Question 277897: I am having great difficulty with this problem.
I will try to explain the picture.
I have a right triangle ACB with (the 90 deg is at point C).
There is a line in the triangle EF parallel to AB.
There is also an altitude CD that starts at point C, passes through line EH, forming point H and hits line AB at point D.
I will try to post the image below:
Here is the question that baffles me.
If EF is parallel to AB, AD = 9, DB = 16, EC = 2(AE), find AE, AC, CF, CB, CD, and EF.
Thanks to anyone who can help.
Answer by edjones(8007)   (Show Source):
You can put this solution on YOUR website! 9/25 * 90 = 32.4 deg for left part of angle c
90-32.4=57.6 deg right part of angle c
angle a=57.6
angle b=32.4
.
sin(32.4)=9/hyp
.5358=9/h
h=9/.5358=16.8 AC
16.8/3=5.6 AE
5.6*2=11.2 CE
.
9^2+a^2=16.8^2
81a^2=282.24
a^2=201.24
a=CD=14.2
.
14.2^2+16^2=BC^2
=483.88
BC=22.0
.
I'll let you find CF and EF
Let me know if you have trouble.
.
Ed
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