Question

unit 7 right triangles & trigonometry homework 2 special right triangles pls help with 16,18 and 20

Answer 1

The questions are illustrations of Pythagoras theorem and trigonometry ratios.

What is Pythagoras theorem?

Pythagoras theorem is used to determine the lengths of the legs of a right triangle.

It is represented as:

$a^2 = b^2 + c^2$

Where:

b & c are the legs of the triangle, while a is the hypotenuse

Question 16

Start by calculating the value of x, using the sine ratio

$\sin(30) = \frac{x}{16\sqrt 3}$

Evaluate sin(30)

$0.5 = \frac{x}{16\sqrt 3}$

Solve for x

$x = 0.5 * 16\sqrt 3$

$x = 8\sqrt 3$

Next, calculate the length (l) of the boundary between both triangles using Pythagoras theorem

$(16\sqrt 3)^2 = (8\sqrt 3)^2 +l^2$

$768 = 192 +l^2$

Collect like terms

$l^2 = 768 -192$

$l^2 = 576$

Take the square root of both sides

$l = 24$

Given that the angle is 45 degrees, it means that:

$z = y$

So, we have:

$z^2 + y^2 = 24^2$

$z^2 + z^2 = 24^2$

$2z^2 = 576$

Divide through by 2

$z^2 = 288$

Take the square root of both sides

$z= 12\sqrt 2$

Hence, the values of x, y and z are:

$x = 8\sqrt 3$

$y= 12\sqrt 2$

$z= 12\sqrt 2$

Question 18

Start by calculating the value of z, using the sine ratio

$\sin(45) = \frac{z}{20}$

Evaluate sin(45)

$\frac{\sqrt 2}{2} = \frac{z}{20}$

Solve for z

$z = \frac{\sqrt 2}{2} * 20$

$z = 10\sqrt 2$

Next, calculate the value of y using tangent ratio

$\tan(30) = \frac{y}{10\sqrt 2}$

Solve for y

$y = \tan(30) * 10\sqrt 2$

Evaluate tan(30)

$y = \frac{\sqrt 3}{3} * 10\sqrt 2$

$y = \frac{10\sqrt 6}{3}$

Next, calculate the value of x using sine ratio

$\sin(30) = \frac{10/3\sqrt 6}{x}$

Solve for x

$x = \frac{10/3\sqrt 6}{\sin(30)}$

Evaluate sin(30)

$x = \frac{10/3\sqrt 6}{1/2}$

$x = \frac{20\sqrt 6}{3}$

Hence, the values of x, y and z are:

$x = \frac{20\sqrt 6}{3}$

$y = \frac{10\sqrt 6}{3}$

$z = 10\sqrt 2$

Question 20

Start by calculating the value of z, using the sine ratio

$\sin(45) = \frac{z}{10\sqrt 6}$

Evaluate sin(45)

$\frac{\sqrt 2}{2} = \frac{z}{10\sqrt 6}$

Solve for z

$z = \frac{\sqrt 2}{2} * 10\sqrt 6$

$z = 5\sqrt {12$

Simplify

$z = 10\sqrt {3$

Next, calculate the value of y using tangent ratio

$\tan(30) = \frac{10\sqrt 3}{y}$

Solve for y

$y = \frac{10\sqrt 3}{\tan(30)}$

Evaluate tan(30)

$y = \frac{10\sqrt 3}{1/\sqrt 3}$

Simplify

$y = 10\sqrt 3 * \sqrt 3$

$y = 30$

Next, calculate the value of x using sine ratio

$\sin(30) = \frac{10\sqrt 3}{x}$

Solve for x

$x = \frac{10\sqrt 3}{\sin(30)}$

Evaluate sin(30)

$x = \frac{10\sqrt 3}{1/2}$

$x = 20\sqrt 3$

Hence, the values of x, y and z are:

$x = 20\sqrt 3$

$y = 30$

$z = 10\sqrt {3$

Read more about Pythagoras theorem and trigonometry ratios at:

brainly.com/question/6241673