How do I solve for the variables of the figures below here?

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1 Answer
Jan 23, 2018

See below.

Explanation:

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From the diagram above we have the trigonometrical ratios:

#sin(x)=0/H#

#cos(x)=0/H#

#tan(x)=0/A#

These will help us in solving the given triangles.

I have turned and flipped the triangles as to make it easier to explain.

Figure 1

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We have the length of the hypotenuse H, so we can use this to find #y# as follows:

#sin(30)=y/11#

Rearranging:

#y=11sin(30)#

#sin(30)=1/2# so #11sin(30)=11/2#

#y=11/2#

We can find #x# by using:

#cos(30)=x/11#

Rearranging:

#x=11cos(30)

#cos(30)=sqrt(3)/2#

#x=11cos(30)=(11sqrt(3))/2#

So for figure 1:

#color(blue)(y=11/2)#

#color(blue)(x=(11sqrt(3))/2)#

Figure 2

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Using the same reasoning as before:

#sin(45)=(8sqrt(3))/x#

#x=(8sqrt(3))/sin(45)#

#sin(45)=sqrt(2)/2#

#x=(8sqrt(3))/(sqrt(2)/2)=(16sqrt(3))/sqrt(2)=(8sqrt(6))/2#

#x=8sqrt(6)#

For #y# we use the #tan(45)#

#tan(45)=(8sqrt(2))/y#

#y=(8sqrt(2))/tan(45)#

#tan(45)=1#

#y=(8sqrt(2))/1=8sqrt(2)#

So for figure 2

#color(blue)(x=8sqrt(6))#

#color(blue)(y=8sqrt(2))#