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Question Number 38593 by ajfour last updated on 27/Jun/18

Commented by ajfour last updated on 27/Jun/18

$$Find\theta intermsof\mathit{c}.$$

Commented by MrW3 last updated on 27/Jun/18

$$\frac{c}{\tan \theta }=1+\tan \theta$$$${\tan }^2\theta +\tan \theta -c=0$$$$\tan \theta =\frac{\sqrt{1+4c}-1}{2}$$$$\Rightarrow \theta ={\tan }^{-1}\frac{\sqrt{1+4c}-1}{2}$$

Commented by ajfour last updated on 27/Jun/18

$$thankyouSir.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 27/Jun/18

$$tan\theta =\frac{c}{1+b}b=perpendicularofsmalright$$$$angledtriangle$$$$tan\theta =\frac{b}{1}$$$$\frac{c}{1+b}=\frac{b}{1}$$$$b^{}+b^2=c$$$$b^2+b-c=0$$$$b=\frac{-1+\sqrt{1^2-4\times 1\times (-c)}}{2}=\frac{-1+\sqrt{1+4c}}{2}$$$$tan\theta =\frac{b}{1}=\frac{\sqrt{1+4c}-1}{2}$$$$plscheck$$

Commented by ajfour last updated on 27/Jun/18

$$ThankyouSir,butithinkthe$$$$nextquestionisbitchallenging!$$

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