Find the measure of the largest angle of a triangle formed by the reciprocals of the altitudes of a given triangle For reference:

The measures of the sides of a triangle
are $3$, $5$ and $7$. Calculate the measure of the largest angle
of a triangle whose sides are the inverses of the heights of the first triangle (Answer:$120^\circ$)

*I checked in geogebra and the answer is correct.
My progress: I tried but I don't know if that's the way to go.
$p = \frac{3+5+7}{2}=\frac{15}{2}\\p(p-a)p(p-b)(p-c)=\frac{15}{2}(\frac{9.}{2}\frac{5}{2}.\frac{1}{2}) = \frac{675}{16}$
Heron's Formula: $
h_b=\frac{2}{7}\sqrt{\frac{675}{16}}=\frac{2.15\sqrt3} {7.4}=\frac{15\sqrt3}{14}$
Similarly:
$h_a = \frac{2}{5}\cdot\frac{15\sqrt3}{4} = \frac{3\sqrt3}{2}\\h_c = \frac{2}{3}\cdot\frac{15\sqrt3}{4}=\frac{5\sqrt3}{2} \\
\frac{1}{h_a}=\frac{2}{3\sqrt3}\implies (\frac{1}{h_c})^2 = \frac{4}{27}\\
 \frac{1}{h_b} = \frac{14}{15\sqrt3}\implies (\frac{1}{h_b})^2=\frac{196}{675}\\
\frac{1}{h_c} = \frac{2}{5\sqrt3} \implies (\frac{1}{h_c})^2=\frac{4}{75}$
By Law of Cosines:
$\frac{196}{675} = \frac{4}{27}+\frac{4}{75} - 2\cdot\frac{2}{3\sqrt3}\frac{2}{5\sqrt3}\cdot\cos B\implies\\
\cos B=-0.5\\ \therefore B = 120^\circ$

Elaborating @Stinking Bishop's hint, consider triangle $ABC$ with sides $a$, $b$ and $c$ (in standard notation). Let altitudes from each vertex to the opposite side be $h_a$, $h_b$ and $h_c$. Now, if the area of the triangle is $A$, we get, $$\frac1{h_a}=\frac{a}{2A}$$ $$\frac1{h_b}=\frac{b}{2A}$$ $$\frac1{h_c}=\frac{c}{2A}$$
It is clear that the triangle formed by the inverses of the heights is similar to the original triangle with similarity ratio $\frac1{2A}$.
Now we can find the required angle considering the above fact.

The angle opposite to the longest side is the largest angle. Now using law of cosines, $$\cos\theta=\frac{3^2+5^2-7^2}{2\cdot3\cdot5}=-\frac12$$ $$\implies\theta=120^\circ$$

