Question-125195

Question Number 125195 by Eric002 last updated on 09/Dec/20

Answered by mr W last updated on 09/Dec/20

((AD)/(sin 58))=((DC)/(sin x)) ...(i) ((AD)/(sin ∠B))=((BD)/(sin 32)) ∠B=180−58−x+32=90−x ((AD)/(cos x))=((DC)/(sin 32)) ...(ii) (ii)÷(i): ((cos x)/(sin 58))=((sin 32)/(sin x)) ⇒sin x cos x=sin 32 cos 32=sin 58 cos 58 ⇒x=32° or 58°

$$\frac{{AD}}{\mathrm{sin}\:\mathrm{58}}=\frac{{DC}}{\mathrm{sin}\:{x}}\:\:\:…\left({i}\right) \\ $$$$\frac{{AD}}{\mathrm{sin}\:\angle{B}}=\frac{{BD}}{\mathrm{sin}\:\mathrm{32}}\: \\ $$$$\angle{B}=\mathrm{180}−\mathrm{58}−{x}+\mathrm{32}=\mathrm{90}−{x} \\ $$$$\frac{{AD}}{\mathrm{cos}\:{x}}=\frac{{DC}}{\mathrm{sin}\:\mathrm{32}}\:\:\:…\left({ii}\right) \\ $$$$\left({ii}\right)\boldsymbol{\div}\left({i}\right): \\ $$$$\frac{\mathrm{cos}\:{x}}{\mathrm{sin}\:\mathrm{58}}=\frac{\mathrm{sin}\:\mathrm{32}}{\mathrm{sin}\:{x}} \\ $$$$\Rightarrow\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}=\mathrm{sin}\:\mathrm{32}\:\mathrm{cos}\:\mathrm{32}=\mathrm{sin}\:\mathrm{58}\:\mathrm{cos}\:\mathrm{58} \\ $$$$\Rightarrow{x}=\mathrm{32}°\:{or}\:\mathrm{58}° \\ $$

Commented by Eric002 last updated on 09/Dec/20