Question Number 201882 by hardmath last updated on 14/Dec/23
Commented by hardmath last updated on 14/Dec/23
$$ \\ $$If angleABC=45°, angleADC=60°,DC = 2BD, then find the value of angle ACB
Answered by mr W last updated on 14/Dec/23
$${say}\:{AE}={h}={height} \\ $$$${DE}=\frac{{h}}{\mathrm{tan}\:\mathrm{60}°}=\frac{{h}}{\:\sqrt{\mathrm{3}}} \\ $$$${BE}=\frac{{h}}{\mathrm{tan}\:\mathrm{45}°}={h} \\ $$$${BD}={h}−\frac{{h}}{\:\sqrt{\mathrm{3}}} \\ $$$${DC}=\mathrm{2}{h}\left(\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right) \\ $$$${EC}=\mathrm{2}{h}\left(\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)−\frac{{h}}{\:\sqrt{\mathrm{3}}}={h}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right) \\ $$$$\mathrm{tan}\:{C}=\frac{{h}}{{h}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)}=\frac{\mathrm{1}}{\mathrm{2}−\sqrt{\mathrm{3}}}=\mathrm{2}+\sqrt{\mathrm{3}} \\ $$$${C}=\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}+\sqrt{\mathrm{3}}\right)=\mathrm{75}° \\ $$
Commented by hardmath last updated on 15/Dec/23
$$\mathrm{perfect}\:\mathrm{dear}\:\mathrm{professor}\:\mathrm{thankyou} \\ $$