Question-170323

Question Number 170323 by cortano1 last updated on 20/May/22

Answered by mr W last updated on 20/May/22

\frac{A C}{\sin 18 °} = \frac{A D}{\sin 12 °} \dots (i)

\frac{\sin (α + 18 °)}{D B} = \frac{\sin α}{A D} \dots (i i)

(i) \times (i i) :

\frac{\sin (α + 18 °)}{\sin 18 °} = \frac{\sin α}{\sin 12 °}

\frac{\sin α \cos 18 ° + \cos α \sin 18 °}{\sin 18 °} = \frac{\sin α}{\sin 12 °}

\frac{1}{\tan 18 °} + \frac{1}{\tan α} = \frac{1}{\sin 12 °}

\tan α = \frac{1}{\frac{1}{\sin 12 °} - \frac{1}{\tan 18 °}} = \frac{1}{\sqrt{3}}

\Rightarrow α = 30 °

Commented by learner22 last updated on 21/May/22

N i c e

Commented by cortano1 last updated on 22/May/22

h o w t o s h o w \frac{1}{\sin 12 °} - \frac{1}{\tan 18 °} = \sqrt{3} ?

Commented by Tawa11 last updated on 08/Oct/22

Great sir

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