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Question Number 84441 by Power last updated on 13/Mar/20

Commented by Power last updated on 13/Mar/20

$sir please solution$
	Answered by mr W last updated on 13/Mar/20

	Commented by mr W last updated on 13/Mar/20

	$o n l y f o r r h o m b u s α = 110 ° . o t h e r w i s e$ $α i s n o t u n i q u e .$
	Commented by mr W last updated on 13/Mar/20

	$\frac{\sin (40 - β)}{2 a} = \frac{\sin β}{a} = \frac{\sin 40}{D E}$ $\frac{\sin 40}{\tan β} - \cos 40 = 2$ $\Rightarrow \tan β = \frac{\sin 40}{2 + \cos 40}$ $D E = \frac{a \sin 40}{\sin β}$ $D F = D E - F E = \frac{a \sin 40}{\sin β} - 2 a \cos β$ $\frac{D F}{\sin (γ - β)} = \frac{D C}{\sin γ}$ $\frac{\sin (γ - β)}{\sin γ} = \frac{D F}{D C} = \frac{\sin 40}{\sin β} - 2 \cos β$ $\cos β - \frac{\sin β}{\tan γ} = \frac{\sin 40}{\sin β} - 2 \cos β$ $3 \cos β - \frac{\sin 40}{\sin β} = \frac{\sin β}{\tan γ}$ $\frac{1}{\tan γ} = \frac{3}{\tan β} - \sin 40 (1 + \frac{1}{\tan^{2} β})$ $\frac{1}{\tan γ} = \frac{1 - \cos 40}{\sin 40}$ $\tan γ = \frac{\sin 40}{1 - \cos 40}$ $\Rightarrow γ = 70 °$ $\Rightarrow α = 180 - γ = 110 °$
	Commented by Power last updated on 13/Mar/20

	$thank you sir$
	Commented by Power last updated on 13/Mar/20

	Commented by Power last updated on 13/Mar/20

	$romb ?$
	Commented by mr W last updated on 13/Mar/20

	${w h a t}^{'} s y o u r c o n c r e t e q u e s t i o n ?$
	Commented by Power last updated on 13/Mar/20

	$for rhombus or parallelogram$
	Commented by Power last updated on 13/Mar/20

	$is it true that in a parallelogram, then angle will be changed .$
	Commented by Power last updated on 13/Mar/20

	$thanks$
	Commented by Power last updated on 13/Mar/20

	$work with the alpha of 40 °$
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