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**katy****Member**- Registered: 2005-12-28
- Posts: 14

Hi...I have a problem that really needs to be solved for my study for the exam please.(it is grade 11 problem.)

1) a) If <C=40degrees, c=35 and b=40 in triangle ABC, then determine the number of triangles that would be possible with these values. Justify your answer by showing your work.

b) Based on your answer to part 1a), solve for <B in your triangle(s).

Many thanks to you guys..!!:P;)

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**irspow****Member**- Registered: 2005-11-24
- Posts: 457

I am a little lost. If everything you typed in a) is what you wished type then there is only one triangle possible. Unless there is an equal sign somewhere that shouldn't be there;

A+B+C = 180°

Using the law of sines;

sin40°/35 = sinB/40

B = arcsin(40sin40°/35) ≈ 47.2746°

If C = 40° and B ≈ 47.2746°, then A ≈ 92.7254°

A = 92.7254°, a = 54.388

B = 47.2746°, b = 40

C = 40°, c = 35

Nothing else is possible unless C, c, or b are allowed to vary.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

There is a second possible triangle.

B can also take the value of (180 - 47.2746)°.

Why did the vector cross the road?

It wanted to be normal.

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**irspow****Member**- Registered: 2005-11-24
- Posts: 457

Good catch Mathsyperson, I should not rely so heavily on the law of sines. Perhaps an expansion of the law of cosines would be more appropriate? I am wondering if the radical sign would produce two simultaneous solutions like we needed here.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

The law of cosines would give this:

35² = 40² + a² - 2*40*a cos 40°.

Rearranging gives a² - 80 cos40° a - 375 = 0, which is a quadratic that can be solved to give the two possible lengths of the third side.

I think the sine law would be simpler to use, you just need to remember to work out all possible values of the arcsin. It's not you relying on the law of sines that's the problem, it's you relying on your calculator.

Sorry, but what's the radical sign?

Why did the vector cross the road?

It wanted to be normal.

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**irspow****Member**- Registered: 2005-11-24
- Posts: 457

I suppose your right.

The radical I was talking about was in thinking;

C = √(A² + B² - 2ABcosc)

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Oh, of course. Sorry, it's just that I don't hear 'radical' being used to mean square root very often, so I forgot about that definition temporarily.

Why did the vector cross the road?

It wanted to be normal.

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**jadasia****Member**- Registered: 2006-01-21
- Posts: 1

how do you solve x - the square root of x = 0 i really need some help with this one.

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**irspow****Member**- Registered: 2005-11-24
- Posts: 457

x - √x = 0

x = √x

x² = x

x = 1

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

...or 0.

Why did the vector cross the road?

It wanted to be normal.

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**irspow****Member**- Registered: 2005-11-24
- Posts: 457

aaaaargh!....yes.

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**kempos****Member**- Registered: 2006-01-07
- Posts: 77

x^2=x

x^2-x=0

x(x-1)=0

x=1 v x=0

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