Two legs of a right triangle are 2 and the square root of 5. Let angle Abe the smallest angle of triangle ABC. What is cos A?

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- May 29th 2007, 02:42 PM #1

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- Apr 2007
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- May 29th 2007, 02:51 PM #2
you should draw a diagram and fill in what you know. ordinarily i would, but i'm not in the mood right now.

let's find the hypotenuse first.

By Pythagoras' Theorem:

$\displaystyle h^2 = 2^2 + \left( \sqrt {5} \right)^2$

where $\displaystyle h$ is the length of the hypotenuse

$\displaystyle \Rightarrow h^2 = 4 + 5 = 9$

$\displaystyle \Rightarrow h = \sqrt {9} = 3$

Now the smallest angle will be opposite the smallest side, that means it will be adjacent to the longest side (other than the hypotenuse). that means $\displaystyle \sqrt {5}$ is the anjacent side to $\displaystyle \angle A$

$\displaystyle \Rightarrow \cos A = \frac {adj}{hyp} = \frac { \sqrt {5}}{3}$