Question Number 213436 by mr W last updated on 05/Nov/24
Answered by A5T last updated on 05/Nov/24
Commented by mr W last updated on 05/Nov/24
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Commented by A5T last updated on 05/Nov/24
![sinθ=((15)/( (√(26^2 +15^2 ))))=((15)/( (√(901))))⇒cosθ=((26)/( (√(901)))) ⇒sin2θ=2sinθcosθ=((780)/(901)) [ABC]=15(((26×2+25×2+2x)/2))=(((26+x)(26+25)sin2θ)/2) ⇒x=27](https://www.tinkutara.com/question/Q213441.png)
$${sin}\theta=\frac{\mathrm{15}}{\:\sqrt{\mathrm{26}^{\mathrm{2}} +\mathrm{15}^{\mathrm{2}} }}=\frac{\mathrm{15}}{\:\sqrt{\mathrm{901}}}\Rightarrow{cos}\theta=\frac{\mathrm{26}}{\:\sqrt{\mathrm{901}}} \\ $$$$\Rightarrow{sin}\mathrm{2}\theta=\mathrm{2}{sin}\theta{cos}\theta=\frac{\mathrm{780}}{\mathrm{901}} \\ $$$$\left[{ABC}\right]=\mathrm{15}\left(\frac{\mathrm{26}×\mathrm{2}+\mathrm{25}×\mathrm{2}+\mathrm{2}{x}}{\mathrm{2}}\right)=\frac{\left(\mathrm{26}+{x}\right)\left(\mathrm{26}+\mathrm{25}\right){sin}\mathrm{2}\theta}{\mathrm{2}} \\ $$$$\Rightarrow{x}=\mathrm{27} \\ $$
Answered by mr W last updated on 05/Nov/24
Commented by mr W last updated on 05/Nov/24
$$\mathrm{tan}\:\alpha=\frac{\mathrm{15}}{{x}} \\ $$$$\mathrm{tan}\:\beta=\frac{\mathrm{15}}{\mathrm{26}} \\ $$$$\mathrm{tan}\:\gamma=\frac{\mathrm{15}}{\mathrm{25}} \\ $$$$\alpha+\beta+\gamma=\frac{\mathrm{180}°}{\mathrm{2}}=\mathrm{90}° \\ $$$$\mathrm{tan}\:\alpha=\frac{\mathrm{1}}{\mathrm{tan}\:\left(\beta+\gamma\right)}=\frac{\mathrm{1}−\mathrm{tan}\:\beta\:\mathrm{tan}\:\gamma}{\mathrm{tan}\:\beta+\mathrm{tan}\:\gamma} \\ $$$$\frac{\mathrm{15}}{{x}}=\frac{\mathrm{1}−\frac{\mathrm{15}}{\mathrm{26}}×\frac{\mathrm{15}}{\mathrm{25}}}{\frac{\mathrm{15}}{\mathrm{26}}+\frac{\mathrm{15}}{\mathrm{25}}}=\frac{\mathrm{26}×\mathrm{25}−\mathrm{15}^{\mathrm{2}} }{\mathrm{15}×\left(\mathrm{25}+\mathrm{26}\right)}=\frac{\mathrm{5}}{\mathrm{9}} \\ $$$$\Rightarrow{x}=\mathrm{27} \\ $$$${A}=\left(\mathrm{25}+\mathrm{26}+\mathrm{27}\right)×\mathrm{15}=\mathrm{1170} \\ $$