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Question Number 191856 by TUN last updated on 02/May/23

Answered by Subhi last updated on 02/May/23

$$$$$$putAD=y$$$$\frac{BD}{sin\left(30\right)}=\frac{y}{sin\left(10\right)}$$$$BD=\frac{sin\left(30\right)}{sin\left(10\right)}yby\left[sinlaw\right]$$$$\frac{CD}{sin\left(50\right)}=\frac{y}{sin\left(x\right)}.henceCD=\frac{sin\left(50\right)}{sin\left(x\right)}y$$$$intriangleBDC$$$$\frac{BD}{sin(70-x)}=\frac{DC}{sin\left(20\right)}$$$$\frac{sin\left(30\right).y}{sin\left(10\right).sin(70-x)}=\frac{sin\left(50\right).y}{sin\left(20\right).sin\left(x\right)}$$$$(sin\left(30\right).sin\left(20\right))/(sin\left(50\right).sin\left(10\right)).sin\left(x\right)=sin\left(70\right).cos\left(x\right)-cos\left(70\right).sin\left(x\right)$$$$\frac{sin\left(30\right).sin\left(20\right)}{sin\left(50\right).sin\left(10\right).sin\left(70\right)}+cot\left(70\right)=cot\left(x\right)$$$$cot\left(x\right)=\sqrt{3}$$$$x=30$$$$$$

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