Trig help

Apr 2010
40
0
The perimeter of a triangle is 100cm and the angles are in the ratio of 1:3:5. Find the length of each side.
 
Nov 2009
717
133
Wahiawa, Hawaii
first get the angles

divide \(\displaystyle 180^o\) by \(\displaystyle 9\) then apply the ratio's to get the \(\displaystyle 3\) angles

so

\(\displaystyle 20^o(1) = 20^o\)
\(\displaystyle 20^o(3) = 60^o\)
\(\displaystyle 20^o(5) = 100^o\)
 
Last edited:
Apr 2010
40
0
Yeah I had already gotten that but what am I supposed to do with those values?
 
Last edited:
Nov 2009
717
133
Wahiawa, Hawaii
one possible way to get the length of the sides is to temporarily give one of the sides a length of 1 then find the lengths of the other sides by the law of sines in proportion to that then you will have another ratio to apply to the 100cm perimeter.
 
Last edited:
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Jul 2007
894
298
New Orleans
The perimeter of a triangle is 100cm and the angles are in the ratio of 1:3:5. Find the length of each side.
\(\displaystyle x+3x+5x= 180\)

\(\displaystyle x = 20\)

angles are 20,60,100

Now use these formulas

\(\displaystyle \frac{\sin{60}}{a} = \frac{\sin{20}}{b} = \frac{\sin{100}}{c}\)

\(\displaystyle a + b + c = 100\)

Now solve for a variable and sub

\(\displaystyle a = \frac{b\sin{60}}{\sin{20}}\)

\(\displaystyle c = \frac{b\sin{100}}{\sin{20}}\)

so now sub

\(\displaystyle \frac{b\sin{60}}{\sin{20}} + b + \frac{b\sin{100}}{\sin{20}} = 100\)
 
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Apr 2010
40
0
one possible way to get the length of the sides is to temporarily give one of the sides a length of 1 then find the lengths of the other sides by the law of sines
I only understood up to that part.
 
Nov 2009
717
133
Wahiawa, Hawaii
as mentioned you should get

approximately:

15.6 + 44.9 + 39.5 = 100