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:
[tex]a^2 = b^2 + c^2[/tex]
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
[tex]\sin(30) = \frac{x}{16\sqrt 3}[/tex]
Evaluate sin(30)
[tex]0.5 = \frac{x}{16\sqrt 3}[/tex]
Solve for x
[tex]x = 0.5 * 16\sqrt 3[/tex]
[tex]x = 8\sqrt 3[/tex]
Next, calculate the length (l) of the boundary between both triangles using Pythagoras theorem
[tex](16\sqrt 3)^2 = (8\sqrt 3)^2 +l^2[/tex]
[tex]768 = 192 +l^2[/tex]
Collect like terms
[tex]l^2 = 768 -192[/tex]
[tex]l^2 = 576[/tex]
Take the square root of both sides
[tex]l = 24[/tex]
Given that the angle is 45 degrees, it means that:
[tex]z = y[/tex]
So, we have:
[tex]z^2 + y^2 = 24^2[/tex]
[tex]z^2 + z^2 = 24^2[/tex]
[tex]2z^2 = 576[/tex]
Divide through by 2
[tex]z^2 = 288[/tex]
Take the square root of both sides
[tex]z= 12\sqrt 2[/tex]
Hence, the values of x, y and z are:
[tex]x = 8\sqrt 3[/tex]
[tex]y= 12\sqrt 2[/tex]
[tex]z= 12\sqrt 2[/tex]
Question 18
Start by calculating the value of z, using the sine ratio
[tex]\sin(45) = \frac{z}{20}[/tex]
Evaluate sin(45)
[tex]\frac{\sqrt 2}{2} = \frac{z}{20}[/tex]
Solve for z
[tex]z = \frac{\sqrt 2}{2} * 20[/tex]
[tex]z = 10\sqrt 2[/tex]
Next, calculate the value of y using tangent ratio
[tex]\tan(30) = \frac{y}{10\sqrt 2}[/tex]
Solve for y
[tex]y = \tan(30) * 10\sqrt 2[/tex]
Evaluate tan(30)
[tex]y = \frac{\sqrt 3}{3} * 10\sqrt 2[/tex]
[tex]y = \frac{10\sqrt 6}{3}[/tex]
Next, calculate the value of x using sine ratio
[tex]\sin(30) = \frac{10/3\sqrt 6}{x}[/tex]
Solve for x
[tex]x = \frac{10/3\sqrt 6}{\sin(30)}[/tex]
Evaluate sin(30)
[tex]x = \frac{10/3\sqrt 6}{1/2}[/tex]
[tex]x = \frac{20\sqrt 6}{3}[/tex]
Hence, the values of x, y and z are:
[tex]x = \frac{20\sqrt 6}{3}[/tex]
[tex]y = \frac{10\sqrt 6}{3}[/tex]
[tex]z = 10\sqrt 2[/tex]
Question 20
Start by calculating the value of z, using the sine ratio
[tex]\sin(45) = \frac{z}{10\sqrt 6}[/tex]
Evaluate sin(45)
[tex]\frac{\sqrt 2}{2} = \frac{z}{10\sqrt 6}[/tex]
Solve for z
[tex]z = \frac{\sqrt 2}{2} * 10\sqrt 6[/tex]
[tex]z = 5\sqrt {12[/tex]
Simplify
[tex]z = 10\sqrt {3[/tex]
Next, calculate the value of y using tangent ratio
[tex]\tan(30) = \frac{10\sqrt 3}{y}[/tex]
Solve for y
[tex]y = \frac{10\sqrt 3}{\tan(30)}[/tex]
Evaluate tan(30)
[tex]y = \frac{10\sqrt 3}{1/\sqrt 3}[/tex]
Simplify
[tex]y = 10\sqrt 3 * \sqrt 3[/tex]
[tex]y = 30[/tex]
Next, calculate the value of x using sine ratio
[tex]\sin(30) = \frac{10\sqrt 3}{x}[/tex]
Solve for x
[tex]x = \frac{10\sqrt 3}{\sin(30)}[/tex]
Evaluate sin(30)
[tex]x = \frac{10\sqrt 3}{1/2}[/tex]
[tex]x = 20\sqrt 3[/tex]
Hence, the values of x, y and z are:
[tex]x = 20\sqrt 3[/tex]
[tex]y = 30[/tex]
[tex]z = 10\sqrt {3[/tex]
Read more about Pythagoras theorem and trigonometry ratios at:
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