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Question Number 190947 by Shrinava last updated on 15/Apr/23

Answered by mr W last updated on 15/Apr/23

((BD)/(AB))=((sin β)/(sin 30)) ...(i) ((AB)/(AC))=((sin (β+45))/(sin 45)) ...(ii) (i)×(ii): 1=((sin β sin (β+45))/(sin 30 sin 45)) 1=2 sin β(sin β+cos β) 1=1−cos 2β+sin 2β tan 2β=1 ⇒2β=180+45 ⇒β=112.5° ✓

$$\frac{{BD}}{{AB}}=\frac{\mathrm{sin}\:\beta}{\mathrm{sin}\:\mathrm{30}}\:\:\:...\left({i}\right) \\ $$$$\frac{{AB}}{{AC}}=\frac{\mathrm{sin}\:\left(\beta+\mathrm{45}\right)}{\mathrm{sin}\:\mathrm{45}}\:\:\:...\left({ii}\right) \\ $$$$\left({i}\right)×\left({ii}\right): \\ $$$$\mathrm{1}=\frac{\mathrm{sin}\:\beta\:\mathrm{sin}\:\left(\beta+\mathrm{45}\right)}{\mathrm{sin}\:\mathrm{30}\:\mathrm{sin}\:\mathrm{45}} \\ $$$$\mathrm{1}=\mathrm{2}\:\mathrm{sin}\:\beta\left(\mathrm{sin}\:\beta+\mathrm{cos}\:\beta\right) \\ $$$$\mathrm{1}=\mathrm{1}−\mathrm{cos}\:\mathrm{2}\beta+\mathrm{sin}\:\mathrm{2}\beta \\ $$$$\mathrm{tan}\:\mathrm{2}\beta=\mathrm{1} \\ $$$$\Rightarrow\mathrm{2}\beta=\mathrm{180}+\mathrm{45} \\ $$$$\Rightarrow\beta=\mathrm{112}.\mathrm{5}°\:\checkmark \\ $$

Commented by Shrinava last updated on 16/Apr/23

thank you professor perfect

$${thank}\:{you}\:{professor}\:{perfect} \\ $$

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