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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

$$\frac{BD}{AB}=\frac{\sin \beta }{\sin 30}...\left(i\right)$$$$\frac{AB}{AC}=\frac{\sin (\beta +45)}{\sin 45}...\left(ii\right)$$$$\left(i\right)\times \left(ii\right):$$$$1=\frac{\sin \beta \sin (\beta +45)}{\sin 30\sin 45}$$$$1=2\sin \beta (\sin \beta +\cos \beta )$$$$1=1-\cos 2\beta +\sin 2\beta$$$$\tan 2\beta =1$$$$\Rightarrow 2\beta =180+45$$$$\Rightarrow \beta =112.5°✓$$

Commented by Shrinava last updated on 16/Apr/23

$$thankyouprofessorperfect$$

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