Hopefully I can explain this without all the fancy text.

I have my 3 corners. A= 30º B=?

I know that B is 60 since both sides add up to 90º

It tells me to find sides a and b.

The hypotenuse c= 25
The opposite side a= ?
The adjacent side b= ?

Now I can sort of follow the Soh Cah and Toa.

The first step says a/25= Sin 30º
(that would be Soh)


then a= 25(Sin 30º) this is because I multiplied each side by 25 and ended up with a=Sin 30*25

I got that.


a=25(.5) = 12.5 I know the .5 is found on chart. Got this.


Now this is where I'm confused. I'm trying to find b.

It says b/25= Cos 30º

Why is this Cos 30º? Do you ever use the other degrees in a problem or it just the one given? W

How do I know when the problem is Sin 30, Cos 30 or Tan 30? What is the relationship between ABC and abc? Can you mix the degrees and angles?

I know the a^2 + b^2= c^2
I know Soh, Cah, Toa
I know < a + < b = 90º

What am I missing?

Do you need the explanation using trig? If not, and any method of solving will work, I can explain that, but I won't if you need trig. I no longer remember trig, and I don't want to add a bunch of facts and figures to confuse you if you need to solve using trig.

An explanation of trig would be great.

Then, unfortunately, I can't help.




But the answers are 12.5 and 12.5(square root of 3), if I remember correctly.

Well, I told you 12.5 was the first answer.
The second isn't 12.5. It's 21.65


I just am not sure why they solved it by b/25= Cos 30º. Why would they multiply the b/25 by Cos 30º. Cos is adjacent divided by hypotenuse. Would that make the letter a on the opposite side a Cosine? I'm confused.

Thanks for the try Dorian.

All 30-60-90° triangles have sides with a ration of 1:2: the square root of 3. The square root of 3 is 1.73205081. That figure multiplied by 12.5 is 21.650635125, which, for your purposes, is 23.65. Memorize the first sentence and memorize the value of the square root of 3. Both will come in handy. Essentially, I am showing you a quick way to solve some problems using ratiios instead of trig. (My father was a carpenter, and didn't know trig, but he could look at problems dealing with angles and give me the answer. He showed my how to do it.)

I just use this.

quote:

Originally posted by clarebear:
The first step says a/25= Sin 30º
(that would be Soh)
..........
Now this is where I'm confused. I'm trying to find b.

It says b/25= Cos 30º

Why is this Cos 30º? Do you ever use the other degrees in a problem or it just the one given? W

How do I know when the problem is Sin 30, Cos 30 or Tan 30? What is the relationship between ABC and abc? Can you mix the degrees and angles?

In the same angle of a right triangle, sin30 = cos60 in the trig table. Also, sin60 = cos30 in the trig table. I think that this is true, unless my memory has failed me. So confirm this to be true in the trig table.

Therefore, given the two smaller angles of a right triangle, when one angle is sin30 (or cos60) the other angle is cos30 (or sin60).

Pick two numbers which add up to 90, for example, 20 and 70. If you use sin20 for one angle, the other angle has to be cos20. If you use cos70 for one angle, the other angle has to be sin70. I did not here change the order of the angles, so I think. (Again, sin20 = cos70 for the first same angle, and cos20 = sin70 for the second same angle. Look at this in the trig table.)

The trig table values for sin and cos are the same, only in opposite directions.

By some miracle, I may have gotten all this correct.

In order to solve a right triangle you need to know 2 values. The length of one of the sides and one of the angles is enough to solve by using the a^2 + b^ = c^2. But if you know one of the angles you automatically know the other angle and you still need to know one of the sides in order to solve the triangle, Obviously a 1,2,sqrt 3, could have any size, so you need at least one length in order to solve for the rest of the measurements.

If you know the smaller angle, you know the other angle and can use either one. If you choose the angle A and you know the opposite side a, you can choose sine A = opposite/hyp. to solve for the hyp dimension, if you choose cos A=adjacent/hypothenuse you can solve for the dimension of the hypothenuse. If you choose the tangent function, tanA=Opp/adj, you can solve for the adjacent side. So as long as you know one of the lengths of any side and one of the angles you can solve for the rest of the right triangle. I think some of the others here may have said the same thing. I hope the cumulative effect proves useful.

Thank you all soooo much for your help.

I passed the test!!