Question

math

Answer 1

You need to use a trigonometric ratio, such as the sine or cosine. Any one you want, really. I'll start with angle C and use the sine ratio.
sine = opposite/hypotenuse
c = sin(11/14)
c = .707
So angle c is approximately 70 degrees. Do you think you can set up a ratio for angle B? You can use sine or cosine or tangent, whatever you want.

Answer 2

Wait, angle A is a right angle, correct?

Answer 3

^ Right.....this is a law of cosines problem to begin with...3 unknown angles....all 3 sides known

Answer 4

I made a mistake, it would be 51.78 degrees. Because we would need to take the arccsin of 11/14, not the sine. And since this is an isosceles triangle we know that two angles (B and C) are congruent, but we can set up this ratio just to be sure.
sin(B) = 11/14
sin(B) = .7857
B = arcsin(.7857)
B = 51.78 degrees. Same as angle C.

Now all three angles are known. Angle A is 90 degrees, since it's a right angle, and the other two are the same. Do you understand how I got this?

Answer 5

11,11,14 are not the sides of a right triangle, but if they are 11,11,11sqrt(2) yes it is a right triangle (<A=90)

Answer 6

The arcsin is just the inverse sine, so you know.

Answer 7

if you want it be a right triangle, take a blade, cut off it into 2 part :v

Answer 8

To solve for angle C we can use the formula

\[\large c^2 = a^2 + b^2 - 2abcos(c)\]

c is the side opposite angle C
a is the side opposite angle A
and b is the side opposite angle B

so we have

\[\large 11^2 = 14^2 + 11^2 - 2(14)(11)cos(C)\]
\[\large 121 = 196 + 121 - 308cos(C)\]
\[\large 121 = 317 - 308cos(C)\]
\[\large -196 = -308cos(C)\]
\[\large cos(C) = 0.636\]
\[\large C = \cos^{-1}(0.636)\]
\[\large C = 50.48^\circ\]

Answer 9

Ah yes, @RadEn is correct. Angle A is not a right angle, which is why my measurements didn't exactly add up. He has the right idea of how to solve this.

Answer 10

Now you can either continue on using this method to solve the other 2 angles...

or you can switch to the law of sines now that you know at least 1 angle

Answer 11

I agree @RadEn but the law of cosines doesn't solve for only right triangles.

Answer 12

I thought sine and cosine ratios only work with right triangles? Am I wrong about that?

Answer 13

How could that be? Here I'll do it to solve using the right triangle method above you had :P

solving for height we get

\[\large 11^2 = a^2 + 7^2\]
\[\large 121 - 49 = a^2\]
\[\large a = \sqrt{72}\]
\[\large a = 8.48\]

If that is true...using sin to solve for angle C we have

\[\large sin(C) = \frac{8.48}{11}\]
\[\large sin(C) = .77138\]
\[\large C = \sin^{-1}(.77138)\]
\[\large C = 50.48^\circ\]

Answer 14

Are you sureeeee @RadEn :)

Answer 15

nove, actually you are right, cosine rule for any triagle not only for right triangle

Answer 16

@Abbles You are correct that sine and cosine ratios only work with right triangles