An isosceles triangle has sides of 17 centimeter and 20 centimeter and 20 centimeter. Find the magnitude of the angles trigonometry

Question

An isosceles triangle has sides of 17 centimeter and 20 centimeter and 20 centimeter. Find the magnitude of the angles

Answer 1

Use cosine rule to find out the angle between two same sides. Rest two angles would be equal. Use the property that sum of all angles is $180$ degrees.

Answer 2

Suggestion:

An isosceles triangle can be broken into two right triangles which share a common side (the altitude of the triangle which is the segment bisecting the angle in between the sides of equal length, and perpendicular to the third side (and bisecting that third side).

Solving for altitude $a$ (shared side of two adjacent right triangles): $$\left(\frac{17}2\right)^2 + a^2 = 20^2$$

Then you can use the Pythagorean Theorem to find the length of that altitude, and from there, use trigonometric identities of right triangles to determine the needed angles.

We can simply determine the measure of the angles $\alpha$ formed by each of the lengths of equal measure and the third side: $$\sin(\alpha) = \dfrac{a}{20}$.

Once you solve for $\alpha$, we know that $2\alpha + \beta = 180$, where $\beta$ is the measure of the angle between the sides of equal length, so you can use $\alpha$ to solve for $\beta$.