Triangles - Trig: for given triangle, find 3 options equal to "a"

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mcarthyryan

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Hi, I've got a question, and if it was asking for one answer, I think I'd be correct, but it requires 3? It is a mock online test I am trying to practise with, but I don't understand how to get more than one answer:

This is the question:


I think the one I've ticked is correct, as tan& = b / a which converts to a = b / tan&

I just can't see how to work out the other two?

I thought cos& = a / c which converts to a = c.cos& might have been a correct answer, but it isn't there?

a^2 + b^2 = c^2 converted to a^2 = c^2 - b^2 isn't on the list either?


Please can you give any pointers?




Thanks a lot
 
mcarthyryan said:
Hi, I've got a question, and if it was asking for one answer, I think I'd be correct, but it requires 3? It is a mock online test I am trying to practise with, but I don't understand how to get more than one answer:

This is the question:


I think the one I've ticked is correct, as tan& = b / a which converts to a = b / tan&

I just can't see how to work out the other two?

I thought cos& = a / c which converts to a = c.cos& might have been a correct answer, but it isn't there?

a^2 + b^2 = c^2 converted to a^2 = c^2 - b^2 isn't on the list either?


Please can you give any pointers?

Thanks a lot
Click to expand...

Hi Ryan:

Your answer is correct, so far.

a = b/tan(θ)

Your work finding a = c*cos(θ) is also correct, but, you're right; that expression is not listed.

Your work using the Pythagorean Theorem is good, but you stopped at a^2. They're asking for a.

a = sqrt(c^2 - b^2)

Too bad, that's not in the list, either.

Here are two hints for you to consider.

What is the angle between sides b and c? (It can be expressed in terms of θ.)

What is the Law of Sines?

Cheers :cool:
 
Hi, Thank you for the reply.


The angle between b and c would = 180 - (90 + θ) OR could it be cosθ = b/c.....sorry, that is far as I know to go?



The law of Sines is a/sin A = b/sin B = c/sin C
- which I think you could express as:
a = sin A x b / sin B
so I think in this question it would be:
a = bsin(90' - θ) / sin θ ?



Thank you
 
Hi Everyone,

Could anyone please point me in the right direction to get the third answer if possible? Not give me the answer, but any more pointers please?


Thank you
 
mcarthyryan said:
Hi Everyone,

Could anyone please point me in the right direction to get the third answer if possible? Not give me the answer, but any more pointers please?


Thank you
Click to expand...
Hint:

sin(90°-Φ) = cos(Φ)
 
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