Question-220995

Question Number 220995 by hardmath last updated on 21/May/25

Answered by Frix last updated on 22/May/25

Let the red line =1 (1/(sin 30°))=(a/(sin 15°)) ⇒ a=(((√6)−(√2))/2) (a/(sin x))=(1/(sin (135°−x)))=((√2)/(cos x +sin x))⇒ a=(((√2)sin x)/(cos x +sin x))=(((√2)tan x)/(1+tan x)) (((√2)tan x)/(1+tan x))=(((√6)−(√2))/2) tan x =((√3)/3) x=30°

$$\mathrm{Let}\:\mathrm{the}\:\mathrm{red}\:\mathrm{line}\:=\mathrm{1} \\ $$$$\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{30}°}=\frac{{a}}{\mathrm{sin}\:\mathrm{15}°}\:\Rightarrow \\ $$$$\:\:\:\:\:{a}=\frac{\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\frac{{a}}{\mathrm{sin}\:{x}}=\frac{\mathrm{1}}{\mathrm{sin}\:\left(\mathrm{135}°−{x}\right)}=\frac{\sqrt{\mathrm{2}}}{\mathrm{cos}\:{x}\:+\mathrm{sin}\:{x}}\Rightarrow \\ $$$$\:\:\:\:\:{a}=\frac{\sqrt{\mathrm{2}}\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}\:+\mathrm{sin}\:{x}}=\frac{\sqrt{\mathrm{2}}\mathrm{tan}\:{x}}{\mathrm{1}+\mathrm{tan}\:{x}} \\ $$$$\frac{\sqrt{\mathrm{2}}\mathrm{tan}\:{x}}{\mathrm{1}+\mathrm{tan}\:{x}}=\frac{\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\mathrm{tan}\:{x}\:=\frac{\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$${x}=\mathrm{30}° \\ $$

Commented by hardmath last updated on 22/May/25

$$ \\ $$Why did it become 1?

Commented by Frix last updated on 22/May/25

Since only angles are given you can assume any length for one of the sides.

$$\mathrm{Since}\:\mathrm{only}\:\mathrm{angles}\:\mathrm{are}\:\mathrm{given}\:\mathrm{you}\:\mathrm{can}\:\mathrm{assume} \\ $$$$\mathrm{any}\:\mathrm{length}\:\mathrm{for}\:\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sides}. \\ $$

Answered by fantastic last updated on 22/May/25