Answer:
Step-by-step explanation:
Question 13.
By applying cosine rule,
cos(45°) = [tex]\frac{\text{Adjacent side}}{\text{Hypotenuse}}[/tex]
[tex]\frac{1}{\sqrt{2} }=\frac{10}{y}[/tex]
y = 10√2
By applying sine rule,
sin(45°) = [tex]\frac{\text{Opposite side}}{\text{Hypotenuse}}[/tex]
[tex]\frac{1}{\sqrt{2} }=\frac{x}{y}[/tex]
[tex]\frac{1}{\sqrt{2} }=\frac{x}{10\sqrt{2} }[/tex]
x = 10
Question 15.
By applying sine rule,
sin(60°) = [tex]\frac{\text{Opposite side}}{\text{Hypotenuse}}[/tex]
[tex]\frac{\sqrt{3} }{2}=\frac{y}{32}[/tex]
y = 16√3
By applying cosine rule,
cos(60°) = [tex]\frac{\text{Adjacent side}}{\text{Hypotenuse}}[/tex]
[tex]\frac{1}{2}=\frac{x}{32}[/tex]
x = 16
Question 17.
By applying sine rule,
sin(60°) = [tex]\frac{\text{Opposite side}}{\text{Hypotenuse}}[/tex]
[tex]\frac{\sqrt{3} }{2}=\frac{11\sqrt{3} }{y}[/tex]
y = 22
By applying cosine rule,
cos(60°) = [tex]\frac{\text{Adjacent side}}{\text{Hypotenuse}}[/tex]
[tex]\frac{1}{2}=\frac{x}{y}[/tex]
[tex]\frac{1}{2}=\frac{x}{22}[/tex]
x = 11
