Question

Sin A + Cos B =1, A=30° and B is an acute angle, then find the value of B.​

Answer 1

Hear,

Sin A + Cos B =1

& A = 30 degree

So,

Sin 30 + Cos B =1

also,

1/2 + Cos B = 1

& Cos B = 1-(1/2)

=> Cos B = 1/2

B = Cos (inverse) 1/2

B=60 degree is the answer

Answer 2

Given: Sin A + Cos B = 1 and A = 30°

To find: The value of B.

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$\underline{\bf{\dag} \:\mathfrak{Putting\; value \; of \; A\: :}}$⠀⠀

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$:\implies\sf Sin\; A + Cos \; B = 1 \\\\\\:\implies\sf Sin~30^{\circ} + Cos~ B = 1 \\\\\\:\implies\sf \dfrac{1}{2} + Cos~B = 1 \qquad\quad\bigg\lgroup\bf Sin\; 30^{\circ} = \dfrac{1}{2} \bigg\rgroup \\\\\\:\implies\sf Cos~ B = 1 - \dfrac{1}{2} \\\\\\:\implies\sf Cos~ B = \dfrac{2 - 1}{2} \\\\\\:\implies\sf Cos~ B = \dfrac{1}{2} \\\\\\:\implies\sf Cos~ B = Cos~ 60^{\circ} \\\\\\:\implies{\underline{\boxed{\frak{\pink{B = 60^{\circ}}}}}}\;\bigstar$

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$\therefore{\underline{\sf{Hence, \;the\; required\; value \; of \; B \;is\; \bf{60^{\circ} }.}}}$

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