x^2-x=0

x(x-1)=0

x=1 v x=0]]>

x = √x

x² = x

x = 1

]]>The radical I was talking about was in thinking;

C = √(A² + B² - 2ABcosc)

]]>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?

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

]]>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.

]]>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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