Solving Right Triangle Given Two Sides 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?
 A: 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 $180^\circ - 90^\circ - 21.53^\circ = 68.47^\circ$.
A: 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.
