Question

1. find sr, st, rt, m∠trs, and m∠rst. round to the nearest tenth, if necessary.

Explanation:

Step1: Determine the coordinates

Assume \(T(-4,-4)\), \(R(-4,4)\), \(S(2,1)\)

Step2: Calculate \(SR\)

Use the distance formula \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\)
For \(S(2,1)\) and \(R(-4,4)\)
\(SR=\sqrt{(2 + 4)^2+(1 - 4)^2}=\sqrt{36 + 9}=\sqrt{45}\approx6.7\)

Step3: Calculate \(ST\)

For \(S(2,1)\) and \(T(-4,-4)\)
\(ST=\sqrt{(2 + 4)^2+(1 + 4)^2}=\sqrt{36+25}=\sqrt{61}\approx7.8\)

Step4: Calculate \(RT\)

For \(R(-4,4)\) and \(T(-4,-4)\)
\(RT=\sqrt{(-4+4)^2+(4 + 4)^2}=\sqrt{0 + 64}=8\)

Step5: Calculate \(m\angle TRS\)

Use the tangent function. Let's consider the right - triangle formed.
\(\tan\angle TRS=\frac{ST}{RT}\) (This is wrong, we should use vectors or the law of cosines. Let's use the law of cosines: \(c^{2}=a^{2}+b^{2}-2ab\cos C\). For \(\triangle TRS\), \(ST^{2}=SR^{2}+RT^{2}-2(SR)(RT)\cos\angle TRS\)
\(\cos\angle TRS=\frac{SR^{2}+RT^{2}-ST^{2}}{2(SR)(RT)}=\frac{45 + 64-61}{2\times\sqrt{45}\times8}=\frac{48}{16\sqrt{45}}=\frac{3}{\sqrt{45}}\approx0.447\)
\(m\angle TRS\approx63.4^{\circ}\)

Step6: Calculate \(m\angle RST\)

Since the sum of angles in a triangle is \(180^{\circ}\) and we know one angle (\(\angle RTS = 56^{\circ}\), assume from the grid - based right - triangle relation (using the fact that in the lower sub - triangle with angle \(56^{\circ}\), and using the law of sines \(\frac{RT}{\sin\angle RST}=\frac{ST}{\sin\angle TRS}=\frac{SR}{\sin\angle RTS}\))
\(\sin\angle RST=\frac{RT\sin\angle RTS}{ST}\)
\(\sin\angle RST=\frac{8\times\sin56^{\circ}}{7.8}\)
\(\sin\angle RST=\frac{8\times0.829}{7.8}\approx0.85\)
\(m\angle RST\approx58.3^{\circ}\)

Answer:

\(SR\approx6.7\), \(ST\approx7.8\), \(RT = 8\), \(m\angle TRS\approx63.4^{\circ}\), \(m\angle RST\approx58.3^{\circ}\)