Trigonometry question- help!!!

I can do trigonometry questions, however when coming across a question which gives all 3 sides of a triangle (instead of just 2) and asks to find an unknown angle- how are you suppose to know whether to use sin/ cos or tan and which 2 lengths to pick?

:confused:

(edited 11 years ago)

Reply 1

Tomcrease

Original post by Lid-the-squid

I can do trigonometry questions, however when coming across a question which gives all 3 sides of a triangle (instead of just 2) and asks to find an unknown angle- how are you suppose to know whether to use sin/ cos or tan and which 2 lengths to pick?

e.g take this question below for example (click to enlarge):

Attachment not found

How would you approach it? Because if you use x = sin-1 x (opp/ adj) it gives a different answer to using x = tan-1 x (opp/ adj) or even x =cos-1 x (adj/ hyp)

:confused:

That would be because it's

sin^{-1}\frac{opp}{hyp}

Edit:

The triangle doesn't work, it can't exist.

by pythagorus theorem

a^2 + b^2 = c^2

well, sub in the values yourself and see.

Or for even more fun, get out a ruler and protractor and try and draw it :tongue:

(edited 11 years ago)

Reply 2

Tulian

Well you can do :

Sinx = Side Opposite (of the angle) / Hyp
Then rearrange for x

So here it'd be , sin x = 4/11.2

Reply 3

MatthewGallagher

sine rule as well! a/sinA = b/sinB

Reply 4

Tomcrease

Ah, if it didn't have that right angle there which you had in the diagram then it's just rearranged cosine rule :smile:

CosA = \frac{b^2 + c^2 - a^2}{2bc}

Reply 5

Horro421

2

Use the cosine rule and rearrange to make 'CosA' the subject :smile:

Reply 6

MrBlueMo0n

10

Original post by Lid-the-squid

I can do trigonometry questions, however when coming across a question which gives all 3 sides of a triangle (instead of just 2) and asks to find an unknown angle- how are you suppose to know whether to use sin/ cos or tan and which 2 lengths to pick?

:confused:

I'm not sure of the question, and I can't remember if this even comes up in GCSE, but I'd use the cosine rule:

a^2 = b^2 + c^2 - 2bccosA

a, b and c are lengths, A is the angle you want. A is the opposite angle to length a. b and c can be either side - it won't make a difference.

Hope this helps, just ask or show me a better version of the question if you would like any more help :h:

Trigonometry question- help!!!

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