Question-179938

Question Number 179938 by Acem last updated on 04/Nov/22

Commented by Acem last updated on 04/Nov/22

{I t}^{'} s c o r r e c t, T h a n k s S i r!

Commented by HeferH last updated on 04/Nov/22

Commented by HeferH last updated on 04/Nov/22

x \sqrt{3} - x = 1

x = \frac{1}{\sqrt{3} - 1} = \frac{\sqrt{3} + 1}{2}

\tan β = \frac{x \sqrt{3}}{2 - x} = \frac{3 + \sqrt{3}}{3 - \sqrt{3}} = 2 + \sqrt{3}

β = \tan^{- 1} (2 + \sqrt{3})

β = 75 °

Answered by mr W last updated on 04/Nov/22

\frac{2 \sin β}{\sin (β + 60)} = \frac{1 \times \sin 45}{\sin 15}

\frac{\sin β}{\sin (β + 60)} = \frac{4}{2 \sqrt{2} (\sqrt{6} - \sqrt{2})}

\frac{\sin 60}{\tan β} + \cos 60 = \sqrt{3} - 1

\tan β = 2 + \sqrt{3}

\Rightarrow β = 75 °

Commented by Acem last updated on 05/Nov/22

V e r y g o o d S i r! t h a n k s

Answered by Acem last updated on 05/Nov/22

Commented by Acem last updated on 05/Nov/22

λ = 60 - 45 = 15

M N = 2 \cos 60 = 1 \Rightarrow M N = M A \Rightarrow α = ϕ = 30

\Rightarrow Υ = λ \Rightarrow N A = N C

α = β_{1} \Rightarrow N A = N B = N C \Rightarrow θ = β_{2} = 45

\Rightarrow β = 75 °

Commented by mr W last updated on 05/Nov/22

n i c e s o l u t i o n!

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