Answer:
Part 1) [tex]x=\frac{15}{2}\ units[/tex]
Part 2) [tex]z=\frac{15\sqrt{3}}{2}\ units[/tex]
Part 3) [tex]y= \frac{15\sqrt{3}}{4}\ units[/tex]
Part 4) [tex]b= \frac{45}{4}\ units[/tex]
Step-by-step explanation:
step 1
Find the value of x
In the large right triangle
[tex]cos(60^o)=\frac{x}{15}[/tex] ----> by CAH (adjacent side divided by the hypotenuse)
Remember that
[tex]cos(60^o)=\frac{1}{2}[/tex]
substitute
[tex]\frac{1}{2}=\frac{x}{15}[/tex]
solve for x
[tex]x=\frac{15}{2}\ units[/tex] ---> improper fraction
step 2
Find the value of z
In the large right triangle
Applying the Pythagorean Theorem
[tex]15^2=x^2+z^2[/tex]
substitute the value of x
[tex]15^2=(\frac{15}{2})^2+z^2[/tex]
solve for z
[tex]z^2=15^2-(\frac{15}{2})^2[/tex]
[tex]z^2=225-\frac{225}{4}[/tex]
[tex]z^2=\frac{675}{4}[/tex]
[tex]z=\frac{\sqrt{675}}{2}\ units[/tex]
simplify
[tex]z=\frac{15\sqrt{3}}{2}\ units[/tex]
step 3
Find the value of y
In the right triangle of the right
[tex]sin(30^o)=\frac{y}{z}[/tex] ---> by SOH (opposite side divided by the hypotenuse)
substitute the given values of y and z
Remember that
[tex]sin(30^o)=\frac{1}{2}[/tex]
so
[tex]\frac{1}{2}=y:\frac{15\sqrt{3}}{2}[/tex]
solve for y
[tex]\frac{1}{2}= \frac{2y}{15\sqrt{3}}[/tex]
[tex]y= \frac{15\sqrt{3}}{4}\ units[/tex]
step 4
Find the value of b
In the right triangle of the right
[tex]cos(30^o)=\frac{b}{z}[/tex] ---> by CAH (adjacent side divided by the hypotenuse)
substitute the given values of y and z
Remember that
[tex]cos(30^o)=\frac{\sqrt{3}}{2}[/tex]
so
[tex]\frac{\sqrt{3}}{2}=b:\frac{15\sqrt{3}}{2}[/tex]
solve for y
[tex]\frac{\sqrt{3}}{2}= \frac{2b}{15\sqrt{3}}[/tex]
[tex]b= \frac{45}{4}\ units[/tex]
