Question

solve the right triangle
write your answers in simplified, rationalized form. do not round
st=
su=
m\angle t=

Explanation:

Step1: Find \(m\angle T\)

In a right - triangle, the sum of angles is \(180^{\circ}\). Since \(\angle U = 90^{\circ}\) and \(\angle S=30^{\circ}\), then \(m\angle T=180^{\circ}-90^{\circ}-30^{\circ}=60^{\circ}\)

Step2: Find \(SU\)

We know that \(\sin S=\dfrac{TU}{ST}\) and \(\cos S = \dfrac{SU}{ST}\), also for a \(30 - 60-90\) triangle, if the side opposite to \(30^{\circ}\) is \(a\), the side opposite to \(60^{\circ}\) is \(a\sqrt{3}\) and the hypotenuse is \(2a\). Here \(TU = 10\) (opposite to \(30^{\circ}\)).
We use the formula \(\tan S=\dfrac{TU}{SU}\), since \(S = 30^{\circ}\) and \(TU = 10\), \(\tan30^{\circ}=\dfrac{1}{\sqrt{3}}=\dfrac{TU}{SU}\), then \(SU=\dfrac{TU}{\tan30^{\circ}}=\dfrac{10}{\dfrac{1}{\sqrt{3}}}= 10\sqrt{3}\div1\times\dfrac{1}{1}\) (rationalizing) \(SU=\dfrac{10\sqrt{3}}{3}\)

Step3: Find \(ST\)

Using the Pythagorean theorem \(ST^{2}=SU^{2}+TU^{2}\), or using the sine formula \(\sin S=\dfrac{TU}{ST}\), since \(S = 30^{\circ}\) and \(TU = 10\), \(\sin30^{\circ}=\dfrac{1}{2}=\dfrac{10}{ST}\), then \(ST = 20\) (wrong, correct way: using \(30 - 60-90\) triangle ratio).
Another way: \(\cos S=\dfrac{SU}{ST}\), \(SU=\dfrac{10\sqrt{3}}{3}\), \(\cos30^{\circ}=\dfrac{\sqrt{3}}{2}\), \(\dfrac{\sqrt{3}}{2}=\dfrac{\dfrac{10\sqrt{3}}{3}}{ST}\), cross - multiply: \(\sqrt{3}\times ST=\dfrac{20\sqrt{3}}{3}\), \(ST=\dfrac{20\sqrt{3}}{3}\)

Answer:

\(m\angle T = 60\)，\(SU = \dfrac{10\sqrt{3}}{3}\)，\(ST = \dfrac{20\sqrt{3}}{3}\)