Question Number 176716 by cortano1 last updated on 25/Sep/22 | ||
Answered by mr W last updated on 26/Sep/22 | ||
Commented by Tawa11 last updated on 26/Sep/22 | ||
$$\mathrm{Great}\:\mathrm{sir} \\ $$ | ||
Commented by mr W last updated on 26/Sep/22 | ||
$$\beta+\mathrm{30}°+\alpha+\mathrm{120}°=\mathrm{180}° \\ $$$$\Rightarrow\beta=\mathrm{30}°−\alpha \\ $$$$\frac{\mathrm{sin}\:\mathrm{2}\alpha}{\:\sqrt{\mathrm{3}}}=\frac{\mathrm{sin}\:\beta}{\mathrm{1}}=\frac{\mathrm{sin}\:\left(\mathrm{30}°−\alpha\right)}{\mathrm{1}} \\ $$$$\mathrm{sin}\:\mathrm{2}\alpha=\sqrt{\mathrm{3}}\:\mathrm{sin}\:\left(\mathrm{30}°−\alpha\right) \\ $$$$\Rightarrow\alpha\approx\mathrm{14}.\mathrm{1328}°\:\Rightarrow\angle{ABC}\approx\mathrm{45}.\mathrm{8672}° \\ $$ | ||