We are asked to find the value of $y$ in two right triangles. a) The right triangle has a leg of length 1, another leg of length 1, and a hypotenuse of length $y$. One angle is $45^\circ$. b) The right triangle has a leg of length 1, another leg of length $y$, and a hypotenuse of length 2. One angle is $60^\circ$.

1. Problem Description

We are asked to find the value of $y$ in two right triangles.

a) The right triangle has a leg of length 1, another leg of length 1, and a hypotenuse of length $y$. One angle is $45^\circ$.

b) The right triangle has a leg of length 1, another leg of length $y$, and a hypotenuse of length

2. One angle is $60^\circ$.

2. Solution Steps

a) We can use the Pythagorean theorem: $a^2 + b^2 = c^2$, where $a$ and $b$ are the legs and $c$ is the hypotenuse. In this case, $1^2 + 1^2 = y^2$, so $1 + 1 = y^2$, which means $y^2 = 2$. Taking the square root of both sides, we get $y = \sqrt{2}$.

Alternatively, since we have a $45^\circ$-$45^\circ$-$90^\circ$ triangle, the ratio of the sides is $1:1:\sqrt{2}$. Since the legs have length 1, the hypotenuse has length $\sqrt{2}$. So $y = \sqrt{2}$.

b) We can use the cosine function. We have $\cos(60^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{1}{2}$.

Therefore, $\cos(60^\circ) = \frac{1}{2}$. Since we are given the hypotenuse is 2 and the adjacent side to the $60^\circ$ angle is 1, we can use the Pythagorean theorem to find $y$:

$1^2 + y^2 = 2^2$, so $1 + y^2 = 4$. Subtracting 1 from both sides, we get $y^2 = 3$. Taking the square root of both sides, we get $y = \sqrt{3}$.

Alternatively, we can use the sine function. We have $\sin(60^\circ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{y}{2}$.

Since $\sin(60^\circ) = \frac{\sqrt{3}}{2}$, we have $\frac{y}{2} = \frac{\sqrt{3}}{2}$. Multiplying both sides by 2, we get $y = \sqrt{3}$.

3. Final Answer

a) $y = \sqrt{2}$

b) $y = \sqrt{3}$