Probably a very easy maths question (trigonometry)

Before I get told to 'go ask your teacher' i'll make sure you know that it's not my work, but a friend asked me to help and since doing GCSE last year i've forgotten how to do this part of trig since starting AS.



'Find angle ABC' says the question.

I know how to find angles and sides with a right angled triangle, but if anyone could quickly run past me the method here then i'm sure i'll be able to remember quite quickly.

its an isoceles, so bisect the trinagle down teh middle to create a right angle triangle, and you should know teh rest then :P

Sim

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And then double whatever i get to find B. thanks

You're looking pretty acute today. Yes, I'm being obtuse.

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ahhh, memories come flooding back. and it's like.. i think i know how to do it, until i try, and then i realise i can't. total of all angles in a triangle = 180 degrees.

i don't see how bisecting the thing helps, because then all you have is 90 + two angles you don't know. only that they too = 90. because 90 + 90 = 180 in the triangles you created.

i am an idiot.

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i am an idiot.

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sasjag schrieb:
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> its an isoceles, so bisect the trinagle down teh
> middle to create a right angle triangle, and you
> should know teh rest then :P

After bisecting angle B, you'll have the triangle ABD, with angles 90º, B/2 and [90º-(B/2)]. Since ABC is isoceles, AD = AC/2 = 0,95. With that, you know that sin (B/2) = 0,95/2,42 = 0,392561... . Looking at a table like this one, you have that B/2 is approximately 23º. Thus, the angle B is approximately 46º.

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The answer is 42. It always is. So move on

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I think the cosine rule (b^2 = a^2 + c^2 - 2acCos(Angle@B)) can be used here...Could be very wrong on that though, maths is not my strong point.

Protractor. Bam, problem solved.

One thing that is amazing is that it doesn't matter how advanced you go with Math, it's easy to forget the simple equations to simple problems. That's why you can't ever stop exercising your brain and doing Math! That said, I have stopped doing that kind of Math and don't feel like going back to it :P

Cosine rule with sides a, b, c; angles A, B, C

a squared = b squared + c squared - 2bc(cosA)
b squared = a squared + c squared - 2ac(cosB)
c squared = a squared + b squared - 2ab(cosC)

Re-arranging those equations to give you angle B should work because you have all 3 sides.

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