Question

Determine the length of the leg of a 45o – 45o – 90o triangle with a hypotenuse length of 15 inches. ( ANSWER NEEDS TO BE IN REDUCED RADICAL FORM )

Answer 1

Answer:

The lenghts of both legs: $\frac{15\sqrt{2}}{2}\ inches$

Step-by-step explanation:

By definition, when a triangle has angles that measures 45°, 45° and 90°, its legs are congruent.

Then, knowing the lenght of the hypotenuse, we can find the lenght (in inches) of  any leg of the given triangle by applying the Trigonometric Identity $sin\alpha=\frac{opposite}{hypotenuse}$:

$sin(45\°)=\frac{leg}{15}\\\\leg=\frac{15}{\sqrt{2}}$

Finally, simplifying, we get:

$leg=\frac{15(\sqrt{2})}{(\sqrt{2})(\sqrt{2})}=\frac{15\sqrt{2}}{2}$