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Mathematics 13 Online
OpenStudy (darkbluechocobo):

Help with law of sines

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):

\[\frac{ 1 }{ \sin30 }=\frac{ 2 }{ sinB }\]

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

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

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?

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

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

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?

OpenStudy (freckles):

if A and B were the same as before then C would have to be the same

OpenStudy (freckles):

|dw:1427493209009:dw| this is the info given

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