Animated Solution for Mathematics - Trigonometry: The sides of a triangle are $3x+4y$, $4x+3y$ and $5x+5y$ where $x,y>0$ then the triangle is

Imagine you are standing in a field, holding three sticks of lengths $a=3x+4y$, $b=4x+3y$, and $c=5x+5y$. You are tasked with determining the nature of the triangle formed by these sticks.

Since $x$ and $y$ are strictly positive, we can immediately identify the longest stick. By comparing the coefficients, we see that $5x+5y$ is strictly greater than $3x+4y$ and $4x+3y$.

This is crucial because the angle opposite the longest side, let's call it $C$, is the only one that could possibly be obtuse. If $C$ is obtuse, the triangle is obtuse; if $C$ is $90^{\circ }$, it is right-angled; if $C$ is acute, the triangle is acute.

The Cosine Rule

To connect these side lengths to the angle $C$, we invoke the powerful Cosine Rule:

This formula is our bridge. It does not matter what $x$ and $y$ are; as long as they are positive, the sign of $\cos C$ will reveal the truth about our triangle.

A Beautiful Collapse

Now, let us perform the substitution. We need to calculate the numerator: $a^2+b^2-c^2$.

First, expand the squares:

Adding these gives:

Now, for the third side:

When we subtract $c^2$ from the sum of $a^2+b^2$, notice the magic:

The complex quadratic terms vanish, leaving us with a simple, elegant $-2xy$.

The Final Verdict

Our expression for the cosine of $C$ is now:

Since $x,y>0$, the numerator $-2xy$ is strictly negative, and the denominator is strictly positive. Therefore, $\cos C<0$.

In the world of trigonometry, a negative cosine value for an angle in a triangle means that the angle $C$ must be greater than $90^{\circ }$ and less than $180^{\circ }$.

Thus, angle $C$ is obtuse, and our triangle is definitively an obtuse-angled triangle. You have successfully navigated the algebra to uncover the geometric reality!