SOLUTION: Find the interior angles of the triangle whose vertices are at (2,0),(-7,-8) and (-2,-7).

Algebra ->  Length-and-distance -> SOLUTION: Find the interior angles of the triangle whose vertices are at (2,0),(-7,-8) and (-2,-7).      Log On


   

Question 815147: Find the interior angles of the triangle whose vertices are at (2,0),(-7,-8) and (-2,-7).
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
I hope you have had or are taking Trig because I do not see another way to solve this.

Let's start by naming the points:
A: (2,0), B: (-7,-8) and C: (-2,-7)

The distances between these points will be the lengths of the sides of the triangle. We'll use the distance formula to find these lengths:
AB = sqrt%28%282-%28-7%29%29%5E2%2B%280-%28-8%29%29%5E2%29
Simplifying:
AB = sqrt%289%5E2%2B8%5E2%29
AB = sqrt%2881%2B64%29
AB = sqrt%28145%29

AC = sqrt%28%282-%28-2%29%29%5E2%2B%280-%28-7%29%29%5E2%29
Simplifying:
AC = sqrt%284%5E2%2B7%5E2%29
AC = sqrt%2816%2B49%29
AC = sqrt%2865%29

BC = sqrt%28%28-7-%28-2%29%29%5E2%2B%28-8-%28-7%29%29%5E2%29
Simplifying:
AC = sqrt%28%28-5%29%5E2%2B%28-1%29%5E2%29
AC = sqrt%2825%2B1%29
AC = sqrt%2826%29

Now that we have the lengths of the three sides of the triangle we can start finding the angles. For this we will use the Law of Cosines (since it connects the three sides to an angle):
c%5E2+=a%5E2%2Bb%5E2-2ab%2Acos%28C%29
First I am going to use this to find the largest angle. (I find the largest angle first so that I can choose to use the Law of Sines to find the second angle without having to worry whether the second angle is obtuse or not. There can be no more than one obtuse (or right) angle in a triangle and it will always be the largest angle (which we will have already found).)

The largest angle is opposite the largest side. Our largest side in AB. The angle opposite AB is C. (Draw a picture if this doesn't make sense.) So the equation we get from the Law of Cosines will be:

Simplifying...
145+=+65+%2B+26-2%2Asqrt%281690%29cos%28C%29
145+=+91-2%2Asqrt%281690%29cos%28C%29
Now we solve for cos(C). Subtract 91:
54+=+-2%2Asqrt%281690%29cos%28C%29
Divide by -2%2Asqrt%281690%29:
54%2F%28-2%2Asqrt%281690%29%29+=+cos%28C%29
Simplifying:
-27%2Fsqrt%281690%29+=+cos%28C%29
Now that we've solved for cos(C) we solve for C. The left side is clearly not a special angle value for cos. So we must get out our calculator. (Note: From this point on we will be using rounded-off decimal approximations)
-27%2F41.11+=+cos%28C%29
-0.6568+=+cos%28C%29
From the inverse cos:
C = 131.06

We could use the Law of Cosines again to find the next angle. But the Law of Sines is a simpler formula (and we've already found an obtuse angle so that is no longer an issue). To find angle A:
sin%28C%29%2FAB+=+sin%28A%29%2FBC
Inserting the known values:
sin%28131.06%29%2Fsqrt%28145%29+=+sin%28A%29%2Fsqrt%2826%29
Solving for sin(A):
sqrt%2826%29%2A%28sin%28131.06%29%2Fsqrt%28145%29%29+=+sin%28A%29
5.10%2A%280.7540%2F12.04%29+=+sin%28A%29
0.3194+=+sin%28A%29
A+=+18.63

We could use either the Law of Cosines or the Law of Sines to find the third angle. But even easier would be to use the fact that the three angles add up to 180:
A + B + C = 180
18.63 + B + 131.06 = 180
B + 149.69 = 180
B = 180 - 149.69
B = 30.31

So the three angles are (approximately): 18.63, 30.31 and 131.06 degrees.