Question-193548

Question Number 193548 by lmcp1203 last updated on 16/Jun/23

Commented by lmcp1203 last updated on 16/Jun/23

$${hint}\:{by}\:{trigonometry}\:\:{x}=\mathrm{10} \\ $$

Answered by Subhi last updated on 16/Jun/23

(a/(sin(30)))=(b/(sin(100))) ⇛ a=(b/(2sin(100))) (i) (b/(sin(x)))=(c/(sin(10))) ⇛ c=((sin(10)b)/(sin(x))) (ii) (a/(sin(20−x)))=(c/(sin(20))) ⇛ (a/c)=((sin(20−x))/(sin(20)))=(i/(ii)) ((sin(x))/(2sin(100).sin(10)))=((sin(20−x))/(sin(20)))=((sin(20)cos(x)−sin(x)cos(20))/(sin(20))) 2sin(20)sin(100)sin(10)cos(x)−2sin(100)sin(10)cos(20)sin(x)=sin(20)sin(x) (sin(20)+(2sin(100)sin(10)cos(20))sin(x)=(2sin(20)sin(100)sin(10))cos(x) x=tan^(−1) (((2sin(20)sin(100)sin(10))/(sin(20)+(2sin(100)sin(10)cos(20)))))=10

$$ \\ $$$$\frac{{a}}{{sin}\left(\mathrm{30}\right)}=\frac{{b}}{{sin}\left(\mathrm{100}\right)}\:\Rrightarrow\:{a}=\frac{{b}}{\mathrm{2}{sin}\left(\mathrm{100}\right)}\:\left({i}\right) \\ $$$$\frac{{b}}{{sin}\left({x}\right)}=\frac{{c}}{{sin}\left(\mathrm{10}\right)}\:\:\Rrightarrow\:{c}=\frac{{sin}\left(\mathrm{10}\right){b}}{{sin}\left({x}\right)}\:\left({ii}\right) \\ $$$$\frac{{a}}{{sin}\left(\mathrm{20}−{x}\right)}=\frac{{c}}{{sin}\left(\mathrm{20}\right)}\:\Rrightarrow\:\frac{{a}}{{c}}=\frac{{sin}\left(\mathrm{20}−{x}\right)}{{sin}\left(\mathrm{20}\right)}=\frac{{i}}{{ii}} \\ $$$$\frac{{sin}\left({x}\right)}{\mathrm{2}{sin}\left(\mathrm{100}\right).{sin}\left(\mathrm{10}\right)}=\frac{{sin}\left(\mathrm{20}−{x}\right)}{{sin}\left(\mathrm{20}\right)}=\frac{{sin}\left(\mathrm{20}\right){cos}\left({x}\right)−{sin}\left({x}\right){cos}\left(\mathrm{20}\right)}{{sin}\left(\mathrm{20}\right)} \\ $$$$\mathrm{2}{sin}\left(\mathrm{20}\right){sin}\left(\mathrm{100}\right){sin}\left(\mathrm{10}\right){cos}\left({x}\right)−\mathrm{2}{sin}\left(\mathrm{100}\right){sin}\left(\mathrm{10}\right){cos}\left(\mathrm{20}\right){sin}\left({x}\right)={sin}\left(\mathrm{20}\right){sin}\left({x}\right) \\ $$$$\left({sin}\left(\mathrm{20}\right)+\left(\mathrm{2}{sin}\left(\mathrm{100}\right){sin}\left(\mathrm{10}\right){cos}\left(\mathrm{20}\right)\right){sin}\left({x}\right)=\left(\mathrm{2}{sin}\left(\mathrm{20}\right){sin}\left(\mathrm{100}\right){sin}\left(\mathrm{10}\right)\right){cos}\left({x}\right)\right. \\ $$$${x}={tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{sin}\left(\mathrm{20}\right){sin}\left(\mathrm{100}\right){sin}\left(\mathrm{10}\right)}{{sin}\left(\mathrm{20}\right)+\left(\mathrm{2}{sin}\left(\mathrm{100}\right){sin}\left(\mathrm{10}\right){cos}\left(\mathrm{20}\right)\right)}\right)=\mathrm{10} \\ $$

Commented by Subhi last updated on 16/Jun/23

Commented by lmcp1203 last updated on 16/Jun/23

$${thanks}. \\ $$

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