Right-angled isosceles triangles If a right-angled triangle is isosceles then the other two angles must be equal to $45^\circ$ ?
Is this always the case or are there other possible right-angled isosceles triangles? 
 A: From this, we know in isosceles triangles, the angles at the base are equal to each other. 
If each is $\theta,2\theta<180^\circ$ as the other angle $>0$
$\implies \theta<90^\circ$
So, the other unequal angle $=90^\circ$ and we have $90^\circ+2\theta=180^\circ$
A: In a right angle triangle two sides which are equal will base and perpendicular because hypotaneous is always greatest side in right angle triangle.
so one angle is $90^\circ$ and sum of other two is also $90^\circ$ since sides are equal so there angle made with other side is also same so each will be $45^\circ$.
A: Here's our isosceles right triangle.

Because it's a right triangle, we know Pythagorean theorem holds true, and so:
$$c^2 = a^2 + b^2$$
But, this is an isosceles triangle, and so by definition, $a = b$, and so:
$$c^2 = a^2 + a^2$$
$$c^2 = 2a^2$$
$$a = \frac{c}{\sqrt 2}$$
Moreover, we know that $\sin A = \frac{a}{c}$, and so:
$$\sin A = \frac{a}{c}$$
$$\sin A = \frac{\frac{c}{\sqrt 2}}{c}$$
$$\sin A = \frac{c}{c\sqrt 2}$$
$$\sin A = \frac{1}{\sqrt 2}$$
$$A = \sin^{-1}(\frac{1}{\sqrt 2})$$
$$A = 45^o$$
(Note that $A$ is technically an infinite amount of answers because of the nature of $\sin^{-1}$, but because in this case it cannot be negative, and the angles in a triangle cannot add up to more than $180^o$, this is the only possibility).
We can then find $B$, as we know all angles in a triangle in Euclidean space add up to $180^o$, and we find that $B = 45^o$ too.
Thus, all right angle isosceles triangles have two angles of $45^o$.
