Calculate the angles of a isosceles triangle

Question

In the triangle below, is there a way to calculate the $x$ and $y$?

To be more specific, $b = 12.8\rm\,cm\ $ and $h = 10\rm\,cm$, hence $a = 11.87\rm\,cm$.

I don't know what to do from here.

Answer 1

If $h$ bisects angle $x^\circ$ (hope it does!), then we can apply right-angle trigonometry: \begin{align} \sin y^\circ&= \frac{\text{opposite}}{\text{hypotenuse}}=\frac{h}{a}=\frac{10}{11.87}=0.842 \\ y^\circ&=\sin^{-1}(0.842) \\ y^\circ &\approx \boxed{57.4^\circ} \end{align} As the sum of all angles in a triangle is $180^\circ$, \begin{align} x^\circ+y^\circ+y^\circ&=180^\circ \\ x^\circ+(57.4)^\circ+(57.4)^\circ&=180^\circ \\ x^\circ&\approx\boxed{65.2^\circ} \end{align}

Answer 2

Note that the upper half of the diagram is a right triangle, so $\tan \frac x2=\frac b{2a}$

Answer 3

Well, heres a few hints:

*The height of a isoceles triangle to its base is also the median of its base.

*The tangent of an angle of a right angled triangle is the ratio of the length of the cathetus opposite the angle to the cathetus adjacent to the angle.

*the arctangent function returns an angle given its tangent.

Answer 4

Law of cosines will work to find the angles:

$c^2 = a^2 + b^2 - 2abcos(C)$, where C is the angle opposite to side c

So $b^2 = 2a^2 - 2a^2cos(X)$ and $a^2 = a^2 + b^2 - 2abcos(Y)$