In right triangle ABC, we have AB = 10, BC = 20, and angle ABC = 90 degrees. If M is on AC such that BM is an altitude of triangle ABC, then what is cos angle ABM?
+9519 Note that by the Pythagorean theorem,
\(AC = \sqrt{10^2 + 20^2} = 10\sqrt5\)
Now, consider the interior angle sum of \(\triangle ABM\). We have \(\angle ABM + 90^\circ + \angle BAC = 180^\circ\). That means \(\angle ABM = 90^\circ - \angle BAC\)
Therefore,
\(\begin{array}{rcl} \cos \angle ABM &=& \sin \angle BAC\\ &=& \dfrac{\boxed{\phantom{\text{aa}}}}{\boxed{\phantom{\text{aa}}}}\\ &=& \boxed{\phantom{\text{aaaa}}} \end{array}\)
Fill in the blanks. You should be able to arrive at the conclusion with the help of SOHCAHTOA.
