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Question Number 6852 by Tawakalitu. last updated on 31/Jul/16

Commented by Tawakalitu. last updated on 31/Jul/16

$P l e a s e i n e e d s o l u t i o n h e r e . . . . t o s o l v e f o r x$
	Answered by Rasheed Soomro last updated on 03/Aug/16

	Commented by Rasheed Soomro last updated on 05/Aug/16

	$B y G e o G e b r a x = 20 °$ $W i t h g i v e n a n g l e s x {c a n}^{'} t b e$ $d e t e r m i n e d .$ $I t h i n k o n e s h o u l d c o n s i d e r$ $m e a s u r e s o f t h e s i d e s i n o r d e r$ $t o d e t e r m i n e t h e r e q u i r e d a n g l e .$
	Answered by Rasheed Soomro last updated on 07/Aug/16

	$F o r d i a g r a m s e e c o m m e n t b e l o w .$ $I n △ A C B$ $∠ C A B = 10 + 70 = 80 a n d ∠ C B A = 20 + 60 = 80$ $∵ ∠ C A B = ∠ C B A = 80$ $∴ A C = B C [O p p o s i t e s i d e s t o c o n g r u e n t a n g l e s]$ $A l s o ∠ C = 180 - 2 \times 80 = 20$ $I n △ A F B, ∠ A F B = 180 - 70 - 60 = 50$ $S o, ∠ E F B = 180 - 50 = 130$ $a n d i n △ E F B ∠ F E B = 180 - 20 - 130 = 30$ $L e t A C = B C = 1$ $I n △ A E C, A C = 1, ∠ C = 20, ∠ C A E = 10$ $S o, ∠ A E C = 180 - 20 - 10 = 150$ $B y S i n e l a w$ $\frac{C E}{\sin 10} = \frac{A C}{\sin 150} = \frac{1}{\sin 150}$ $C E = \frac{\sin 10}{\sin 150}$ $S i m i l a r l y i n △ B D C, B C = 1, ∠ C = 20, ∠ C B D = 20 & ∠ B D C = 140$ $\frac{C D}{\sin 20} = \frac{B C}{\sin 140} = \frac{1}{\sin 140}$ $C D = \frac{\sin 20}{\sin 140}$ $N o w i n △ C D E, A c c o r d i n g t o S i n e l a w$ $\frac{C E}{\sin (160 - y)} = \frac{C D}{\sin y}$ $\frac{\sin (160 - y)}{\sin y} = \frac{C E}{C D} = \frac{\sin 10}{\sin 150} / \frac{\sin 20}{\sin 140}$ $\frac{\sin 160 \cos y - \cos 160 \sin y}{\sin y} = \frac{\sin 10 \sin 140}{\sin 20 \sin 150}$ $\sin 160 \cot y = \frac{\sin 10 \sin 140}{\sin 20 \sin 150} + \cos 160$ $= \frac{\sin 10 \sin 140 + \sin 20 \sin 150 \cos 160}{\sin 20 \sin 150}$ $\cot y = \frac{\sin 10 \sin 140 + \sin 20 \sin 150 \cos 160}{\sin 20 \sin 150 \sin 160}$ $\tan y = \frac{\sin 20 \sin 150 \sin 160}{\sin 10 \sin 140 + \sin 20 \sin 150 \cos 160}$ $y = \tan^{- 1} (\frac{\sin 20 \sin 150 \sin 160}{\sin 10 \sin 140 + \sin 20 \sin 150 \cos 160}) = 130 (I f 0 < y < 180)$ $[W i t h t h e h e l p o f c a l c u l a t o r]$ $x = 180 - y - 30 = 150 - 130 = 20$
	Commented by Rasheed Soomro last updated on 05/Aug/16

	Commented by Tawakalitu. last updated on 20/Aug/16

	$W o w, t h a n k s s o m u c h . i d i d n o t s e e i t b e f o r e . I r e a l l y a p p r e c i a t e$ $. T h a n k s a g a i n .$
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