Question-220243

Question Number 220243 by Spillover last updated on 10/May/25

Answered by mr W last updated on 11/May/25

Commented by mr W last updated on 11/May/25

y=CD=TB=(√2) ((sin α)/1)=((sin 45°)/( (√2)))=(1/2) ⇒α=30° [BCD]=((1×(√2)×sin 75°)/2) =(((√2)+(√6))/(4(√2)))=((1+(√3))/4)

$${y}={CD}={TB}=\sqrt{\mathrm{2}} \\ $$$$\frac{\mathrm{sin}\:\alpha}{\mathrm{1}}=\frac{\mathrm{sin}\:\mathrm{45}°}{\:\sqrt{\mathrm{2}}}=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\alpha=\mathrm{30}° \\ $$$$\left[{BCD}\right]=\frac{\mathrm{1}×\sqrt{\mathrm{2}}×\mathrm{sin}\:\mathrm{75}°}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\sqrt{\mathrm{2}}+\sqrt{\mathrm{6}}}{\mathrm{4}\sqrt{\mathrm{2}}}=\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{4}} \\ $$

Commented by Spillover last updated on 11/May/25