We know $ an 60^\circ = \sqrt{3} $, so:
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- Question: Compute $ an 120^\circ $ using the unit circle and trigonometric identities.
- Solution: The angle $ 120^\circ $ lies in the second quadrant. Using the identity $ an(180^\circ - heta) = - an heta $, we write:
- an 120^\circ = an(180^\circ - 60^\circ) = - an 60^\circ
- an 120^\circ = -\sqrt{3}
- Final answer: $ oxed{-\sqrt{3}} $
- Question: In a triangle with side lengths 13, 14, and 15 units, find the length of the longest altitude.
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