Triangle abc is such that ab=4 and ac=8 if m is the midpoint of bc and am=3, what is the length of bc?
1 answer:
This is an interesting question. I chose to tackle it using the Law of Cosines.
AC² = AB² + BC² - 2·AB·BC·cos(B)
AM² = AB² + MB² - 2·AB·MB·cos(B)
Subtracting twice the second equation from the first, we have
AC² - 2·AM² = -AB² + BC² - 2·MB²
We know that MB = BC/2. When we substitute the given information, we have
8² - 2·3² = -4² + BC² - BC²/2
124 = BC² . . . . . . . . . . . . . . . . . . add 16, multiply by 2
2√31 = BC ≈ 11.1355
AC² = AB² + BC² - 2·AB·BC·cos(B)
AM² = AB² + MB² - 2·AB·MB·cos(B)
Subtracting twice the second equation from the first, we have
AC² - 2·AM² = -AB² + BC² - 2·MB²
We know that MB = BC/2. When we substitute the given information, we have
8² - 2·3² = -4² + BC² - BC²/2
124 = BC² . . . . . . . . . . . . . . . . . . add 16, multiply by 2
2√31 = BC ≈ 11.1355
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Answer:
Step-by-step explanation:
If the roots are 1 + 5i and 1 - 5i, then you need the factors that result from those roots. They are (x - 1 + 5i) and (x - 1 - 5i). Now what you do with those is FOIL them out. Doing that gives you the following:
(what a mess, huh?)
The good thing is that several of those terms cancel each other out. +5ix cancels out the -5ix; -5i cancels out the 5i; and the 2 -x terms combine to -2x. That leaves you with:
Obviously you're in the section in math that deals with complex (imaginary) numbers so you should know that i-squared is equal to -1. Making that replacement:
a = 1, b = -2, c = 25