Question

find the sine, cosine, and tangent of ∠t.

right triangle tus with right angle at u, tu = 2√3, ts = 5

write your answer in simplified, rationalized form. do not round.

sin(t) =
cos(t) =
tan(t) =

Explanation:

Step1: Find the length of US

In right triangle \( \triangle TUS \), by the Pythagorean theorem \( a^2 + b^2 = c^2 \), where \( c = 5 \) (hypotenuse), \( b = 2\sqrt{3} \) (leg \( TU \)), and \( a = US \) (leg \( US \)). So, \( US^2 + (2\sqrt{3})^2 = 5^2 \).
\( US^2 + 12 = 25 \)
\( US^2 = 25 - 12 = 13 \)
\( US = \sqrt{13} \) (since length is positive)

Step2: Find \( \sin(T) \)

\( \sin(T) = \frac{\text{opposite}}{\text{hypotenuse}} \). For \( \angle T \), opposite side is \( US = \sqrt{13} \), hypotenuse is \( TS = 5 \). So, \( \sin(T) = \frac{\sqrt{13}}{5} \).

Step3: Find \( \cos(T) \)

\( \cos(T) = \frac{\text{adjacent}}{\text{hypotenuse}} \). For \( \angle T \), adjacent side is \( TU = 2\sqrt{3} \), hypotenuse is \( TS = 5 \). So, \( \cos(T) = \frac{2\sqrt{3}}{5} \).

Step4: Find \( \tan(T) \)

\( \tan(T) = \frac{\text{opposite}}{\text{adjacent}} \). For \( \angle T \), opposite side is \( US = \sqrt{13} \), adjacent side is \( TU = 2\sqrt{3} \). Rationalize the denominator: \( \tan(T) = \frac{\sqrt{13}}{2\sqrt{3}} = \frac{\sqrt{13} \times \sqrt{3}}{2\times 3} = \frac{\sqrt{39}}{6} \).

Answer:

\( \sin(T) = \boldsymbol{\frac{\sqrt{13}}{5}} \)
\( \cos(T) = \boldsymbol{\frac{2\sqrt{3}}{5}} \)
\( \tan(T) = \boldsymbol{\frac{\sqrt{39}}{6}} \)