Question

In an obtuse-angled triangle, the obtuse angle is $\frac{3\pi }{4}$ and the other two angles are equal to two values of $\theta$ satisfying $a\tan \theta +b\sec \theta =c$, where $\left|b\right|\le \sqrt{a^2+c^2}$, then $a^2-c^2=kac$, where $k=$

Answer 1

We have,

$a\tan \theta +b\sec \theta =c\Rightarrow b^2{\sec }^2\theta ={(c-a\tan \theta )}^2$

$\Rightarrow b^2(1+{\tan }^2\theta )=c^2-2ca\tan \theta +a^2{\tan }^2\theta$

$\Rightarrow (a^2-b^2){\tan }^2\theta -2ac\tan \theta +c^2-b^2=0$ ...(1)

Roots of equation (1) are $\tan \alpha$ and $\tan \beta ,$ where $\alpha$and $\beta$ are the two angles of the triangle.

We have, $\tan \alpha +\tan \beta =\frac{2ca}{a^2-b^2}$

and, $\tan \alpha .\tan \beta =\frac{c^2-b^2}{a^2-b^2}$

$\therefore \tan (\alpha +\beta )=\frac{\frac{2ca}{a^2-b^2}}{1-\frac{c^2-b^2}{a^2-b^2}}=\frac{2ca}{a^2-c^2}$

$\therefore \tan (\pi -\frac{3\pi }{4})=\frac{2ca}{a^2-c^2}\Rightarrow a^2-c^2=2ca.$

$\therefore k=2.$