# 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?

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 $180^\circ - 90^\circ - 21.53^\circ = 68.47^\circ$.

• Ok thanks the height is now 15.779.... lets say 15.78. – david Nov 28 '12 at 5:00
• @david One of your equations will be $15.78^2 = 40^2 + 43^2 - 2 \cdot 40 \cdot 43 \cos \alpha$ and one will be $40^2 = 15.78^2 + 43^2 - 2 \cdot 15.78 \cdot 43 \cos \beta$, where $\alpha$ is the angle opposite the side of length $15.78$ and $\beta$ is the angle opposite the side of length $40$. – Austin Mohr Nov 28 '12 at 5:24
• @david The typical way to state the Law of Cosines uses $a,b,c$. I switched to $x,y,z$ because I didn't want to imply that the $c$ in Law of Cosines had to represent the hypotenuse. – Austin Mohr Nov 28 '12 at 5:26
• @mohr Dont you think the sine rule would be better in this case? – Sawarnik Oct 14 '13 at 21:09
• @Sawarnik Simple trigonometry would be even better. See my edit. – Austin Mohr Oct 16 '13 at 22:45

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.

• I'm not entirely sure why I went to Law of Cosines first. Your approach is certainly much cleaner. – Austin Mohr Nov 28 '12 at 6:10