Question

if cos (A+B)=sin(A-B)=1/2,0<A+B<₩ 90 degree and A>B, then find the values A and B?​

Answer 1

$\begin{gathered}\begin{gathered}\bf\: Given-\begin{cases} &\sf{0 < B < A < 90\degree } \\ \\ &\sf{cos(A + B) = \dfrac{1}{2} } \\ \\ &\sf{sin(A - B) = \dfrac{1}{2} } \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \: To \: Find - \begin{cases} &\sf{A \: and \: B}\end{cases}\end{gathered}\end{gathered}$

Values Used :-

$\rm :\longmapsto\:sin30\degree = \dfrac{1}{2}$

$\rm :\longmapsto\:cos60\degree = \dfrac{1}{2}$

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

Given that,

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

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

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

$\bf\implies \:A + B = 60\degree - - - (1)$

Also,

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

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

$\bf\implies \:A - B = 30\degree - - - (2)$

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

$\rm :\longmapsto\:A + B + A - B = 60\degree + 30\degree$

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

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

On substituting the value of A in equation (1), we get

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

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

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

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

Additional Information :-

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