Question

What is the answer to the question I have posted

Answer 1

$\large{\underline{\underline{\red{\tt{\purple{\leadsto } GiveN:-}}}}}$

• $\sf sin ( A - B ) = \dfrac{1}{2}$

• $\sf cos ( A + B ) = \dfrac{1}{2}$

• $\sf 0° < ( A + B ) < 90° .$

$\large{\underline{\underline{\red{\tt{\purple{\leadsto } To\:FinD:-}}}}}$

• $\sf The\: value\:of\:A.$
• $\sf The\: value\:of\:B.$

$\large{\underline{\underline{\red{\tt{\purple{\leadsto } AnsweR:-}}}}}$

$\underline{\green{\sf\orange{\mapsto} According\:to\: first\: Condition:-}}$

$\tt:\implies sin(A-B)=\dfrac{1}{2}$

$\tt:\implies sin(A-B)=sin30^{\circ}$

$\underline{\boxed{\blue{\tt{\longmapsto A-B=30^{\circ}}}}}$

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$\underline{\green{\sf\orange{\mapsto} According\:to\: second\: Condition:-}}$

$\tt:\implies cos(A+B)=\dfrac{1}{2}$

$\tt:\implies cos(A+B)=sin60^{\circ}$

$\underline{\boxed{\blue{\tt{\longmapsto A+B=60^{\circ}}}}}$

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$\tt \green{ \orange{\mapsto}Add\:these\:two\: equation}$

$\tt:\implies A-B + ( A + B) = 30^{\circ}+60^{\circ}$

$\tt:\implies A +\cancel B + A \cancel - B = 90^{\circ}$

$\tt:\implies 2A = 90^{\circ}$

$\tt:\implies A=\bigg\lgroup\dfrac{90^{\circ}}{2}\bigg\rgroup$

$\underline{\boxed{\red{\tt{\longmapsto A=45^{\circ}}}}}$

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$\tt\green{\orange{\mapsto} Put\: this\:value\:in\:(i):-}$

$\tt:\implies A - B = 30^{\circ}$

$\tt:\implies 45^{\circ} - B = 30^{\circ}$

$\tt:\implies B = (45 -30)^{\circ}$

$\underline{\boxed{\red{\tt{\longmapsto B=15^{\circ}}}}}$

$\boxed{\green{\pink{\dag}\bf Hence\:value\:of\:A\:is\:45^{\circ}\&\:B\:is\:15^{\circ}.}}$

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$\large{\underline{\underline{\red{\tt{\purple{\leadsto } More\:To\: Know:-}}}}}$

$\Large{ \begin{tabular}{|c|c|c|c|c|c|} \cline{1-6} \theta & \sf 0^{\circ} & \sf 30^{\circ} & \sf 45^{\circ} & \sf 65^{\circ} & \sf 90^{\circ} \\ \cline{1-6} \sin & 0 & \dfrac{1}{2 } & \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} & 1 \\ \cline{1-6} \cos & 1 & \dfrac{ \sqrt{ 3 }}{2} } & \dfrac{1}{ \sqrt{2} } & \dfrac{ 1 }{ 2 } & 0 \\ \cline{1-6} \tan & 0 & \dfrac{1}{ \sqrt{3} } & 1 & \sqrt{3} & \infty \\ \cline{1-6} \cot & \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } &0 \\ \cline{1 - 6} \sec & 1 & \dfrac{2}{ \sqrt{3}} & \sqrt{2} & 2 & \infty \\ \cline{1-6} \csc & \infty & 2 & \sqrt{2 } & \dfrac{ 2 }{ \sqrt{ 3 } } & 1 \\ \cline{1 - 6}\end{tabular}}$