Since here ABC is a isosceles right triangle, its height is half the size of the hypotenuse.

We just need to apply the Pythagorean Theorem to get the length of BC, and divide this length by two.

 $\displaystyle BC^{2}=AB^{2}+AC^{2}$, so $\displaystyle BC=\sqrt{50}$.

Therefore, $\displaystyle BC=5\sqrt{2}$ and the final answer is  $\displaystyle \frac{5\sqrt{2}}{2}$.

Let one of the angles measure $\displaystyle x$; then the other angle measures $\displaystyle 2x - 15$. The sum of the measures of the acute angles of a triangle is $\displaystyle 90^{\circ }$, so we can set up and solve this equation:

$\displaystyle x+ 2x - 15 = 90$

$\displaystyle 3x - 15 = 90$

$\displaystyle 3x =105$

$\displaystyle x = 35$

$\displaystyle 2x - 15 = 2 \cdot 30 - 15 = 55 ^{\circ }$

The acute angles measure $\displaystyle 35^{\circ },55 ^{\circ }$; since we want the smaller of the two, $\displaystyle 35^{\circ }$ is the correct choice.

By definition, one of the angles of a right triangle must be 90 degrees. We are given the measure of another angle in the problem, so we now know the measure of two angles in the triangle. Because the sum of the angles of any triangle is 180 degrees, we can then solve for the measure of the only remaining unknown angle:

$\displaystyle 90^{\circ}+36^{\circ}+x=180^{\circ}$

$\displaystyle x=180^{\circ}-90^{\circ}-36^{\circ}$

$\displaystyle x=54^{\circ}$

For any right triangle, whose sides are in ratio $\displaystyle n, \sqrt{3}n, 2n$, where $\displaystyle n$ is a constant, its angles must be $\displaystyle 60^{\circ}$ and $\displaystyle 30^{\circ}$. Here the triangle has its sides in that ratio with $\displaystyle n=\sqrt{3}$. Therefore, angle B must be the smallest angle, $\displaystyle 30^{\circ}$, since it is the angle between the two longest sides. This rule is really useful on the test, and it is advised to memorize it!

Triangle ABC is an isosceles right triangle. Therefore, its angles at the basis BC will always be $\displaystyle 45^{\circ}$.

This stems from the fact that the sums of the angles of a triangle are $\displaystyle 180^{\circ}$ and in our case with ABC a right and isosceles triangle, $\displaystyle 180-90=90^{\circ}$, therefore for the two remaining angles are equal.

There are 90 degrees left, therefore to find the measure of each angle we do the following,

 $\displaystyle \frac{90}{2}=45^{\circ}$. 

Here, we can tell the size of the angles by recognizing the length of the sides indicative of a right triangle with angles $\displaystyle 30^{\circ}-60^{\circ}-90^{\circ}$.

Indeed, the length of the sides are $\displaystyle \sqrt{3}$ and $\displaystyle 3$. Any triangle with sides $\displaystyle n,n\sqrt{3},2n$, where $\displaystyle n$ is a constant, will have angles $\displaystyle 30^{\circ}-60^{\circ}-90^{\circ}$.

In our case $\displaystyle n=\sqrt3$. Therefore, angle $\displaystyle \widehat{CBA}$ will be the smallest of the three possible angles, since it is between the two longest sides ( the hypotenuse and AB, which is longer than AC). Therefore the larger angle $\displaystyle \widehat{ACB}$ will be $\displaystyle 60^\circ$ thus arriving at our final answer.

Let $\displaystyle X$ be the measure of $\displaystyle \angle A$. The sum of the measures of the acute angles of a right triangle, $\displaystyle \angle A$ and $\displaystyle \angle C$, is $\displaystyle 90^{\circ }$, so

$\displaystyle m \angle A + m \angle C = 90^{\circ }$

$\displaystyle X + m \angle C = 90^{\circ }$

$\displaystyle X + m \angle C -X = 90^{\circ } -X$

$\displaystyle m \angle C = 90^{\circ } -X$

Assume Statement 1 alone. This can be rewritten:

$\displaystyle m \angle A - m \angle C = 10^{\circ }$

$\displaystyle X - (90^{\circ } -X) = 10^{\circ }$

$\displaystyle 2 X - 90^{\circ }= 10^{\circ }$

$\displaystyle 2 X - 90^{\circ }+ 90^{\circ } = 10^{\circ } + 90^{\circ }$

$\displaystyle 2 X = 10 0^{\circ }$

$\displaystyle 2 X \div 2 = 10 0^{\circ } \div 2$

$\displaystyle X = 50^{\circ }$

Assume Statement 2 alone. This can be rewritten:

$\displaystyle 4 \cdot m \angle A = 5 \cdot m \angle C$

$\displaystyle 4 X = 5 (90^{\circ }-X)$

$\displaystyle 4 X = 450 ^{\circ }-5X$

$\displaystyle 4 X+ 5X = 450 ^{\circ }-5X + 5X$

$\displaystyle 9X = 450 ^{\circ }$

$\displaystyle 9X\div 9 = 450 ^{\circ } \div 9$

$\displaystyle X = 50^{\circ }$

An acute angle must have measure less than $\displaystyle 90^{\circ }$.

Assume Statement 1 alone. We are looking for a whole number that is a multiple of both 2 and 7, and is less than 90. There are several such numbers - 14 and 28, for example. There is no way of eliminating any of them, so the question is left unanswered.

Statement 2 alone provides insufficient information for similar reasons, since there are several whole numbers less than 90 that are multiples of 3 and 4 - 12 and 24, for example - with no way of eliminating any of them.

Now assume both statements to be true. 3, 4, and 7 are relatively prime - the greatest common factor of the four is 1 - so in order to find the least common multiple of the four, we need to multiply them. This product is

$\displaystyle 3 \cdot 4 \cdot 7 = 84$,

which is also a multiple of 2. Any other multiple of all four numbers must be a  multiple of 84, but any other positive multiple of 84 is greater than 90. Therefore, from the two statements together, it can be deduced that $\displaystyle \angle A$ has measure $\displaystyle 84^{\circ }$.

The measures of the acute angles of a right triangle have sum $\displaystyle 90^{\circ }$, so

$\displaystyle x+y = 90$

Along with $\displaystyle x - y = 15$, a system of linear equations can be formed and solved as follows:

$\displaystyle x - y = 15$

$\displaystyle \underline{x+y = 90}$

$\displaystyle 2x \; \; \; = 105$

$\displaystyle 2x \div 2 = 105 \div 2$

$\displaystyle x = 52 \frac{1}{2}$

An equilateral triangle has three equal sides; therefore, to find the length of each side, divide the perimeter by $\displaystyle 3$:

$\displaystyle \frac{18}{3}= 6$