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DENIUS [597]
3 years ago
11

I need to know the improper fractions answers for:

Mathematics
1 answer:
zmey [24]3 years ago
7 0

Answer:

Part 1) x=\frac{15}{2}\ units

Part 2) z=\frac{15\sqrt{3}}{2}\ units

Part 3) y= \frac{15\sqrt{3}}{4}\ units

Part 4) b= \frac{45}{4}\ units

Step-by-step explanation:

step 1

Find the value of x

In the large right triangle

cos(60^o)=\frac{x}{15} ----> by CAH (adjacent side divided by the hypotenuse)

Remember that

cos(60^o)=\frac{1}{2}

substitute

\frac{1}{2}=\frac{x}{15}

solve for x

x=\frac{15}{2}\ units ---> improper fraction

step 2

Find the value of z

In the large right triangle

Applying the Pythagorean Theorem

15^2=x^2+z^2

substitute the value of x

15^2=(\frac{15}{2})^2+z^2

solve for z

z^2=15^2-(\frac{15}{2})^2

z^2=225-\frac{225}{4}

z^2=\frac{675}{4}

z=\frac{\sqrt{675}}{2}\ units

simplify

z=\frac{15\sqrt{3}}{2}\ units

step 3

Find the value of y

In the right triangle of the right

sin(30^o)=\frac{y}{z} ---> by SOH (opposite side divided by the hypotenuse)

substitute the given values of y and z

Remember that

sin(30^o)=\frac{1}{2}

so

\frac{1}{2}=y:\frac{15\sqrt{3}}{2}

solve for y

\frac{1}{2}= \frac{2y}{15\sqrt{3}}

y= \frac{15\sqrt{3}}{4}\ units

step 4

Find the value of b

In the right triangle of the right

cos(30^o)=\frac{b}{z} ---> by CAH (adjacent side divided by the hypotenuse)

substitute the given values of y and z

Remember that

cos(30^o)=\frac{\sqrt{3}}{2}

so

\frac{\sqrt{3}}{2}=b:\frac{15\sqrt{3}}{2}

solve for y

\frac{\sqrt{3}}{2}= \frac{2b}{15\sqrt{3}}

b= \frac{45}{4}\ units

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