Question

law of sines
solve $\triangle abc$. give lengths to the nearest tenth and write angles in degrees and minutes.
1.
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$$\begin{tikzpicture} \\coordinate (p) at (-2,0); \\coordinate (h) at (0,0); \\coordinate (k) at (0.5,1); \\draw (p) -- (h) -- (k) -- (p); \ ode at (p) below left {$p$}; \ ode at (h) below right {$h$}; \ ode at (k) above {$k$}; \ ode at (p) below {$113^\\circ$}; \ ode at (h) below {$14$ yd}; \ ode at (k) above {$36^\\circ$}; \\end{tikzpicture}$$

2.
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$$\begin{tikzpicture} \\coordinate (s) at (-2,1); \\coordinate (r) at (0,0); \\coordinate (t) at (2,0); \\draw (s) -- (r) -- (t) -- (s); \ ode at (s) above left {$s$}; \ ode at (r) below left {$r$}; \ ode at (t) below right {$t$}; \ ode at (s) above {$28^\\circ$}; \ ode at (r) below {$10$ yd}; \ ode at (t) below {$25^\\circ$}; \\end{tikzpicture}$$

3.
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$$\begin{tikzpicture} \\coordinate (h) at (-2,0); \\coordinate (k) at (0,0); \\coordinate (p) at (-1,1); \\draw (h) -- (k) -- (p) -- (h); \ ode at (h) below left {$h$}; \ ode at (k) below right {$k$}; \ ode at (p) above {$p$}; \ ode at (h) below {$17$ in}; \ ode at (k) below {$38^\\circ$}; \ ode at (p) above {$108^\\circ$}; \\end{tikzpicture}$$

4.
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$$\begin{tikzpicture} \\coordinate (p) at (0,0); \\coordinate (r) at (2,0); \\coordinate (q) at (-0.5,1); \\draw (p) -- (r) -- (q) -- (p); \ ode at (p) below left {$p$}; \ ode at (r) below right {$r$}; \ ode at (q) above left {$q$}; \ ode at (p) below {$110^\\circ$}; \ ode at (q) left {$13$ ft}; \ ode at (r) right {$30$ ft}; \\end{tikzpicture}$$

5.
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$$\begin{tikzpicture} \\coordinate (p) at (-2,0); \\coordinate (h) at (0,0); \\coordinate (k) at (0.5,1); \\draw (p) -- (h) -- (k) -- (p); \ ode at (p) below left {$p$}; \ ode at (h) below right {$h$}; \ ode at (k) above {$k$}; \ ode at (p) below {$25^\\circ$}; \ ode at (h) below {$20$ m}; \ ode at (k) above {$127^\\circ$}; \\end{tikzpicture}$$

6.
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$$\begin{tikzpicture} \\coordinate (p) at (0,0); \\coordinate (h) at (2,0); \\coordinate (k) at (0.5,1); \\draw (p) -- (h) -- (k) -- (p); \ ode at (p) below left {$p$}; \ ode at (h) below right {$h$}; \ ode at (k) above {$k$}; \ ode at (p) below {$123^\\circ$}; \ ode at (q) left {$27$ mi}; \ ode at (r) right {$44$ mi}; \\end{tikzpicture}$$

Explanation:

Let's solve the first triangle (△PHK) as an example.

Step1: Find the third angle

In a triangle, the sum of angles is \(180^\circ\). So, \(\angle P=180^\circ - 113^\circ - 36^\circ = 31^\circ\)

Step2: Apply the Law of Sines

The Law of Sines states \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\). Let's find \(PH\) and \(PK\).
For \(PH\): \(\frac{PH}{\sin 36^\circ}=\frac{HK}{\sin 31^\circ}\), \(HK = 14\) yd. So, \(PH=\frac{14\sin 36^\circ}{\sin 31^\circ}\)
\(\sin 36^\circ\approx0.5878\), \(\sin 31^\circ\approx0.5150\)
\(PH=\frac{14\times0.5878}{0.5150}\approx\frac{8.2292}{0.5150}\approx15.98\approx16.0\) yd
For \(PK\): \(\frac{PK}{\sin 113^\circ}=\frac{HK}{\sin 31^\circ}\), \(\sin 113^\circ=\sin(90^\circ + 23^\circ)=\cos 23^\circ\approx0.9205\)
\(PK=\frac{14\times0.9205}{0.5150}\approx\frac{12.887}{0.5150}\approx25.0\) yd

Now let's solve the second triangle (△RST):

Step1: Find the third angle

\(\angle R = 180^\circ - 28^\circ - 25^\circ = 127^\circ\)

Step2: Apply the Law of Sines

For \(RS\): \(\frac{RS}{\sin 25^\circ}=\frac{RT}{\sin 28^\circ}\), \(RT = 10\) yd
\(RS=\frac{10\sin 25^\circ}{\sin 28^\circ}\), \(\sin 25^\circ\approx0.4226\), \(\sin 28^\circ\approx0.4695\)
\(RS=\frac{10\times0.4226}{0.4695}\approx\frac{4.226}{0.4695}\approx9.0\) yd
For \(ST\): \(\frac{ST}{\sin 127^\circ}=\frac{RT}{\sin 28^\circ}\), \(\sin 127^\circ\approx0.7986\)
\(ST=\frac{10\times0.7986}{0.4695}\approx\frac{7.986}{0.4695}\approx17.0\) yd

(We can continue this process for other triangles following the same steps: find the missing angle first, then apply the Law of Sines to find the missing sides.)

Answer:

For △PHK: \(\angle P = 31^\circ\), \(PH\approx16.0\) yd, \(PK\approx25.0\) yd
For △RST: \(\angle R = 127^\circ\), \(RS\approx9.0\) yd, \(ST\approx17.0\) yd
(And similar solutions for other triangles by following the Law of Sines steps: find the third angle, then use \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\) to find missing sides.)