Home Ed trig question

[Johnsonsfiat]

Is this sin, cos or tan please? I'm seeing hyp over adj and I presume there's no such thing.

Remember

Some Obnoxious Herbert (Sine - Opposite - Hypoteneuse)
Calculated A Horrible (Cosine - Adjacent - Hypoteneuse)
Type Of Algebra (Tangent - Opposite - Adjacent)

You need to have two of the variables in order to use Trigonometry. Marking off the ones you have will tell you what to use.

You also know what the total of the three angles inside a triangle add up to, so you can work out the third angle.

Thanks Mitzi.
Could someone work through this for me so I can find d.

cosine.

If I've interpreted the picture correctly.

You have an angle (10 deg), an adjacent side (9.2 cm) and you want to know the hypotenuse.

cos(angle) = adj/hyp

So if you rearrange,
hyp = adj/cos(angle)

so p = 9.2/cos(10)

In this example you have an angle x, the length of the adjacent side and you want the hypotenuse.

Therefore use cosine as cosine x = adjacent/hypotenuse

In your example:

Cos 10 = 9.2/d

Multiply both sides by d

d x cos 10 = 9.2

Divide both sides by cos 10

Therefore

d = 9.2 / cos 10

Cos - adj/hypotenuse

Thanks Bimbleberries.
Do you always put the unknown over the known or am I just making that up?

Soh (opp/hyp)
Cah (adj/hyp)
Toa (opp/adj)

Problem is- I'm seeing hypotenuse over adjacent and there's no such thing.

What do you mean no such thing?

You have the hypotenuse d (or p?) and the adjacent is 9.2

Cos 10 = 9.2/d
d = 9.2/cos10

Hypotenuse over adjacent is just 1 / cos.
It used to be called secant, but that name is hardly ever used now.

We have told you which way to put the figures
If its cos you put adjacent over hypotenuse and rearrange to find d

You put the unknown wherever it goes in the formulae listed above. In this case it is the hypotenuse and you know the adjacent side, so you use cosine and the unknown is the denominator (bottom) and you rearrange the equations as Merryoldgoat has described to get the answer.

You have your three trigonometric ratios:

Sin x = opposite/hypotenuse

Cos x = adjacent/ hypotenuse

Tan x = opposite/adjacent

Essentially when solving a trig problem you will have two bits of information and an unknown that you are trying to find.

You look at the thee bits of info you have, and use whichever function refers to ALL THREE your problem does.

In this case you have hyp (the unknown), the angle (10 degrees) and the adjacent side (9.2)

So with the angle, the hypotenuse and the adjacent it’s the cosine function you use.

You feed in the info you have and rearrange algebraically to get your answer.

IIRC this is GCSE maths and Cos comes after Tan and Sine.

Does the ‘pupil’ not understand how to solve it?

the formula always work like that, with the variables in the same place. You then rearrange them to get the one you want on one side.

so for cosine:

cos(angle) = adj/hyp.

If you knew the angle and the adjacent side, but wanted hyp, as in this case, you rearrange to get hyp = adj/cos(angle)

If you knew the angle and hypotenuse, but wanted adj, you would rearrange to get adj = hyp x cos(angle)

If you knew adjacent and hypotenuse and wanted the angle, you'd rearrange to get

cos-1 (adj/hyp) = angle

Then put your actual values into the variables in the equation, and use the calculator to find the answer.

I prefer rearranging with the variables/letters, rather than the numbers, as it's shorter to write down, but you can put the numbers in first if you prefer.

Merryoldgoat you explain things so clearly! Are you a maths teacher? If so you must be a good one!

So it's like algebra. You can rearrange things. I'd just got the hang of, for example, Sin = opp over hyp and was ready to get unmotivated DD to work thru some more of the same when she wandered off because I'd got confused. Aaargh

We were taught it as
Soh cah toa to make it easy to remember

Sin = opp/hyp
Cos = adj/hyp
Tan = opp/adj

Plug in the values you have and rearrange to find the missing value

Yes you can rearrange things in all branches of maths (and physics for that matter), and often need to to get the answer, as long as you follow the rules.

The side you know is 9.2, and is adjacent to the known angle.

The side you have to calculate is the hypotenuse.

The relationship between the hypotenuse and the adjacent side is given by the cosine. (cosine = adjacent/hypotenuse)

So in this case:

cos 10 = 9.2/p

Rearrange the formula as you normally would, remembering that the way to rearrange is to always do the same thing to both sides of the equation.

So you start with cos 10 = 9.2/p as above.

Multiply both sides by p. On the rhs the multiplicand by p cancels out the divide by p, so on that side you now have just 9.2. On the other side you have cos 10 x p. So now the equation is cos 10 x p = 9.2.

You want p on its own, but the side with p in it is also x cos 10. So now you divide both sides by cos 10. As before but in reverse, the division by cos 10 on the lhs cancels out the multiply, leaving p. On the rhs dividing by cos 10 gives 9.2/cos 10.

So voila! p = 9.2/cos 10.

Cos 10 is 0.839, so p = 9.2/0.839, which is 10.965.

These handy triangles make it easy. You are looking for the H in CAH, so the rest of that triangle is an A over a C. This means you do Adjacent divided by Cos of the angle. The triangles mean you don't have to do any rearranging, because they give you all three equations in one go.