Question-193522

Question Number 193522 by Mingma last updated on 16/Jun/23

Answered by Subhi last updated on 16/Jun/23

((x+z)/(sin(100)))=(x/(sin(40))) sin(100)x=sin(40)x+sin(40)z ⇛ x=((sin(40)z)/(sin(10)−sin(40))) (i) (z/(sin(y−40)))=((x+z)/(sin(u))) u+y−40=180−140=40 ⇛ y=80−u (ii) (z/(sin(40−u)))=((x+z)/(sin(u))) sin(40−u)x+sin(40−u)z=zsin(u) sin(40−u).((sin(40)z)/(sin(100)−sin(40)))+sin(40−u)z=zsin(u) (((sin^2 (40))/(sin(100)−sin(40)))+sin(40))cos(u)=(1+cos(40)+((cos(40).sin(40))/(sin(100)−sin(40))))sin(u) u=tan^(−1) ((((sin^2 (40))/(sin(100)−sin(40)))+sin(40)))/(1+cos(40)+(((sin(80))/2)/(sin(100)−sin(40))))=tan^(−1) ((1/( (√3))))=30

$$\frac{{x}+{z}}{{sin}\left(\mathrm{100}\right)}=\frac{{x}}{{sin}\left(\mathrm{40}\right)} \\ $$$${sin}\left(\mathrm{100}\right){x}={sin}\left(\mathrm{40}\right){x}+{sin}\left(\mathrm{40}\right){z}\:\Rrightarrow\:{x}=\frac{{sin}\left(\mathrm{40}\right){z}}{{sin}\left(\mathrm{10}\right)−{sin}\left(\mathrm{40}\right)}\:\left({i}\right) \\ $$$$\frac{{z}}{{sin}\left({y}−\mathrm{40}\right)}=\frac{{x}+{z}}{{sin}\left({u}\right)} \\ $$$${u}+{y}−\mathrm{40}=\mathrm{180}−\mathrm{140}=\mathrm{40}\:\Rrightarrow\:{y}=\mathrm{80}−{u}\:\left({ii}\right) \\ $$$$\frac{{z}}{{sin}\left(\mathrm{40}−{u}\right)}=\frac{{x}+{z}}{{sin}\left({u}\right)} \\ $$$${sin}\left(\mathrm{40}−{u}\right){x}+{sin}\left(\mathrm{40}−{u}\right){z}={zsin}\left({u}\right) \\ $$$${sin}\left(\mathrm{40}−{u}\right).\frac{{sin}\left(\mathrm{40}\right){z}}{{sin}\left(\mathrm{100}\right)−{sin}\left(\mathrm{40}\right)}+{sin}\left(\mathrm{40}−{u}\right){z}={zsin}\left({u}\right) \\ $$$$\left(\frac{{sin}^{\mathrm{2}} \left(\mathrm{40}\right)}{{sin}\left(\mathrm{100}\right)−{sin}\left(\mathrm{40}\right)}+{sin}\left(\mathrm{40}\right)\right){cos}\left({u}\right)=\left(\mathrm{1}+{cos}\left(\mathrm{40}\right)+\frac{{cos}\left(\mathrm{40}\right).{sin}\left(\mathrm{40}\right)}{{sin}\left(\mathrm{100}\right)−{sin}\left(\mathrm{40}\right)}\right){sin}\left({u}\right) \\ $$$${u}={tan}^{−\mathrm{1}} \left(\left(\frac{{sin}^{\mathrm{2}} \left(\mathrm{40}\right)}{{sin}\left(\mathrm{100}\right)−{sin}\left(\mathrm{40}\right)}+{sin}\left(\mathrm{40}\right)\right)\right)/\left(\mathrm{1}+{cos}\left(\mathrm{40}\right)+\frac{\frac{{sin}\left(\mathrm{80}\right)}{\mathrm{2}}}{{sin}\left(\mathrm{100}\right)−{sin}\left(\mathrm{40}\right)}\right)={tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)=\mathrm{30} \\ $$

Commented by Subhi last updated on 16/Jun/23

Commented by Mingma last updated on 16/Jun/23

Perfect ��

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