1.

Let ABCD be a quadrilateral with /_CBD=2/_ADB, /_ABD=2/_CDB, AB=BC, then

Answer» <html><body><p>`AD=CD`<br/>`/_ADB=/_CDB`<br/>`/_CBD=/_ABD`<br/>`/_ADC` is`(2pi)/<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>`</p>Solution :`2x=/_CBD` <br/> `2y=/_ABD` <br/> In `/_\CBD` <br/> `(sin(pi-(2y+x)))/(sinx)=(<a href="https://interviewquestions.tuteehub.com/tag/bd-389836" style="font-weight:bold;" target="_blank" title="Click to know more about BD">BD</a>)/(<a href="https://interviewquestions.tuteehub.com/tag/ba-389206" style="font-weight:bold;" target="_blank" title="Click to know more about BA">BA</a>)=(BD)/(BC)=(sin(pi-(2x+y)))/(<a href="https://interviewquestions.tuteehub.com/tag/siny-3034871" style="font-weight:bold;" target="_blank" title="Click to know more about SINY">SINY</a>)` <br/> `impliessin(2+x)siny=sin(2x+y)sinx` <br/> `=1/2[cos(y+x)-cos(3y+x)]=1/2[cos(x+y)-cos(3x+y)]` <br/> `0ltx+y=1/2ABClt(pi)/2` <br/> `<a href="https://interviewquestions.tuteehub.com/tag/0lt-1749185" style="font-weight:bold;" target="_blank" title="Click to know more about 0LT">0LT</a>(3y+x)+(3x+y)lt2pi` <br/> `=:. 3y+x=3x+yimpliesx=y` <br/> `implies/_ABD=/_CBDimpliesAD=CD` <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/FIT_JEE_FT3_P2_E01_366_S01.png" width="80%"/></body></html>


Discussion

No Comment Found

Related InterviewSolutions