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Question Number 168103 by Tawa11 last updated on 03/Apr/22

Commented by greogoury55 last updated on 03/Apr/22

$F (12, 6), H (0, 2) \Rightarrow \vec{H F} = 12 i + 4 j$ $G (4, 0), E (\frac{17}{2}, 9) \Rightarrow \vec{G E} = \frac{9}{2} i + 9 j$ $\Rightarrow \cos x = \frac{\vec{H F} . \vec{G E}}{∣ \vec{H F} ∣ ∣ \vec{G E} ∣} = \frac{54 + 36}{\sqrt{144 + 16} . \sqrt{\frac{81}{4} + 81}}$ $\Rightarrow \cos x = \frac{90}{\frac{9}{2} \sqrt{5} . 4 \sqrt{10}} = \frac{5}{5 \sqrt{2}}$ $\Rightarrow \cos x = \frac{\sqrt{2}}{2}; x = \frac{π}{4}$
Commented by som(math1967) last updated on 03/Apr/22

$x = 45 °$
Commented by Tawa11 last updated on 03/Apr/22

$Workings sir$
Commented by Tawa11 last updated on 03/Apr/22

$God bless you sir$
	Answered by som(math1967) last updated on 03/Apr/22

	$l e t D C a l o n g X a x i s D A y a x i s$ $D a t o r i g i n$ $∴ G (4, 0) E (8.5, 9)$ $H (0, 2) F (12, 6)$ $s l o p e o f G E = \frac{9 - 0}{8.5 - 4} = 2$ $s l o p e o f H F = \frac{6 - 2}{12 - 0} = \frac{1}{3}$ $∴ a n g l e b e t w e e n G E a n d H F$ $x = \tan^{- 1} \frac{2 - \frac{1}{3}}{1 + 2 \times \frac{1}{3}}$ $x = \tan^{- 1} 1 = 45 °$
	Commented by Tawa11 last updated on 03/Apr/22

	$God bless you sir$
	Answered by mr W last updated on 03/Apr/22

	Commented by mr W last updated on 03/Apr/22

	$[E F F H] = 12 \times 9 - \frac{1}{2} (2 \times 4 + 8 \times 6 + 3 \times 3.5 + 8.5 \times 7)$ $= 45$ $H F = \sqrt{12^{2} + {(6 - 2)}^{2}} = 4 \sqrt{10}$ $G E = \sqrt{9^{2} + {(8.5 - 4)}^{2}} = \frac{9 \sqrt{5}}{2}$ $[E F G H] = \frac{H F \times G E \times \sin x}{2}$ $45 = \frac{4 \sqrt{10} \times 9 \sqrt{5} \times \sin x}{2 \times 2}$ $\Rightarrow \sin x = \frac{\sqrt{2}}{2}$ $\Rightarrow x = 45 °$
	Commented by Tawa11 last updated on 03/Apr/22

	$God bless you sir .$
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