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# right triangle

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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?

May 20, 2022

#1
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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.

May 20, 2022