Question

Find, to the nearest degree, the measure of each angle of the triangle with the given measures of the sides $5\sqrt{5},10,15$

Answer 1

Finding the measure of each angle of the triangle, to the nearest degree.

The given measures of the sides are $5\sqrt{5},10,15$

$\text{Opposite=5}\sqrt{5}\text{cm, hypotanuse=15 cm, and adjacent=10 cm}$

From the trigonometric identities, we can write it as,

$$\begin{gathered} \Rightarrow \sin \theta =\frac{\text{opposite}}{\text{hypotenuse}} \\ \Rightarrow \sin \theta =\frac{5\sqrt{5}}{15} \\ \Rightarrow \sin \theta =\frac{\sqrt{5}}{3} \\ \Rightarrow \theta ={\sin }^{-1}\left(\frac{\sqrt{5}}{3}\right) \\ \Rightarrow \theta =47.{73}^o \end{gathered}$$

Then,

$$\begin{gathered} \Rightarrow \cos \theta =\frac{\text{adjacent}}{\text{hypotenuse}} \\ \Rightarrow \cos \theta =\frac{10}{15} \\ \Rightarrow \cos \theta =0.66 \\ \Rightarrow \theta ={\cos }^{-1}\left(0.66\right) \\ \Rightarrow \theta =53.54° \end{gathered}$$

And

$$\begin{gathered} \Rightarrow \tan \theta =\frac{\text{opposite}}{\text{adjacent}} \\ \Rightarrow \tan \theta =\frac{5\sqrt{5}}{10} \\ \Rightarrow \tan \theta =\frac{\sqrt{5}}{2} \\ \Rightarrow \theta ={\tan }^{-1}\left(\frac{\sqrt{5}}{2}\right) \\ \Rightarrow \theta =47.{98}^o \end{gathered}$$

Therefore, the measure of each angle of the triangle is $47.73°,47.{98}^o,\text{and}53.54°$.