Question

1. The hypotenuse of a right triangle measties 3V15 centimeters and its shorter legmeasures 3v6 centimeters. What is the measure of the larger acute angle of the triangle?Round your answer to the nearest tenth of a degree.

Answer 1

we have the following information from a right triangle

hypotenuse, h =

$h=3\sqrt{15}$

and, short leg, a =

$a=3\sqrt{6}$

we need o calculate the larger acute angle

1. Let's draw this triangle and identify the angle in question

Angle Θ is the largest acute angle because it's opposite to the larger leg

We can calculate this angle using the following

$\cos \theta=\frac{adjacent}{hypotenuse}=\frac{a}{h}=\frac{3\sqrt{6}}{3\sqrt{15}}$

then

$\theta=\arccos (\frac{\sqrt[]{6}}{\sqrt[]{15}})=50.77$

Rounding to the nearest tenth of a degree:

larger acute angle of the triangle, Θ = 50.8°