Math help?

Question

if measure AC is 12, i need to find out measure AB and CB.

how do i do that with knowing angle A is 90 degrees and B is 60 and C is 30???

Answer 1

The law of sines states length A/sin(A) = length B/sin(B) = length C/sin(C)

Answer 2

The answer is: AC=12, AB=4rt3, BC= 8rt3
rt is short-hand for "the square root of"

Here's how it works- get a pencil and paper and follow the instructions step by step
1) Draw an equilateral triangle. That's a triangle with 3- 60 degree angles in it. And, the sides are also equal in length.

2) Mark one side "2", to indicate it is 2 units(ft, inches,meters. furlongs, I don't care) in length. Of course the other two sides must also be "2". Please mark them as well. NOTE: I could have used "1", but I'm going to divide by 2 later on, and I would much prefer a whole number after the division as opposed to a stupid fraction.

3)You should now have an equilateral triangle with each side marked 2. Now, from the vertex(that's the top angle) draw a line to the mid-point of the base. You now have 2 triangles, and trust me , each triangle is a 30-60-90 degree triangle . Ignore one of the triangles you've just created and concentrate on the one that you elect to keep. It's base is now "1" unit long, and the Pythagorean Theorem says the other side (that's the line you drew from the vertex) is rt3 units.

What we have shown here is that a 30-60-90- triangle has sides in the ratio of 1:2:rt3. Lesson ends.

In your triangle, AC=12. Ac is the vertical line to the vertex. AB is the base, and BC is the side of the equilateral triangle. It is also the hypoteneuse, since it's opposite the 90 degree angle.

Comparing similar triangles ABC and the one you sketched, these ratios hold true:
12/rt3 =BC/2, from which BC= 24/rt3, =24rt3/3=8rt3
12/rt3=AB/1, AB=12rt3/3 =4rt3

For an isosceles triangle 45-45-90 degrees, the ratios are 1:1:rt2. But that's for another problem!

I hope this is of use to you. You asked a good question. Good luck

Answer 3

Hii, I'm not sure if I'm right but here is how I did it
I draw the picture so you have AC as a straight line and one of the sides and AB as another straight line and one of the side, meaning that CB or BC is the hyphotenuse. So then I used Sin, I have Sin (angle) = Opp/hyp

ok, I first used angle 60 you have Sin (60) = AC/BC and you know AC = 12
then solve for BC = 13.85

then I put Sin(30) = AB/BC, you solved for BC so now solve for AB = 6.93

and if you used the a^2 + b^2 = c^2 you have that (12^2 + 6.93^2) = (13.85^2)

Answer 4

AB = 12/sqrt(3) = 4sqrt (3)
BC = 2AB = 8sqrt(3)

In a 30-60-90 degree triangle, the hypotenuse is twice the shorter leg and the longer leg is = shorter leg *sqrt(3).

Answer 5

ok well the angle directly across from the line is like "angle A" for example right? well the name of the line right across from "angle A" is called "a".
so like you hav "angle B" and "side b" (or as you put it, "AC") and "angle A" but no "side a" you would do this:

a/sineA = b/sineB = c/sineC

just choose two:

a/sineA = b/sineB

and plug nunmbers in

a/sine90 = 12/sine60
a=44. soething i think, there was a neg. sign when i calculated it for some reason, but i know the formula is right
do it, then do it backwards to see if you get the right angles
hope i helped

Answer 6

Use AC which you know.

AC/AB = tan(B) = tan(60)
AC/BC = sin(B) = sin(60)
Also AC/BC = cos(C) = cos(30)

So...

AB = AC/tan(60) = 12/1.73205 = 6.92820
BC = AC/sin(60) = 12/0.86603 = 13.8564

You can confirm this answer by noting that
BC^2 (hypoteneuse) = AB^2 + AC^2 (Pythagoras' theorem)