Construct the triangle yourself!

From point $P$ drop perpendiculars to the three sides as shown. Name the points of intersection with the sides as $D$ , $E$ and $F$ as shown. Let $BE$ be $x$ , $FC$ be $y$ and $AD$ be $z$ . Considering the triangles with angle $\theta$ , we have

$tan~\theta=\frac{PD}{z}$ $\color{#D61F06}\implies$ $PD = z~tan\theta$

$tan~\theta=\frac{FP}{y}$ $\color{#D61F06}\implies$ $FP = y~tan\theta$

$tan~\theta=\frac{EP}{x}$ $\color{#D61F06}\implies$ $EP = x~tan\theta$

The area of triangle $ABC$ can be solve in two ways

$(1.)$

$A_{ABC}=A_{APB}+A_{APC}+A_{BPC}=\frac{1}{2}(41)(z~tan\theta)+\frac{1}{2}(89)(y~tan\theta)+\frac{1}{2}(50)(x~tan\theta)=\frac{1}{2}tan\theta(41z+89y+50x$ )

$(2.)$ By Heron's Formula

$s = \frac{41+50+89}{2} = 90$

$A_{ABC} = \sqrt{90(90-41)(90-50)(90-89)} = 420$

$\text{By Pythagorean Theorem, we have}$

$x^2 + x^2tan^2\theta = z^2tan^2\theta + (41-z)^2$ $\color{#3D99F6}(1)$

$y^2 + y^2tan^2\theta = x^2tan^2\theta + (50-x)^2$ $\color{#3D99F6}(2)$

$z^2 + z^2tan^2\theta = y^2tan^2\theta + (89-y)^2$ $\color{#3D99F6}(3)$

$\text{Adding the three equations, we get}$

$41z + 50x + 89y = 6051$ $\color{#3D99F6}(4)$

$\text{Going back to the area of triangle}$ $ABC$ , $\text{substitute}$ $\color{#3D99F6}(4)$ $\text{and}$ $A_{ABC}=420$

$A_{ABC}=\frac{1}{2}tan\theta(41z+89y+50x$ ) $\color{#D61F06}\implies$ $420 = \frac{1}{2}tan\theta(6051)$ $\color{#D61F06}\implies$ $\boxed{tan\theta = \large\dfrac{280}{2017}}$

$\huge\therefore$ $\boxed{\huge\color{#20A900}k = 280}$

In general, in any $\Delta ABC$ , the angle $\theta$ as per given condition, is given by general formula

$\boxed{\tan\theta=\frac{\sin A\sin B\sin C}{1+\cos A\cos B\cos C}}$

As per given question, $a=50, b=89, c=41$

$\cos A=\frac{b^2+c^2-a^2}{2bc}=\frac{3551}{3649}, \sin A=\sqrt{1-\cos^2A}=\frac{840}{3649}$

$\cos B=-\frac{187}{205}, \sin B=\sqrt{1-\cos^2B}=\frac{84}{205}$

$\cos C=\frac{437}{445}, \sin C=\sqrt{1-\cos^2C}=\frac{84}{445}$

setting all the values in above formula & simplifying, we get

$\tan\theta=\frac{\frac{840}{3649}\cdot\frac{84}{205}\cdot\frac{84}{445} }{1+\frac{3551}{3649}\cdot\left(-\frac{187}{205}\right)\cdot\frac{437}{445}}$

$\tan\theta=\frac{280}{2017}$