Question

In a $△ABC,AB=AC,P$ and $Q$ are points on $AC$ and $AB$ respectively such that $CB=BP=PQ=QA.$ If $\angle AQP=\theta ,$then ${\tan }^2\theta$ is a root of the equation

• A. $y^3+21y^2-35y-12=0$
• B. $y^3-21y^2+35y-12=0$
• C. $y^3-21y^2+35y-7=0$
• D. $12y^3-35y^2+35y-12=0$

Answer 1

The correct option is C $y^3-21y^2+35y-7=0$


Given, $\angle AQP=\theta$
$\angle QAP=\angle QPA=90-\frac{\theta }{2}\angle PQB=\angle PBQ=180-\theta$
In isosceles
$△ABC:\angle BCA=\angle ABC=\frac{1}{2}[{180}^{\circ }-\angle A]$
$=45+\frac{\theta }{4}$
Also in isosceles $△PBQ:\angle PQB={180}^{\circ }-\theta$
and $\angle BPQ={180}^{\circ }-2\angle PQB=(2\theta -{180}^{\circ })$
Now sum of all angles at point $P$ on line $CA={180}^{\circ }$,
$(90-\frac{\theta }{2})+(2\theta -180)+(45+\frac{\theta }{4})=180$
$\Rightarrow \theta =\frac{5\pi }{7}\Rightarrow 4\theta =5\pi -3\theta \Rightarrow \tan 4\theta =-\tan 3\theta \Rightarrow \frac{2\tan 2\theta }{1-{\tan }^{2}2\theta }=-\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }\Rightarrow \frac{2\cdot \frac{2t}{1-t^2}}{1-{\left(\frac{2t}{1-t^2}\right)}^2}=-\frac{3t-t^3}{1-3t^2}$
where $t=\tan \theta ,$ on simplifying
we get, $t^6-21t^4+35t^2-7=0$
So, ${\tan }^2\theta$ is the root of the equation $y^3-21y^2+35y-7=0$