Question Number 212720 by Spillover last updated on 22/Oct/24
Answered by A5T last updated on 22/Oct/24
![AB=(√(2(3(√2))^2 ))=6; Let ∠OAB=θ ((sinθ)/r)=((sin(180−2θ))/(AB))⇒cosθ=(3/r)⇒sin2θ=((6(√(r^2 −9)))/r^2 ) (4(√2))^2 =r^2 +r^2 −2r^2 cos(90−2θ)=2r^2 −12(√(r^2 −9)) ⇒r=(√(58)); ∠HAB=45°⇒∠OAH=cos^(−1) (3/( (√(58))))−45° ⇒OH=(√(18+58−2×3(√(2×58))cos(cos^(−1) (3/( (√(58))))−45))) ⇒OH=4; ((sin(cos^(−1) (3/( (√(58))))−45))/4)=((sinAOH)/(3(√2))) ⇒sin(AOH)=((3(√2)×sin(cos^(−1) (3/( (√(58))))−45))/4) ⇒sin(90−AOH)=cos(AOH)=(√(1−sin^2 (AOH))) ⇒[HOC]=(1/2)×4×(√(58))×(√(1−sin^2 (AOH)))=14](https://www.tinkutara.com/question/Q212728.png)
$${AB}=\sqrt{\mathrm{2}\left(\mathrm{3}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} }=\mathrm{6};\:{Let}\:\angle{OAB}=\theta \\ $$$$\frac{{sin}\theta}{{r}}=\frac{{sin}\left(\mathrm{180}−\mathrm{2}\theta\right)}{{AB}}\Rightarrow{cos}\theta=\frac{\mathrm{3}}{{r}}\Rightarrow{sin}\mathrm{2}\theta=\frac{\mathrm{6}\sqrt{{r}^{\mathrm{2}} −\mathrm{9}}}{{r}^{\mathrm{2}} } \\ $$$$\left(\mathrm{4}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{2}{r}^{\mathrm{2}} {cos}\left(\mathrm{90}−\mathrm{2}\theta\right)=\mathrm{2}{r}^{\mathrm{2}} −\mathrm{12}\sqrt{{r}^{\mathrm{2}} −\mathrm{9}} \\ $$$$\Rightarrow{r}=\sqrt{\mathrm{58}};\:\angle{HAB}=\mathrm{45}°\Rightarrow\angle{OAH}={cos}^{−\mathrm{1}} \frac{\mathrm{3}}{\:\sqrt{\mathrm{58}}}−\mathrm{45}° \\ $$$$\Rightarrow{OH}=\sqrt{\mathrm{18}+\mathrm{58}−\mathrm{2}×\mathrm{3}\sqrt{\mathrm{2}×\mathrm{58}}{cos}\left({cos}^{−\mathrm{1}} \frac{\mathrm{3}}{\:\sqrt{\mathrm{58}}}−\mathrm{45}\right)} \\ $$$$\Rightarrow{OH}=\mathrm{4};\:\frac{{sin}\left({cos}^{−\mathrm{1}} \frac{\mathrm{3}}{\:\sqrt{\mathrm{58}}}−\mathrm{45}\right)}{\mathrm{4}}=\frac{{sinAOH}}{\mathrm{3}\sqrt{\mathrm{2}}} \\ $$$$\Rightarrow{sin}\left({AOH}\right)=\frac{\mathrm{3}\sqrt{\mathrm{2}}×{sin}\left({cos}^{−\mathrm{1}} \frac{\mathrm{3}}{\:\sqrt{\mathrm{58}}}−\mathrm{45}\right)}{\mathrm{4}} \\ $$$$\Rightarrow{sin}\left(\mathrm{90}−{AOH}\right)={cos}\left({AOH}\right)=\sqrt{\mathrm{1}−{sin}^{\mathrm{2}} \left({AOH}\right)} \\ $$$$\Rightarrow\left[{HOC}\right]=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{4}×\sqrt{\mathrm{58}}×\sqrt{\mathrm{1}−{sin}^{\mathrm{2}} \left({AOH}\right)}=\mathrm{14} \\ $$
Commented by A5T last updated on 22/Oct/24
![cos(cos^(−1) (3/( (√(58))))−45)=(3/( (√(58))))×(1/( (√2)))+(√(1−((3/( (√(58)))))^2 ))×(1/( (√2))) =((5(√(29)))/(29)) [cos(A−B)=cosAcosB+sinAsinB] ⇒sin(cos^(−1) (3/( (√(58))))−45)=(√(1−(((5(√(29)))/(29)))^2 ))=((2(√(29)))/(29))](https://www.tinkutara.com/question/Q212729.png)
$${cos}\left({cos}^{−\mathrm{1}} \frac{\mathrm{3}}{\:\sqrt{\mathrm{58}}}−\mathrm{45}\right)=\frac{\mathrm{3}}{\:\sqrt{\mathrm{58}}}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\sqrt{\mathrm{1}−\left(\frac{\mathrm{3}}{\:\sqrt{\mathrm{58}}}\right)^{\mathrm{2}} }×\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$$=\frac{\mathrm{5}\sqrt{\mathrm{29}}}{\mathrm{29}}\:\left[{cos}\left({A}−{B}\right)={cosAcosB}+{sinAsinB}\right] \\ $$$$\Rightarrow{sin}\left({cos}^{−\mathrm{1}} \frac{\mathrm{3}}{\:\sqrt{\mathrm{58}}}−\mathrm{45}\right)=\sqrt{\mathrm{1}−\left(\frac{\mathrm{5}\sqrt{\mathrm{29}}}{\mathrm{29}}\right)^{\mathrm{2}} }=\frac{\mathrm{2}\sqrt{\mathrm{29}}}{\mathrm{29}} \\ $$
Commented by Spillover last updated on 23/Oct/24
$${great}\:{work}.{thanks} \\ $$