Animated Solution for Mathematics - Trigonometry: If the lengths of the sides of triangle are 3, 5, 7 then the largest angle of the triangle is

Welcome, future engineer! Today, we are not just solving a problem; we are exploring the rigid, beautiful constraints of geometry. Imagine you are holding three rods of lengths $3$, $5$, and $7$ units.

You are tasked with pinning them together to form a triangle. Have you ever wondered how the lengths of these rods dictate the 'spread' of the corners? This is the essence of the Law of Cosines.

Before we touch a single equation, let us visualize the triangle. We have sides $a=3$, $b=5$, and $c=7$.

In the world of triangles, there is a beautiful, proportional relationship: the largest angle is always 'looking' at the longest side. Think of it as a lever; the longer the side, the wider the opening it creates at the opposite vertex.

Since $7>5>3$, our largest angle must be the one opposite the side of length $7$. Let us call this angle $C$. Our mission is clear: find $C$.

Our Mathematical Bridge

When we are trapped with only side lengths and no angles, the Law of Cosines is our best friend. It is the generalized version of the Pythagorean theorem. While Pythagoras only works for right-angled triangles, the Law of Cosines works for every triangle.

The formula is given by:

This equation is a masterpiece of balance. It tells us that the cosine of an angle is determined entirely by the squares of the sides.

If $a^2+b^2$ is exactly equal to $c^2$, the cosine is zero, and we have a $90^{\circ }$ angle. If $c^2$ is larger, the cosine becomes negative, pushing the angle into the obtuse territory. Let us see what happens when we plug in our values.

A Moment of Truth

Substituting our values $a=3$, $b=5$, and $c=7$, we get:

Let us break this down step-by-step. Squaring the sides gives us $9$, $25$, and $49$. Now, look at the numerator: $9+25-49$.

Adding the first two gives us $34$. Subtracting $49$ from $34$ leaves us with $-15$.

Do not be alarmed by this negative sign! In the context of our triangle, this is a profound revelation. It confirms that our triangle is obtuse.

The denominator is simply $2\times 3\times 5=30$. Thus, we arrive at:

We are now at the finish line. We need to find the angle $C$ such that $\cos C=-1/2$.

We know from our trigonometric unit circle that $\cos (60^{\circ })=1/2$. Since we need a negative value, we look to the second quadrant, where $180^{\circ }-60^{\circ }=120^{\circ }$.

To express this in the language of radians, we multiply by $\frac{\pi }{180^{\circ }}$:

And there it is! The largest angle of our triangle is $\frac{2\pi }{3}$. You have successfully navigated the relationship between side lengths and angular geometry.