The problem asks us to find the value of $y$ in two right triangles. In the first triangle, the angle is $45^\circ$, and the sides are 1, 1, and $y$. In the second triangle, the angle is $60^\circ$, and the sides are 1, $y$, and 2.

1. Problem Description

The problem asks us to find the value of $y$ in two right triangles. In the first triangle, the angle is $45^\circ$, and the sides are 1, 1, and $y$. In the second triangle, the angle is $60^\circ$, and the sides are 1, $y$, and

2.

2. Solution Steps

a) For the first triangle, we can use the Pythagorean theorem, which states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. In this case, the hypotenuse is $y$, and the other two sides are 1 and

1. Therefore, we have:

$y^2 = 1^2 + 1^2$

$y^2 = 1 + 1$

$y^2 = 2$

Since $y$ must be positive, we take the positive square root of both sides:

$y = \sqrt{2}$

b) For the second triangle, we again use the Pythagorean theorem. Here, the hypotenuse is 2, and the other two sides are 1 and $y$. Therefore, we have:

$2^2 = 1^2 + y^2$

$4 = 1 + y^2$

$y^2 = 4 - 1$

$y^2 = 3$

Since $y$ must be positive, we take the positive square root of both sides:

$y = \sqrt{3}$

3. Final Answer

a) $y = \sqrt{2}$

b) $y = \sqrt{3}$