Animated Solution for Mathematics - Trigonometry: In a $△PQR$, if $3\sin P+4\cos Q=6$ and $4\sin Q+3\cos P=1$, then the angle $R$ is equal to

Welcome, future engineer. Today, we are going to dive into a problem that, at first glance, looks like a tangled mess of trigonometric variables.

We are given a triangle $PQR$ and two equations:

Our mission is to find the angle $R$. When you see a problem like this, do not panic. The key is to look for the hidden symmetry.

Notice how the coefficients $3$ and $4$ are swapped between the sine and cosine terms. This is a massive hint that the problem is designed to be solved by squaring and adding the equations to invoke the Pythagorean identity, ${\sin }^2\theta +{\cos }^2\theta =1$.

Let us perform the squaring. When we square the first equation, $(3\sin P+4\cos Q{)}^2=6^2$, we expand it to get:

Similarly, squaring the second equation, $(4\sin Q+3\cos P{)}^2=1^2$, gives us:

Now, watch what happens when we add these two results together. We group the terms with the same coefficients:

This is the moment of elegance. The terms in the parentheses become $1$, and the cross-terms form the expansion of the compound angle formula, $\sin (P+Q)=\sin P\cos Q+\cos P\sin Q$.

Our equation simplifies to:

This further reduces to:

Solving for $\sin (P+Q)$ is now straightforward:

Since we are in a triangle, $P+Q+R=\pi$, which means $P+Q=\pi -R$. Therefore, $\sin (P+Q)=\sin (\pi -R)=\sin R$.

We have found that $\sin R=\frac{1}{2}$. This gives us two candidates: $R=\frac{\pi }{6}$ or $R=\frac{5\pi }{6}$.

But as a true JEE aspirant, you know that we must check for validity. If $R=\frac{5\pi }{6}$, then $P+Q=\frac{\pi }{6}$.

If you test these values in the original equations, you will find they cannot sum to $6$. Thus, $R=\frac{5\pi }{6}$ is an extraneous solution.

The only valid answer is $R=\frac{\pi }{6}$. Keep practicing this kind of rigorous verification; it is what separates the good from the great.