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Question Number 123708 by mnjuly1970 last updated on 27/Nov/20

Commented by mnjuly1970 last updated on 27/Nov/20

$$pleaseprove\Uparrow \Uparrow$$

Answered by ajfour last updated on 27/Nov/20

$$\frac{a\tan \theta }{\cos \theta }=h$$$$\frac{b}{\sin \theta \tan \theta }=h$$$$\frac{a}{\cos \theta }+\frac{b}{\sin \theta }=c$$$$\Rightarrow a\sin \theta =h\cos {}^2\theta$$$$b\cos \theta =h\sin {}^2\theta$$$$\Rightarrow a\sin \theta +b\cos \theta =c\sin \theta \cos \theta =h$$$$\mathrm{\&}ab=h^2\sin \theta \cos \theta =\frac{h^3}{c}$$$$\Rightarrow abc=h^3$$

Commented by mnjuly1970 last updated on 27/Nov/20

$$bravobravomasterajfor.$$$$thankyou$$$$$$

Answered by ajfour last updated on 27/Nov/20

$$h^2=\left(\frac{a}{\cos \theta }\right)\left(\frac{b}{\sin \theta }\right)...\left(i\right)$$$$h(\frac{\sin \theta }{\cos \theta }+\frac{\cos \theta }{\sin \theta })=c$$$$\Rightarrow \sin \theta \cos \theta =\frac{h}{c}...\left(ii\right)$$$$\left(i\right)\times \left(ii\right)\Rightarrow h^3=abc.$$

Commented by mnjuly1970 last updated on 27/Nov/20

$$godkeepyouyoungmaster$$$${}^{″}mrajfor{}^{″}..$$

Commented by ajfour last updated on 27/Nov/20

$$thanks,Iwishthesameforyou.$$

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