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Question Number 88525 by jagoll last updated on 11/Apr/20

Commented by jagoll last updated on 11/Apr/20

$$\mathrm{i}\mathrm{got}{20}^{\mathrm{o}}$$

Commented by mr W last updated on 11/Apr/20

$$x=80°$$

Commented by jagoll last updated on 11/Apr/20

$$\mathrm{why}?$$

Commented by mahdi last updated on 11/Apr/20

$$\mathrm{in}\mathrm{\Delta }\mathrm{AOC}:\frac{\mathrm{sinx}}{\mathrm{AO}}=\frac{\sin 20}{\mathrm{OC}}$$$$\mathrm{in}\mathrm{\Delta }\mathrm{OBC}:\frac{\sin (110-\mathrm{x})}{\mathrm{OB}}=\frac{\sin 10}{\mathrm{OC}}$$$$\Rightarrow \frac{\mathrm{sinx}}{\mathrm{AO}}=\frac{\sin 20}{\frac{\mathrm{OB}.\sin 10}{\sin (110-\mathrm{x})}}\Rightarrow (\mathrm{OB}=\mathrm{OA}=\mathrm{r})$$$$\frac{\mathrm{sinx}}{\sin (110-\mathrm{x})}=\frac{\sin 20}{\sin 10}=2\cos 10=\frac{\sin 80}{\sin 30}$$$$\Rightarrow \mathrm{x}=80$$

Commented by jagoll last updated on 11/Apr/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}.$$

Commented by jagoll last updated on 11/Apr/20

$$\mathrm{sir}.\mathrm{how}\mathrm{to}\mathrm{get}110-\mathrm{x}?\mathrm{sir}$$

Answered by mr W last updated on 11/Apr/20

Commented by mr W last updated on 11/Apr/20

$$\frac{AD}{\sin 70}=\frac{CA}{\sin 110}=\frac{CA}{\cos 20}$$$$\frac{CA}{\sin 30}=\frac{AB}{\sin 70}$$$$AB=2OA\cos 20$$$$$$$$AD=\frac{\sin 70}{\cos 20}\times CA=\frac{\sin 70}{\cos 20}\times \frac{\sin 30}{\sin 70}\times AB$$$$=\frac{\sin 70}{\cos 20}\times \frac{\sin 30}{\sin 70}\times 2\cos 20\times OA$$$$=2\sin 30\times OA$$$$=OA$$$$\Rightarrow \mathrm{\Delta }OAD=isosceles$$$$\Rightarrow x=\frac{180-20}{2}=80°$$

Commented by mr W last updated on 11/Apr/20

$$sorry,\mathrm{\angle }ADC=70°$$$$110=70+40$$

Commented by jagoll last updated on 11/Apr/20

$$\mathrm{sir}\mathrm{W}.\mathrm{if}\mathrm{\angle }\mathrm{ACD}={70}^{\mathrm{o}}$$$$\mathrm{\angle }\mathrm{ADC}={110}^{\mathrm{o}}$$$$\mathrm{\angle }\mathrm{CAD}={40}^0$$$$\mathrm{how}\mathrm{it}?{70}^{\mathrm{o}}+{110}^0+{40}^{\mathrm{o}}>{180}^{\mathrm{o}}$$$$$$

Commented by jagoll last updated on 11/Apr/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

Commented by jagoll last updated on 11/Apr/20

Commented by jagoll last updated on 11/Apr/20

$$\mathrm{like}\mathrm{this}?$$

Commented by mr W last updated on 11/Apr/20

Commented by jagoll last updated on 11/Apr/20

$$\mathrm{ok}\mathrm{sir}.\mathrm{thank}\mathrm{you}$$

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