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I have a right triangle whose base has length 40 cm and whose hypotenuse has length 43 cm.

How can I determine the height and the measures of the remaining two angles?

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The lengths of the sides of a right triangle are related by the Pythagorean Theorem, which states $a^2 + b^2 = c^2$, where $a$ and $b$ are the lengths the two legs and $c$ is the length of the hypotenuse. Using the information you have, you want to solve $ 40^2 + b^2 = 43^2 $ to get $b = \sqrt{249} \approx 15.78$.

Once you have the lengths of all three sides, you can use the Law of Cosines to figure out the missing angles, which states $ x^2 = y^2 + z^2 - 2yz\cos \alpha, $ where $x,y,z$ are the lengths of the legs of the triangle and $\alpha$ is the angle opposite $x$. You'll need to plug in all the side lengths (making sure that $x$ is the length of the side opposite the angle you want to find) and solve for $\alpha$ (using $\arccos$ at the final step).

EDIT: As pointed out in Marty's answer, the Law of Cosines is not needed for a right triangle. Let's call the angle between the base and the hypotenuse $\alpha$. We can use the simpler relationship $\cos \alpha = \frac{40}{43}$, which rearranges to $\alpha = \arccos \frac{40}{43} \approx 21.53^\circ$. Since a triangle has $180^\circ$ in total, the remaining angle must have approximate measure $180^\circ - 90^\circ - 21.53^\circ = 68.47^\circ$.

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    @mohr Oh, yes! As simple as it can be! Although sine rules directly and quite simply leads to the same thing.2013-10-17
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Once you have the three sides (a, b, and hypotenuse c), the angle opposite side a has sine of a/c and cosine b/c and the angle opposite side b has sine of b/c and cosine of a/c.

Use of the cosine formula for a right triangle seems like overkill.

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    I'm not entirely sure why I went to Law of Cosines first. Your approach is certainly much cleaner.2012-11-28