Question Number 165057 by MathsFan last updated on 25/Jan/22
Commented by MathsFan last updated on 25/Jan/22
$${how}\:{do}\:{i}\:{solve}\:{for}\:{x}? \\ $$
Answered by mr W last updated on 25/Jan/22
$${say}\:{side}\:{length}\:{of}\:{kare}\:=\mathrm{1} \\ $$$${BD}={BE}=\sqrt{\mathrm{2}} \\ $$$$\frac{{BE}}{{BC}}=\frac{\mathrm{sin}\:\left({x}+\mathrm{15}\right)}{\mathrm{sin}\:{x}} \\ $$$$\sqrt{\mathrm{2}}=\frac{\mathrm{sin}\:{x}\:\mathrm{cos}\:\mathrm{15}+\mathrm{cos}\:{x}\:\mathrm{sin}\:\mathrm{15}}{\mathrm{sin}\:{x}} \\ $$$$\sqrt{\mathrm{2}}=\mathrm{cos}\:\mathrm{15}+\frac{\mathrm{sin}\:\mathrm{15}}{\mathrm{tan}\:{x}} \\ $$$$\mathrm{tan}\:{x}=\frac{\mathrm{sin}\:\mathrm{15}}{\:\sqrt{\mathrm{2}}−\mathrm{cos}\:\mathrm{15}}=\frac{\mathrm{sin}\:\left(\mathrm{45}−\mathrm{30}\right)}{\:\sqrt{\mathrm{2}}−\mathrm{cos}\:\left(\mathrm{45}−\mathrm{30}\right)} \\ $$$$=\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\:\mathrm{4}−\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)}=\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{3}−\sqrt{\mathrm{3}}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$${x}=\mathrm{30}° \\ $$
Commented by MathsFan last updated on 25/Jan/22
$${thank}\:{very}\:{much}\:{sir} \\ $$
Commented by Tawa11 last updated on 25/Jan/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$