Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

Question Number 89580 by jagoll last updated on 18/Apr/20

Answered by $@ty@m123 last updated on 18/Apr/20

((BD)/(sin x))=((AD)/(sin {180−(x+84+42)})) ((BD)/(sin x))=((AD)/(sin (54−x))) ...(1) ((BD)/(sin 84))=((BC)/(sin {180−(84+42)})) ((BD)/(sin 84))=((AD)/(sin 54)) ...(2) From (1) &(2), ((sin (54−x))/(sin x))=((sin 54)/(sin 84)) ((sin (54−x)+sin x)/(sin (54−x)−sin x))=((sin 54+sin 84)/(sin 54−sin 84)) ((2sin 27cos (27−x))/(2cos 27sin (27−x)))=((2sin 69cos 15)/(2cos 69(−sin 15))) ((tan 27)/(tan (27−x)))=−((tan 69)/(tan 15)) tan (x−27)=((tan 27)/(tan 69))×tan 15 tan (x−27)≈tan 3^o x≈30^o

$$\frac{{BD}}{\mathrm{sin}\:{x}}=\frac{{AD}}{\mathrm{sin}\:\left\{\mathrm{180}−\left({x}+\mathrm{84}+\mathrm{42}\right)\right\}} \\ $$$$\frac{{BD}}{\mathrm{sin}\:{x}}=\frac{{AD}}{\mathrm{sin}\:\left(\mathrm{54}−{x}\right)}\:...\left(\mathrm{1}\right) \\ $$$$\frac{{BD}}{\mathrm{sin}\:\mathrm{84}}=\frac{{BC}}{\mathrm{sin}\:\left\{\mathrm{180}−\left(\mathrm{84}+\mathrm{42}\right)\right\}} \\ $$$$\frac{{BD}}{\mathrm{sin}\:\mathrm{84}}=\frac{{AD}}{\mathrm{sin}\:\mathrm{54}}\:...\left(\mathrm{2}\right) \\ $$$${From}\:\left(\mathrm{1}\right)\:\&\left(\mathrm{2}\right), \\ $$$$\frac{\mathrm{sin}\:\left(\mathrm{54}−{x}\right)}{\mathrm{sin}\:{x}}=\frac{\mathrm{sin}\:\mathrm{54}}{\mathrm{sin}\:\mathrm{84}} \\ $$$$\frac{\mathrm{sin}\:\left(\mathrm{54}−{x}\right)+\mathrm{sin}\:{x}}{\mathrm{sin}\:\left(\mathrm{54}−{x}\right)−\mathrm{sin}\:{x}}=\frac{\mathrm{sin}\:\mathrm{54}+\mathrm{sin}\:\mathrm{84}}{\mathrm{sin}\:\mathrm{54}−\mathrm{sin}\:\mathrm{84}} \\ $$$$\frac{\mathrm{2sin}\:\mathrm{27cos}\:\left(\mathrm{27}−{x}\right)}{\mathrm{2cos}\:\mathrm{27sin}\:\left(\mathrm{27}−{x}\right)}=\frac{\mathrm{2sin}\:\mathrm{69cos}\:\mathrm{15}}{\mathrm{2cos}\:\mathrm{69}\left(−\mathrm{sin}\:\mathrm{15}\right)} \\ $$$$\frac{\mathrm{tan}\:\mathrm{27}}{\mathrm{tan}\:\left(\mathrm{27}−{x}\right)}=−\frac{\mathrm{tan}\:\mathrm{69}}{\mathrm{tan}\:\mathrm{15}} \\ $$$$\mathrm{tan}\:\left({x}−\mathrm{27}\right)=\frac{\mathrm{tan}\:\mathrm{27}}{\mathrm{tan}\:\mathrm{69}}×\mathrm{tan}\:\mathrm{15} \\ $$$$\mathrm{tan}\:\left({x}−\mathrm{27}\right)\approx\mathrm{tan}\:\mathrm{3}^{\mathrm{o}} \\ $$$${x}\approx\mathrm{30}^{\mathrm{o}} \\ $$$$ \\ $$

Commented by jagoll last updated on 18/Apr/20

thank you sir

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com