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bob bundy
2013-03-24 05:09:28

Two triangles with sides in the same ratio are similar.  So the angles are the same.  Therefore you can just call t = 1 without losing any of the problem's conditions.

That's why it cancels out of the cosine rule step.

Bob

Agnishom
2013-03-24 01:29:38

Yeah Thanks! I had difficulty understanding the cancellation

bob bundy
2013-03-23 22:07:06

hi Agnishom,

Glad you understand it.  Your method and mine are essentially the same; I used k; you have t; and then I let k=1.  In yours the t cancels out.

Bob

Agnishom
2013-03-23 19:56:16

Okay thanks I managed to understand this

Agnishom
2013-03-23 19:38:29

#### bob bundy wrote:

hi Agnishom

SINE RULE is:

So I put

and re-arranged to get

I have the following doubts:

I. How did you do the rearrangement?
II. How do you know what to put for a?

bob bundy
2013-03-23 19:20:56

hi Agnishom

How did you get that?  It's not what I'm getting.

Here's my method.

SINE RULE is:

So I put

and re-arranged to get

Different values of k will make different sized circles but all the resulting triangles will be similar => same three angles.

So it is ok to take a = √6,   b = 2√2,     c = 3√3

Use the cosine rule to get cosA and hence sinA.

Bob

Agnishom
2013-03-23 16:48:35

If in a triangle,

Then,
What is the value of sin angle A?

Is it 8/27?