MHB Can This Triangle Configuration Prove a Right Angle?

[anemone]

Here is this week's POTW:

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Given a triangle $ABC$ such that angles $A,\,B$ and $C$ satisfy $\dfrac{\cos A}{20}+\dfrac{\cos B}{21}+\dfrac{\cos C}{29}=\dfrac{29}{420}$. Prove that $ABC$ is a right-angled triangle.

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[anemone]

No one answered POTW #494.

Spoiler

However, you can refer to the solution of other, as shown below:

Let $t=\arcsin \dfrac{21}{29}$ so that we have $\sin t=\dfrac{21}{29}$ and $\cos t=\dfrac{20}{29}$.

Writing $C=\pi-A-B$, the given equation becomes $\dfrac{\cos A}{\cos t}+\dfrac{\cos B}{\sin t}-\cos(A+B)=\dfrac{1}{\sin t\cos t}$, or can be simplified to

$\left(\dfrac{1}{\sin t}-\cos A\right)\left(\dfrac{1}{\cos t}-\cos B\right)=\sin A \sin B$.

Since $\sin A\sin B\ne 0$ so

$\dfrac{1}{\sin t}-\cos A=\alpha \sin A$ and $\dfrac{1}{\cos t}-\cos B=\dfrac{1}{\alpha}\sin B$ for some $\alpha >0$.

Writing $u=\arcsin \dfrac{1}{\sqrt{\alpha^2+1}}$, we get

$\dfrac{\sin u}{\sin t}=\sin (A+u)$ and $\dfrac{\cos u}{\cos t}=\cos B-u)$

$0<\sin u \le \sin t$ and $0<\cos u \le \cos t$ which implies $u=t$ and so $(A,\,B,\,C)=\left(\dfrac{\pi}{2}-t,\,t,\,\dfrac{\pi}{2}\right)$.