Find AB. Pre calculus

Question

In an isosceles triangle ABC, CA = CV= 14and the median from A has a length 9 Find AB.

Answer 1

First of all, I assume you meant CA = CB = 14.

 

If you draw this triangle, and then add the median, let the point where the median hits BC be called M.  So  CM = MB = 7.

 

So now you have a triangle AMC with sides 9, 14, and 7.  You can use the law of cosines to find angle C:

 

c² = a² + b² -2ab cos C

9² = 14² + 7² - 2*14*7 cos C

81 = 196 + 49 - 196 cos C

81 = 245 - 196 cos C

-164 = -196 cos C

-164 / -196 = cos C

 

We don't need to figure out what C is.  We're going to use the law of cosines again, so all we really need is cos C as is.  Let's leave it as a fraction rather than a decimal, so we have the exact number.  And let's reduce it to make it a little easier to use:

 

cos C = -164 / -196 = 41/49

 

Now let's look at the original triangle.  Again we can use the law of cosines:

 

c² = a² + b² -2ab cos C

 

This time c is AB, and a and b are both 14, so we get

 

c² = 14² + 14² -2*14*14 cos C

c² = 196 + 196 - 2*14*14* 41/49   -- remember, we got cos C from the inside triangle before

c² = 392 - 392 * 41 / 49 = 392 - 328 = 64

 

So c = 8