Question

Q52. Sin45° + Cos30°​

Answer 1

Step-by-step explanation:

$\sin(45 \:) \: + \cos(30 ) \\ 1 \div \sqrt{2} + \sqrt{3} \div 2 \\ = 1.00515497$

Answer 2

$\huge\tt{Answer:-}$

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$\sf sin45° + cos30°$

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$\sf Now, \: we \: have \: value \: of \: sin45° = \dfrac{1}{\sqrt{2}}$

$\sf and \: cos30° = \dfrac{\sqrt{3}}{2}$

$\sf \implies sin45° + cos30° = \dfrac{1}{\sqrt{2}} + \dfrac{\sqrt{3}}{2}$

$\sf = \dfrac{1}{\sqrt{2}} + \dfrac{\sqrt{3}}{2}$

$\sf = \dfrac{1×2}{\sqrt{2}×2} + \dfrac{\sqrt{3}×\sqrt{2}}{2×\sqrt{2}}$

$\sf = \dfrac{2 + \sqrt{6}}{2\sqrt{2}}$

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$\sf Now, \: by \: rationalisation$

$\sf = \dfrac{2 + \sqrt{6}}{2\sqrt{2}} × \dfrac{2\sqrt{2}}{2\sqrt{2}}$

$\sf = \dfrac{(2 + \sqrt{6})(2\sqrt{2})}{(2\sqrt{2})(2\sqrt{2})}$

$\sf = \dfrac{(2 + \sqrt{6})(2\sqrt{2})}{(2\sqrt{2})²}$

$\sf = \dfrac{4\sqrt{2} + 2\sqrt{12}}{2 × 2×\sqrt{2}×\sqrt{2}}$

$\sf = \dfrac{4\sqrt{2} + 2\sqrt{12}}{4 × 2}$

$\sf = \dfrac{4\sqrt{2} + 2\sqrt{12}}{8}$

$\sf = \dfrac{2(2\sqrt{2} + \sqrt{12})}{8}$

$\sf = \dfrac{\cancel{2}^{1}(2\sqrt{2} + \sqrt{12})}{\cancel{8}_{4}}$

$\sf = \dfrac{2\sqrt{2} + \sqrt{12}}{4}$

$\sf = \dfrac{2\sqrt{2} + \sqrt{2×2×3}}{4}$

$\sf = \dfrac{2\sqrt{2} + 2\sqrt{3}}{4}$

$\sf = \dfrac{2(\sqrt{2} + \sqrt{3})}{4}$

$\sf = \dfrac{\cancel{2}^{1}(\sqrt{2} + \sqrt{3})}{\cancel{4}_{2}}$

$\sf = \dfrac{\sqrt{2} + \sqrt{3}}{2}$

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$\sf \color{fuchsia} Answer = \dfrac{\sqrt{2} + \sqrt{3}}{2}$

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$\huge\tt{Know \: More:-}$

$\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf 0° & \bf 30° & \bf 45° & \bf 60° & \bf 90° \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3} }{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }& 1 & \sqrt{3} & \rm Not \: Defined \\ \\ \rm cosec A & \rm Not \: Defined & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm Not \: De fined \\ \\ \rm cot A & \rm Not \: Defined & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0 \end{array}}}\end{gathered}$