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Trignonometry/Pythagoras Help:

1) When trying to find a side of a right angled triangle how do you know when to use Pythagoras or Trigonometry (SOH CAH TOA).

2) Open the attachment and you'll understand.

Maths Help.doc261.6KB

Reply 1

jonny23563

Pythagoras = 2 sides and you want the third side.
Trigonometry = angles.

And I don't understand your second question.

Reply 2

Kingbradley6

Pythagoras is used only when there is a right angle in the triangle whereas trignometry is used on any triangle subject to which angles and lengths you know.

From the attachment, example one is neither pythagoras nor trignometry (well it's not part of the SOH CAH TOA). It is the sine rule:

sin(A) / a = sin(B) / b = sin(C) / c

Thus you can't subtract sin(B) from the angles as it isn't an angle. Once you have sin(B) = 0.375..., on your calculator to find the angle you would type:

sin-1(0.375) = 22.1

For example two, you use the sine rule to find angle B and NOT T which the question asks for. Thus, you have two angles and you need the find the third thus you subtract them BOTH from 180.

If you still don't understand please say, I'll try and explain better.

Reply 3

sheldonite

Basically what the above said, if you can use Pythag then use it.

However, for Pythag you need 2 sides in a right angled triangle.

If you dont then you use Trig - whether it be SinCosTan, the Sine Rule or the Cosine rule depends on different circumstances.

Reply 4

Unkempt_One

GFX-K

1) When trying to find a side of a right angled triangle how do you know when to use Pythagoras or Trigonometry (SOH CAH TOA).

Pythagoras is used to find one of the sides under the event that you have two of the sides and need to find the other, for example:

A squared + B squared = hyp. squared.

Which can be rearranged to give:

hyp. squared - B squared = A squared OR

hyp. squared - A squared = B sq.

You use Trigonometry to find a missing side when you have an angle and a side. Or to find an angle when you have two sides. For example, in the case of sine:

Opposite/Hypotenuse = Sine of the angle. (Simply use the sin-1 function to convert it into the actual angle.)

HypotenuseXSine of angle = Opposite

As for your document:

"Why not subtract this from 180 and then 110?" - This is because that is the sine of the angle, not the actual value in degrees of the angle. Before you can use subtracting from 180 degrees, you need to convert it into the actual angle using the sine-1 function. eg. sin-1(0.58) = 35.5 degrees. BESIDES that, I just realised you only need to find that angle in the first place! So you don't really need to use the 180 degrees thing, since you don't have to find angle C!

So as for "Why do you subtract 180 from 98 and then from 39.53882171? Because in example one you don’t have to subtract from 180 and 110", it's because you actually need to find C in this case, therefore, you will need to subtract 98 and 39 from 180 in order to find the angle in question.