Question-168103

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)⇒HF^(→) = 12i+4j G(4,0), E(((17)/2),9)⇒GE^(→) = (9/2)i+9j ⇒cos x = ((HF^(→) .GE^(→) )/(∣HF^(→) ∣ ∣GE^(→) ∣)) = ((54+36)/( (√(144+16)) .(√(((81)/4)+81)))) ⇒cos x=((90)/((9/2)(√5) .4(√(10))))=(5/(5(√2))) ⇒cos x=((√2)/2) ; x=(π/4)

$${F}\left(\mathrm{12},\mathrm{6}\right),{H}\left(\mathrm{0},\mathrm{2}\right)\Rightarrow\overset{\rightarrow} {{HF}}\:=\:\mathrm{12}{i}+\mathrm{4}{j} \\ $$$${G}\left(\mathrm{4},\mathrm{0}\right),\:{E}\left(\frac{\mathrm{17}}{\mathrm{2}},\mathrm{9}\right)\Rightarrow\overset{\rightarrow} {{GE}}=\:\frac{\mathrm{9}}{\mathrm{2}}{i}+\mathrm{9}{j} \\ $$$$\Rightarrow\mathrm{cos}\:{x}\:=\:\frac{\overset{\rightarrow} {{HF}}\:.\overset{\rightarrow} {{GE}}}{\mid\overset{\rightarrow} {{HF}}\mid\:\mid\overset{\rightarrow} {{GE}}\mid}\:=\:\frac{\mathrm{54}+\mathrm{36}}{\:\sqrt{\mathrm{144}+\mathrm{16}}\:.\sqrt{\frac{\mathrm{81}}{\mathrm{4}}+\mathrm{81}}} \\ $$$$\: \\ $$$$\Rightarrow\mathrm{cos}\:{x}=\frac{\mathrm{90}}{\frac{\mathrm{9}}{\mathrm{2}}\sqrt{\mathrm{5}}\:.\mathrm{4}\sqrt{\mathrm{10}}}=\frac{\mathrm{5}}{\mathrm{5}\sqrt{\mathrm{2}}}\: \\ $$$$\Rightarrow\mathrm{cos}\:{x}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:;\:{x}=\frac{\pi}{\mathrm{4}} \\ $$

Commented by som(math1967) last updated on 03/Apr/22

$$\boldsymbol{{x}}=\mathrm{45}° \\ $$

Commented by Tawa11 last updated on 03/Apr/22

$$\mathrm{Workings}\:\mathrm{sir} \\ $$

Commented by Tawa11 last updated on 03/Apr/22

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Answered by som(math1967) last updated on 03/Apr/22

let DC along Xaxis DA y axis D at origin ∴G(4,0) E(8.5,9) H(0,2) F(12,6) slope of GE=((9−0)/(8.5−4))=2 slope of HF=((6−2)/(12−0))=(1/3) ∴ angle between GE and HF x=tan^(−1) ((2−(1/3))/(1+2×(1/3))) x=tan^(−1) 1=45°

$${let}\:{DC}\:{along}\:{Xaxis}\:{DA}\:{y}\:{axis} \\ $$$${D}\:{at}\:{origin} \\ $$$$\therefore{G}\left(\mathrm{4},\mathrm{0}\right)\:{E}\left(\mathrm{8}.\mathrm{5},\mathrm{9}\right) \\ $$$${H}\left(\mathrm{0},\mathrm{2}\right)\:{F}\left(\mathrm{12},\mathrm{6}\right) \\ $$$${slope}\:{of}\:\:{GE}=\frac{\mathrm{9}−\mathrm{0}}{\mathrm{8}.\mathrm{5}−\mathrm{4}}=\mathrm{2} \\ $$$${slope}\:{of}\:{HF}=\frac{\mathrm{6}−\mathrm{2}}{\mathrm{12}−\mathrm{0}}=\frac{\mathrm{1}}{\mathrm{3}}\:\: \\ $$$$\therefore\:{angle}\:{between}\:{GE}\:{and}\:{HF} \\ $$$$\:{x}=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{2}−\frac{\mathrm{1}}{\mathrm{3}}}{\mathrm{1}+\mathrm{2}×\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$${x}=\mathrm{tan}^{−\mathrm{1}} \mathrm{1}=\mathrm{45}° \\ $$

Commented by Tawa11 last updated on 03/Apr/22

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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

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

[EFFH]=12×9−(1/2)(2×4+8×6+3×3.5+8.5×7) =45 HF=(√(12^2 +(6−2)^2 ))=4(√(10)) GE=(√(9^2 +(8.5−4)^2 ))=((9(√5))/2) [EFGH]=((HF×GE×sin x)/2) 45=((4(√(10))×9(√5)×sin x)/(2×2)) ⇒sin x=((√2)/( 2)) ⇒x=45°

$$\left[{EFFH}\right]=\mathrm{12}×\mathrm{9}−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}×\mathrm{4}+\mathrm{8}×\mathrm{6}+\mathrm{3}×\mathrm{3}.\mathrm{5}+\mathrm{8}.\mathrm{5}×\mathrm{7}\right) \\ $$$$=\mathrm{45} \\ $$$${HF}=\sqrt{\mathrm{12}^{\mathrm{2}} +\left(\mathrm{6}−\mathrm{2}\right)^{\mathrm{2}} }=\mathrm{4}\sqrt{\mathrm{10}} \\ $$$${GE}=\sqrt{\mathrm{9}^{\mathrm{2}} +\left(\mathrm{8}.\mathrm{5}−\mathrm{4}\right)^{\mathrm{2}} }=\frac{\mathrm{9}\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\left[{EFGH}\right]=\frac{{HF}×{GE}×\mathrm{sin}\:{x}}{\mathrm{2}} \\ $$$$\mathrm{45}=\frac{\mathrm{4}\sqrt{\mathrm{10}}×\mathrm{9}\sqrt{\mathrm{5}}×\mathrm{sin}\:{x}}{\mathrm{2}×\mathrm{2}} \\ $$$$\Rightarrow\mathrm{sin}\:{x}=\frac{\sqrt{\mathrm{2}}}{\:\mathrm{2}} \\ $$$$\Rightarrow{x}=\mathrm{45}° \\ $$

Commented by Tawa11 last updated on 03/Apr/22

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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