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OpenStudy (darkbluechocobo):
The information in the following problem allows you to construct 1 triangle. A = 30°, a = 1 m, b = 2 m
OpenStudy (darkbluechocobo):
This is a true or false, but looking at it you should still be able to
OpenStudy (darkbluechocobo):
you have all the points to get B
OpenStudy (darkbluechocobo):
|dw:1427485351203:dw|
OpenStudy (darkbluechocobo):
\[\frac{ 1 }{ \sin30 }=\frac{ 2 }{ sinB }\]
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OpenStudy (campbell_st):
so find B using the law of sines...
find C using angle sum of a triangle...
see if they work...
smallest angle opposite shortest side etc
if that all works find c
OpenStudy (darkbluechocobo):
\[B=\sin ^{-1}\frac{ 2\sin30 }{ 1}\]
OpenStudy (darkbluechocobo):
B=30 degrees
OpenStudy (darkbluechocobo):
180-30-30=120
OpenStudy (darkbluechocobo):
C=120
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OpenStudy (darkbluechocobo):
|dw:1427485660697:dw|
OpenStudy (darkbluechocobo):
\[\frac{ 1 }{ \sin30 }=\frac{ c }{ \sin120 }\]
OpenStudy (campbell_st):
so you have 2 angles of 30.. with sides of differing lengths... think about that
OpenStudy (darkbluechocobo):
hmm Well that is 2 acute angles and one obtuse
OpenStudy (campbell_st):
ok... 2 equal angles... the sides opposite would need to be equal..?
check your calculations as I don't think B is 30
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OpenStudy (darkbluechocobo):
Sorry Back openstudy isnt loading
OpenStudy (darkbluechocobo):
B=90?
OpenStudy (darkbluechocobo):
I just realized since it was over 1 it is basically saying \[\sin ^{-1}(2\sin30)\]
OpenStudy (darkbluechocobo):
C=60?
OpenStudy (darkbluechocobo):
@campbell_st is 90 correctn?
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OpenStudy (campbell_st):
90 is correct... so the triangle is right angled... the missing side can be found using pythagoras... and the missing angle using angle sum of a triangle
OpenStudy (darkbluechocobo):
180-90-30=60 C=60
OpenStudy (darkbluechocobo):
\[\frac{ 1 }{ \sin30 }=\frac{ c }{ \sin60 }\]
OpenStudy (darkbluechocobo):
\[c=\frac{ \sin30 }{ \sin60 }\]
OpenStudy (campbell_st):
and use pythagoras 2^2 = 1^2 + c^2
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OpenStudy (darkbluechocobo):
so 4=1+c^2 c=sqrt3
OpenStudy (darkbluechocobo):
So how could you see if you can only construct one or two triangles?
OpenStudy (darkbluechocobo):
@freckles could you assist?
OpenStudy (freckles):
what was the first angle you solved for?
OpenStudy (freckles):
looks like it was angle B
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OpenStudy (darkbluechocobo):
I solved for angle B: which was 90
OpenStudy (darkbluechocobo):
then got 60 for C
OpenStudy (freckles):
so you can see if there is a second traingle by doing \[B_2=180-B=180-90=90\]
But we see here we will have the same triangle if we check for the second triangle
so you have one triangle
OpenStudy (darkbluechocobo):
So that means you cannot make another triangle since they have both the same angle?
OpenStudy (darkbluechocobo):
when you double it?
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OpenStudy (freckles):
if A and B were the same as before then C would have to be the same