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Is there a way to do this problem without splitting into two

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Senior Manager
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Is there a way to do this problem without splitting into two [#permalink] New post 21 Aug 2007, 19:08
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Is there a way to do this problem without splitting into two triangles?

A certain triangle has two angles measuring 45 and 75 degrees. the side opposite 45 deg has length 6. what is the length of the side opposite the 75 deg angle?

a. 2+3 SQRT3
b. 3(SQRT 3+1)
c. 6
d. 6 SQRT3
e. 2+6 SQRT2
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Re: PS- triangles [#permalink] New post 21 Aug 2007, 19:33
r019h wrote:
Is there a way to do this problem without splitting into two triangles?

A certain triangle has two angles measuring 45 and 75 degrees. the side opposite 45 deg has length 6. what is the length of the side opposite the 75 deg angle?

a. 2+3 SQRT3
b. 3(SQRT 3+1)
c. 6
d. 6 SQRT3
e. 2+6 SQRT2


I don't see any other way to do this without splitting in half.
You can estimate, but that only eliminate a and c.
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Re: PS- triangles [#permalink] New post 21 Aug 2007, 21:41
r019h wrote:
Is there a way to do this problem without splitting into two triangles?

A certain triangle has two angles measuring 45 and 75 degrees. the side opposite 45 deg has length 6. what is the length of the side opposite the 75 deg angle?

a. 2+3 SQRT3
b. 3(SQRT 3+1)
c. 6
d. 6 SQRT3
e. 2+6 SQRT2


I can elminate C, but Im not sure what to do after this. I split the the triangle into 2 and got an isosceles triangle (75,75,30) where two of the sides are 6... and another triangle (30,45,105).

I don't know what to do from here. plz help :oops:
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Re: PS- triangles [#permalink] New post 21 Aug 2007, 23:44
r019h wrote:
Is there a way to do this problem without splitting into two triangles?

A certain triangle has two angles measuring 45 and 75 degrees. the side opposite 45 deg has length 6. what is the length of the side opposite the 75 deg angle?

a. 2+3 SQRT3
b. 3(SQRT 3+1)
c. 6
d. 6 SQRT3
e. 2+6 SQRT2


I am able to eliminate only C. How do you solve this by splitting the triangle??
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 [#permalink] New post 22 Aug 2007, 05:12
You have 45-60-75 triangle.
Know that 75 = 45 + 30 and so split that angle and draw a perpendicular line to the base. Then you have:
45-45-90 and 30-60-90
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 [#permalink] New post 22 Aug 2007, 09:05
bkk145 wrote:
You have 45-60-75 triangle.
Know that 75 = 45 + 30 and so split that angle and draw a perpendicular line to the base. Then you have:
45-45-90 and 30-60-90



Wonderful!!! its so easy!!! I was so fixated on splitting the triangles by dividing the angles by 2...

So i never thought to split 75 to 30 and 45 :roll:
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 [#permalink] New post 22 Aug 2007, 09:23
hope it'll help
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 [#permalink] New post 22 Aug 2007, 11:59
this is probably stupid but im going to say it anyways.

why couldnt i do
6/45 = x/75 and solve that for x?

i thought if i know two of the angles and one of the sides opposite one of them, then they have the same ratio.

or is there something i hvae to assume in order for me to be able to do that? like do i have to assume theyre two similar triangles and only then.
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Re: [#permalink] New post 18 Mar 2008, 09:23
jimjohn wrote:
this is probably stupid but im going to say it anyways.

why couldnt i do
6/45 = x/75 and solve that for x?

i thought if i know two of the angles and one of the sides opposite one of them, then they have the same ratio.

or is there something i hvae to assume in order for me to be able to do that? like do i have to assume theyre two similar triangles and only then.

that's also what i did....can anyone give insight on why not?
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Re: PS- triangles [#permalink] New post 19 Mar 2008, 00:25
b.
if X=the length of the side opposite the 75 deg angle
then 6/sin45(deg)=X/sin75(deg)
sin75(deg)=sin(45+30)deg=sin45*cos30+cos45*sin30
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Re: PS- triangles [#permalink] New post 25 Mar 2008, 09:01
can u explain it more briefly pls :?:
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Re: PS- triangles [#permalink] New post 25 Mar 2008, 12:09
courant wrote:
b.
if X=the length of the side opposite the 75 deg angle
then 6/sin45(deg)=X/sin75(deg)
sin75(deg)=sin(45+30)deg=sin45*cos30+cos45*sin30

for an triangle with sides a,b,c and angles A,B,C :
a/sinA = b/sinB=c/sinC
here 6/sin45 = x/sin75
=> x = 6* sin75/sin45
sin75 = (sqrt3+1)/2sqrt2
sin45 = 1/sqrt2
so sin75/sin45 = (sqrt3+1)/2

Solving for x => 3(sqrt3+1)

Ans B
Re: PS- triangles   [#permalink] 25 Mar 2008, 12:09
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