Question

It sin A=1/2 and
cos B=1/2, then A+B=?​

Answer 1

Answer:

I think it is right answer

Answer 2

Question:

• If Sin A = 1/2 and Cos B = 1/2, then what's the value of A + B.

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$:\implies\sf Sin A = \dfrac{1}{2}$

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$\underline{\bf{\dag} \:\mathfrak{As\: we \; know \; that \: :}}$

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• Sin A = Sin 30

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Therefore,

$:\implies\sf A = 30^{\circ} \qquad\bigg\lgroup\bf\: Equation (I)\: \bigg\rgroup$

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Similarly,

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$:\implies\sf Cos B = \dfrac{1}{2}$

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$\underline{\bf{\dag} \:\mathfrak{As\: we \; know \; that \: :}}$

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• Cos B = Cos 60

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$\underline{\bf{\dag} \:\mathfrak{From\: both\: equations\: \: :}}$

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$:\implies\sf B = 60^{\circ}\qquad\bigg\lgroup\bf Equation (II)\;\bigg\rgroup$

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Now,

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$:\implies\sf A + B \\\\\\:\implies\sf 30 + 60 \\\\\\:\implies{\underline{\boxed{\frak{\pink{A + B = 90^{\circ}}}}}}\:\bigstar$

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$\therefore\:{\underline{\sf{Hence, \ the \ value \: of \: (A + B) \ is\: \bf{90^{\circ}}.}}}$

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$\bigstar\:\sf Trigonometric\:Values :\\\begin{tabular}{|c|c|c|c|c|c|}\cline{1-6}Radians/Angle & 0 & 30 & 45 & 60 & 90\\\cline{1-6}Sin \theta & 0 & \dfrac{1}{2} & \dfrac{1}{\sqrt{2}} & \dfrac{\sqrt{3}}{2} & 1\\\cline{1-6}Cos \theta & 1 & \dfrac{\sqrt{3}}{2}& \dfrac{1}{\sqrt{2}}& \dfrac{1}{2}& 0\\\cline{1-6}Tan \theta& 0& \dfrac{1}{\sqrt{3}}& 1& \sqrt{3}& Not D \hat{e} fined \\\cline{1-6}\end{tabular}$