A - B) = sin30\degree$

$\rm\implies \:\boxed{\tt{ \: A - B = 30\degree}} - - - (1)$

Also, given that

$\rm :\longmapsto\:cos(A + B) = \dfrac{1}{2}$

$\rm :\longmapsto\:cos(A + B) = cos60\degree$

$\rm\implies \:\boxed{\tt{ \: A + B = 60\degree}} - - - (2)$

On adding equation (1) and (2), we get

$\rm :\longmapsto\:2A = 90\degree$

$\bf\implies \:A = 45\degree$

On substituting the value of A in equation (2), we get

$\rm :\longmapsto\:45\degree + B = 60\degree$

$\rm :\longmapsto\:B = 60\degree - 45\degree$

$\bf\implies \:B = 15\degree$

Hence,

$\red{\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:\begin{cases} &\sf{A = 45\degree} \\ \\ &\sf{B = 15\degree} \end{cases}\end{gathered}\end{gathered}}$

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