Question

37.
38.
39.
40.
41.
42.
43.
x =
y =
z =
44.
x =
y =
45.
x =
y =

Explanation:

Problem 37: 30-60-90 triangle, hypotenuse=17

Step1: Find short leg $x$

In a 30-60-90 triangle, the short leg (opposite 30°) is $\frac{1}{2}$ the hypotenuse.
$x = \frac{1}{2} \times 17 = \frac{17}{2}$

Step2: Find long leg $y$

The long leg is $\sqrt{3}$ times the short leg.
$y = \frac{17}{2} \times \sqrt{3} = \frac{17\sqrt{3}}{2}$

Problem 38: 45-45-90 triangle, leg=12

Step1: Find equal leg $x$

In a 45-45-90 triangle, legs are equal.
$x = 12$

Step2: Find hypotenuse $y$

Hypotenuse is $\sqrt{2}$ times a leg.
$y = 12 \times \sqrt{2} = 12\sqrt{2}$

Problem 39: 30-60-90 triangle, hypotenuse=$18\sqrt{5}$

Step1: Find short leg $x$

Short leg is $\frac{1}{2}$ the hypotenuse.
$x = \frac{1}{2} \times 18\sqrt{5} = 9\sqrt{5}$

Step2: Find long leg $y$

Long leg is $\sqrt{3}$ times the short leg.
$y = 9\sqrt{5} \times \sqrt{3} = 9\sqrt{15}$

Problem 40: Use 30-60-90 triangles

Step1: Find $x$ from 30° triangle

The side of length 12 is the long leg of the small 30-60-90 triangle. Short leg $= \frac{12}{\sqrt{3}} = 4\sqrt{3}$, so $x$ (hypotenuse of the left triangle) equals this short leg:
$x = 4\sqrt{3}$

Step2: Find $y$ from 60° triangle

$y$ is the long leg of the triangle with hypotenuse 12.
$y = 12 \times \frac{\sqrt{3}}{2} = 6\sqrt{3}$

Problem 41: Use 30-60-90 triangles

Step1: Find $y$ from 60° triangle

The side of length 6 is the hypotenuse of the small 60° triangle. Long leg $y = 6 \times \frac{\sqrt{3}}{2} = 3\sqrt{3}$

Step2: Find $x$ from 30° triangle

$x$ is the hypotenuse of the large triangle, where the short leg is 6.
$x = 2 \times 6 = 12$

Problem 42: Use 45-45-90 and 30-60-90 triangles

Step1: Find $x$ from 45° triangle

The side of length 2 is a leg of the 45-45-90 triangle, so legs are equal:
$x = 2$

Step2: Find $y$ from 60° triangle

$y$ is the long leg of the 30-60-90 triangle with short leg $x=2$.
$y = 2 \times \sqrt{3} = 2\sqrt{3}$

Step3: Find $z$ from hypotenuse of large triangle

The short leg of the large 30-60-90 triangle is $y=2\sqrt{3}$, so hypotenuse:
$z = 2 \times 2\sqrt{3} = 4\sqrt{3}$

Problem 43: Use 45-45-90 and 30-60-90 triangles

Step1: Find $x$ from 45° triangle

The side of length 44 is a leg of the 45-45-90 triangle, so legs are equal:
$x = 44$

Step2: Find $y$ from 60° triangle

$y$ is the short leg of the 30-60-90 triangle with hypotenuse $x=44$.
$y = \frac{1}{2} \times 44 = 22$

Step3: Find $z$ from long leg of 60° triangle

$z$ is the long leg of the 30-60-90 triangle.
$z = 22 \times \sqrt{3} = 22\sqrt{3}$

Problem 44: Use 30-60-90 triangle

Step1: Find $y$ from 30° triangle

The side of length 16 is the hypotenuse, so short leg $y = \frac{1}{2} \times 16 = 8$

Step2: Find $x$ from 45° triangle

$x$ is the hypotenuse of the 45-45-90 triangle with leg $y=8$.
$x = 8 \times \sqrt{2} = 8\sqrt{2}$

Problem 45: Use 45-45-90 and 30-60-90 triangles

Step1: Find $y$ from 45° triangle

The side of length 12 is a leg of the 45-45-90 triangle, so legs are equal:
$y = 12$

Step2: Find $x$ from 60° triangle

$x$ is the hypotenuse of the 30-60-90 triangle with long leg $y=12$.
$x = \frac{12}{\sqrt{3}} \times 2 = 8\sqrt{3}$

Answer:

1. $x=\frac{17}{2}$, $y=\frac{17\sqrt{3}}{2}$
2. $x=12$, $y=12\sqrt{2}$
3. $x=9\sqrt{5}$, $y=9\sqrt{15}$
4. $x=4\sqrt{3}$, $y=6\sqrt{3}$
5. $x=12$, $y=3\sqrt{3}$
6. $x=2$, $y=2\sqrt{3}$, $z=4\sqrt{3}$
7. $x=44$, $y=22$, $z=22\sqrt{3}$
8. $x=8\sqrt{2}$, $y=8$
9. $x=8\sqrt{3}$, $y=12$