2021 Fall AMC 12B Problem 8

Let the congruent sides have length $s,$ the base be $b,$ and the height to the base be $h.$ The given condition is $s^2 = 2bh.$ The area equals $\tfrac12 bh$ and also $\tfrac12 s^2 \sin\theta,$ where $\theta$ is the vertex angle. So $bh = s^2 \sin\theta.$ Substituting $s^2 = 2bh$ gives $bh = 2bh\sin\theta,$ so $\sin\theta = \tfrac12.$ Since the triangle is obtuse, $\theta = 150^\circ.$ Thus, the correct answer is D.