1.

If sin^2A + cos B = 1, where angleA = 45°, find angleB.​

Answer»

sin²A + cos B = 1

Substitute the VALUE of A as 45°

➥ sin² 45° + cos B = 1

Transposing sin² 45° to the Right HAND SIDE,

➥ cos B = 1 - sin² 45°      ----- [EQUATION 1]

\rightarrow \: \sf{cos B=1 - \bigg(\dfrac{1}{\sqrt{2}}\bigg)^{2} }\\\\

\rightarrow \: \sf{cosB=1 - \dfrac{1}{2}}\\\\

\implies \sf{cosB = \dfrac{1}{2}}\\\\

\rightarrow \: \sf{cosB = cos \: 60^{\circ}}\\\\

\therefore \boxed{\bf{B = 60^{\circ}}}

Know more:

\bullet\:\bf Trigonometric\:Values :\\\\\boxed{\begin{tabular}{c|c|c|c|c|c}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{e}fined\end{tabular}}


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