# How do I find the base angles without a vertex angle in a isosceles triangle?

How can I find the base angles in a isosceles triangle if the vertex angle is missing? Normally, I would go: 2x + vertex angle = 180, but now even the vertex is missing, the only thing I have is a line in the middle, with 90 degree angle.

Maybe the 90 degree should be used to find one of the base angles and one of the vertex angles, any help is appreciated, thanks a lot, I am still learning this, any tips are very useful. All you need to do is find one base angle, since in an isosceles triangle, both those angles are opposite equal sides, and are therefore equal.

Edit: Now that I see your image, You have enough information to find everything.

• You can use the Pythagorean Theorem to obtain the altitude (the unknown side in the middle.)
• The vertex, B, is split into two equal angles, but you can easily find the vertex angle by first finding one of the base angles.
• You can find the base angle B by using the fact that $$\cos B = \dfrac{\text{Adjacent}}{\text{Hypotenuse}}\implies B = \cos^{-1}\left(\dfrac{\text{Adjacent}}{\text{Hypotenuse}}\right) = \cos^{-1}\left(\frac{10.9}{25.6}\right) = 64.8$$

Each base angle, let's call them each of measure $x$, plus the vertex angle add up to $180$.

So if you need to solve for the vertex angle $\theta$, we have that $\theta = 180 - 2x$.

Also note, to find an angle, given none of the angles are known, you can to use the lengths of the sides of the triangle to solve for the base angles and/or vertex, and can use the Law of Cosines to do that.

• Yes, I know that, but the Q is how, I do not have the vertex angle either, I only have a 90 degree angle, I can subtract 90 degree from 180, but that would still not allow me to find the one of the base angles or the vertex angle. Oct 27 '13 at 18:52
• Yes, John, you can use the Pythagorean Theorem to obtain the altitude (the unknown side in the middle.) The vertex, B, is split into two equal angles. But you can find the base angle B by using the fact that $$\cos B = \dfrac{\text{Adjacent}}{\text{Hypotenuse}}\implies B = \cos^{-1}\left(\dfrac{\text{Adjacent}}{\text{Hypotenuse}}\right)$$ Oct 27 '13 at 19:53
• Exactly! That's correct! Oct 28 '13 at 19:05
• No, the base angles are both $64.8$, the vertex of the largest triangle is $50.4$, and half of that at the top is the angle $25.2.$ The line in the middle of the largest triangle divided the big triangle exactly in half. Oct 28 '13 at 19:09
• Not currently, but I have taught both at a community college and at a university. Oct 28 '13 at 19:13