Question Number 92269 by Power last updated on 05/May/20
Answered by mr W last updated on 05/May/20
$${let}\:{BD}=\mathrm{1} \\ $$$$\frac{{AB}}{\mathrm{sin}\:\mathrm{51}}=\frac{{BD}}{\mathrm{sin}\:\left(\mathrm{25}+\mathrm{51}\right)} \\ $$$$\Rightarrow{AB}=\frac{\mathrm{sin}\:\mathrm{51}}{\mathrm{sin}\:\mathrm{76}} \\ $$$$\frac{{BC}}{\mathrm{sin}\:\mathrm{99}}=\frac{{BD}}{\mathrm{sin}\:\left(\mathrm{15}+\mathrm{99}\right)} \\ $$$$\Rightarrow{BC}=\frac{\mathrm{sin}\:\mathrm{99}}{\mathrm{sin}\:\mathrm{114}} \\ $$$${AC}^{\mathrm{2}} =\left(\frac{\mathrm{sin}\:\mathrm{51}}{\mathrm{sin}\:\mathrm{76}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{sin}\:\mathrm{99}}{\mathrm{sin}\:\mathrm{114}}\right)^{\mathrm{2}} −\mathrm{2}\left(\frac{\mathrm{sin}\:\mathrm{51}}{\mathrm{sin}\:\mathrm{76}}\right)\left(\frac{\mathrm{sin}\:\mathrm{99}}{\mathrm{sin}\:\mathrm{114}}\right)\mathrm{cos}\:\left(\mathrm{24}+\mathrm{15}\right) \\ $$$${AC}\approx\mathrm{0}.\mathrm{681} \\ $$$${BF}/{AC}\approx\mathrm{1}.\mathrm{467} \\ $$$$\frac{\mathrm{sin}\:{x}}{{BC}}=\frac{\mathrm{sin}\:\left(\mathrm{24}+\mathrm{15}\right)}{{AC}} \\ $$$$\mathrm{sin}\:{x}=\frac{\mathrm{sin}\:\mathrm{99}}{\mathrm{sin}\:\mathrm{114}}×\frac{\mathrm{sin}\:\mathrm{39}}{{AC}}=\mathrm{1} \\ $$$$\Rightarrow{x}=\mathrm{90}° \\ $$
Commented by Power last updated on 05/May/20
$$\mathrm{thanks} \\ $$
Commented by otchereabdullai@gmail.com last updated on 05/May/20
$$\mathrm{powerful} \\ $$