As we have previously seen, the height of a right triangle divides a it into two similar triangles with sides of same proportion.

Therefore, we can set up the following equality:  or .

By plugging in the numbers, we get,  or .

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.

 , so .

Therefore,  and the final answer is  .

Let one of the angles measure ; then the other angle measures . The sum of the measures of the acute angles of a triangle is , so we can set up and solve this equation:

The acute angles measure ; since we want the smaller of the two,  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:

For any right triangle, whose sides are in ratio , where  is a constant, its angles must be  and . Here the triangle has its sides in that ratio with . Therefore, angle B must be the smallest angle, , 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 .

This stems from the fact that the sums of the angles of a triangle are  and in our case with ABC a right and isosceles triangle, , 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,

 . 

Here, we can tell the size of the angles by recognizing the length of the sides indicative of a right triangle with angles .

Indeed, the length of the sides are  and . Any triangle with sides , where  is a constant, will have angles .

In our case . Therefore, angle  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  will be  thus arriving at our final answer.

Let  be the measure of . The sum of the measures of the acute angles of a right triangle,  and , is , so

Assume Statement 1 alone. This can be rewritten:

Assume Statement 2 alone. This can be rewritten:

An acute angle must have measure less than .

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

,

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  has measure .

The measures of the acute angles of a right triangle have sum , so

Along with , a system of linear equations can be formed and solved as follows: