Question

1. determine the measure of ∠q to the nearest tenth of a degree.
2. determine the length of rs to the nearest tenth of a centimetre.
3. determine the length of mn to the nearest tenth of an centimetre.

Explanation:

5.

Step1: Identify trig - ratio

In right - triangle \(PQR\) with right - angle at \(P\), we use the sine ratio \(\sin Q=\frac{opposite}{hypotenuse}\). Here, the side opposite to \(\angle Q\) is \(PR\) and the hypotenuse is \(QR = 19\), and the side adjacent to \(\angle Q\) is \(PQ = 7\). So, \(\sin Q=\frac{PR}{\sqrt{PR^{2}+PQ^{2}}}\), but we can also use \(\cos Q=\frac{PQ}{QR}\) since \(\cos Q=\frac{7}{19}\).

Step2: Calculate the angle

\(Q=\cos^{- 1}(\frac{7}{19})\). Using a calculator, \(Q=\cos^{-1}(\frac{7}{19})\approx68.4^{\circ}\).

Step1: Use the tangent function

In right - triangle \(QRT\), \(\tan41^{\circ}=\frac{QR}{QT}\). In right - triangle \(RST\), \(\tan56^{\circ}=\frac{RS}{ST}\). First, find \(QR\) in \(\triangle QRT\). We know that in \(\triangle QRT\), if we consider the relationship between the angles and sides. Let's use the fact that we can find \(RS\) using the right - triangle \(RST\). We know that \(\tan56^{\circ}=\frac{RS}{ST}\). Also, from the figure, we can use the angle - side relationships. We know that \(\sin56^{\circ}=\frac{RS}{RT}\). First, find \(RT\) in \(\triangle QRT\). In \(\triangle QRT\), \(\cos41^{\circ}=\frac{QR}{RT}\), and \(QR = 8.9\) cm. So, \(RT=\frac{8.9}{\cos41^{\circ}}\). Then in \(\triangle RST\), \(\sin56^{\circ}=\frac{RS}{RT}\), so \(RS = RT\times\sin56^{\circ}\). Since \(RT=\frac{8.9}{\cos41^{\circ}}\), then \(RS=\frac{8.9\times\sin56^{\circ}}{\cos41^{\circ}}\).

Step2: Calculate the length

\(RS=\frac{8.9\times\sin56^{\circ}}{\cos41^{\circ}}\approx\frac{8.9\times0.829}{0.755}\approx9.8\) cm.

Step1: Use the sine function

In right - triangle \(LMN\) with right - angle at \(M\), we know that \(\sin N=\frac{LM}{LN}\). Given \(LN = 15.8\) cm and \(\angle N=54^{\circ}\), and we want to find \(LM\). The formula for \(\sin N\) is \(\sin N=\frac{LM}{LN}\), so \(LM = LN\times\sin N\).

Step2: Calculate the length

Substitute \(LN = 15.8\) cm and \(\sin N=\sin54^{\circ}\approx0.809\) into the formula. \(LM=15.8\times0.809 = 12.8\) cm.

Answer:

b. \(68.4^{\circ}\)

6.