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The Angle Between Vectors - Problem 3

Teacher/Instructor Norm Prokup
Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

Let’s try a slightly harder problem. Here I have triangle CAT, find the measure of angle C. All I’m given are the coordinates of the 3 vertices of the triangle. I want to find the measure of this angle here. I can do that using the angle between the vectors, if I define 2 vectors to find the angle between. I’ll define vectors CA and CT.

Let’s look at CA first. In components, CA would be 1 minus -5 or 6 and 8 minus 0, 8. And CT is going to be 10 minus -5, 15 and 8 minus 0, 8. In the formula for the angle between vectors, I need the magnitudes of both of these. So magnitude of CA is the square root of 6² plus 8², 36 plus 64. That’s 100, so the square root of 100 is 10. For the magnitude of CT, I need the square root of 15² or 225 plus 8² or 64. 225 plus 64, that’s 289. 289 is a perfect square. It’s 17², so the square root is 17.

I have everything I need to find the measure of angle C. The cosine of C is going to be CA.CT. The dot product of these two vectors, over the product of the magnitudes; CA and CT. The dot product is 6 times 15, that’s 60 plus 30, 90. Plus 8 times 8, 64. Over CA which is 10, times CT, which is 17. That’s going to be 154 over 170. Angle C is going to be the inverse cosine of 154 divided by 170. Let’s use our calculators to approximate that quickly. Inverse cosine, 154 divided by 170 and I get 25.1 degrees. That’s the measure of angle C; 25.1 degrees. You could find the measures of angle A, and angle T the same way.

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