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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

$$sayAE=h=height$$$$DE=\frac{h}{\tan 60°}=\frac{h}{\sqrt{3}}$$$$BE=\frac{h}{\tan 45°}=h$$$$BD=h-\frac{h}{\sqrt{3}}$$$$DC=2h(1-\frac{1}{\sqrt{3}})$$$$EC=2h(1-\frac{1}{\sqrt{3}})-\frac{h}{\sqrt{3}}=h(2-\sqrt{3})$$$$\tan C=\frac{h}{h(2-\sqrt{3})}=\frac{1}{2-\sqrt{3}}=2+\sqrt{3}$$$$C={\tan }^{-1}(2+\sqrt{3})=75°$$

Commented by hardmath last updated on 15/Dec/23

$$\mathrm{perfect}\mathrm{dear}\mathrm{professor}\mathrm{thankyou}$$

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