First note that [latex]\triangle ABC[/latex] is not a right triangle, so we cannot use the trig ratios directly to find the sides of this triangle.
The unknown distance [latex]d[/latex] is the side opposite [latex]\angle B = 79.4°{.}[/latex] In order to use the law of sines, we must know another angle and the side opposite that angle. We know that side [latex]c = 400{,}[/latex] and we can compute the angle at the ship, [latex]\angle C{.}[/latex] [latex]\angle C = 180° -(79.4° + 83.2°) = 17.4°[/latex]. Now we apply the law of sines, using angles [latex]B[/latex] and [latex]C{.}[/latex] [latex]\dfrac{b}{\sin \angle B} = \dfrac{c}{\sin \angle C}[/latex]. Substitute the given values. [latex]\dfrac{d}{sin 79.4°} = \dfrac{400}{sin 17.4°}[/latex]. Evaluate sines. [latex]\dfrac{d}{0.9829} = \dfrac{400}{0.2990}[/latex]. Multiply both sides by 0.9829. [latex]d = 1315[/latex]

The ship is about 1315 yards from the observer at [latex]A{.}[/latex]

We use the law of sines with [latex]a[/latex] and angle [latex]\angle A[/latex] to find [latex]b{.}[/latex]
[latex]\dfrac{a}{\sin \angle A} = \dfrac{b}{\sin \angle B}  \ \ \  \text{Substitute the given values.} \\ \dfrac{11}{\sin 37°} = \dfrac{b}{\sin 54°} \ \ \   \text{Multiply both sides by sin 54°.} \\ \dfrac{11}{\sin 37°} \cdot \sin 54° = b[/latex]

Evaluating this expression with a calculator, we find that [latex]b \approx 14.79{.}[/latex]
The angle [latex]\angle C = 180° -(37° + 54°) = 89°{.}[/latex] Now we use the law of sines to find side [latex]c{.}[/latex] Note that it is safer to use the given side, [latex]a{,}[/latex] rather than the value we calculated for [latex]b{.}[/latex]
[latex]\dfrac{a}{\sin \angle A} = \dfrac{c}{\sin \angle C} \ \ \text{Substitute the known values.}\\ \dfrac{11}{\sin 37°} = \dfrac{c}{\sin 89°} \ \ \text{Multiply both sides by} \sin 89°.\\ \dfrac{11}{\sin 37°}\cdot \sin 89° = c[/latex]

so [latex]c \approx 18.28.[/latex]

We must find the three remaining parts of the triangle, [latex]\angle A[/latex], [latex]\angle C{,}[/latex] and [latex]c{.}[/latex] First, we use the law of sines to find [latex]\sin \angle A{.}[/latex]
[latex]\dfrac{\sin \angle A}{a} = \dfrac{\sin \angle B}{b} \ \text{Substitute the known values.} \\ \dfrac{\sin \angle A}{5} = \dfrac{\sin 55°}{11} \ \text{Multiply both sides by 5.} \\ \sin \angle A = 5 \cdot \dfrac{\sin 55°}{11} \approx 0.3723[/latex]

So [latex]\angle A = \sin^{-1}(0.3723) \approx 21.9°{.}[/latex] Now we know two angles, and we can find angle [latex]\angle C{.}[/latex]
[latex]\angle C = 180° - (\angle B + \angle A) = 180° - 55° - 21.9° = 103.1°[/latex]
Finally, we use the law of sines to find side [latex]c{.}[/latex]
[latex]\dfrac{b}{\sin \angle B} = \dfrac{c}{\sin \angle C} \ \text{Substitute the known values.}\\ \dfrac{11}{\sin 55°} = \dfrac{c}{\sin 103.1°} \ \text{Multiply both sides by sin 103.1°.}\\ \dfrac{11}{\sin 55°} \cdot \sin 103.1° = c[/latex]

Using the law of sines, we have

[latex]\dfrac{\sin \angle A}{a} = \dfrac{\sin \angle B}{b} \ \text{Substitute the known values.} \\ \dfrac{\sin \angle A}{8} = \dfrac{\sin 14.4°}{3} \ \text{Multiply both sides by} 8.\\ \sin \angle A = 8 \cdot \dfrac{\sin 14.4°}{3} \approx 0.6632[/latex]

There are two angles with sine [latex]0.6632{:}[/latex]
the acute angle [latex]\angle A = \sin^{-1}(0.6632) = 41.5°,[/latex]
or its supplement, [latex]\angle A^{ \prime} = 180° - 41.5° = 138.5°{.}[/latex]
Each of these angles produces a different solution triangle, because the angle [latex]\angle C[/latex] and side [latex]c[/latex] will be different also.
In the first case, angle [latex]\angle C = 180° - (14.4° + 138.5°) = 27.1°{,}[/latex] and we have the triangle shown in figure [latex]a[/latex].
In the second case, [latex]\angle C = 180° - (14.4° + 41.5°) = 124.1°{,}[/latex] which gives us the triangle shown in figure [latex]b[/latex].

Notice that [latex]h[/latex] is one side of the right triangle [latex]ADC{.}[/latex] If we can find its hypotenuse, labeled [latex]r[/latex] in the figure, we can use the sine ratio to find [latex]h{.}[/latex] To find [latex]r{,}[/latex] we consider a second triangle, [latex]\triangle ABC{,}[/latex] as shown below.
In this triangle, we know side [latex]BC = 20[/latex] and would like to find side [latex]AC = r{.}[/latex] We can use the law of sines to find [latex]r{,}[/latex] but first we must calculate the other angles of the triangle.
Now, the angle opposite [latex]r, \angle ABC{,}[/latex] is the complement of [latex]18.5°{,}[/latex] so
[latex]\angle ABC = 180° - 18.5° = 161.5°[/latex]
The angle opposite the 20-yard side, [latex]\angle BAC{,}[/latex] is
[latex]\angle BAC = 180° - (161.5° + 15.9°) = 2.6°[/latex]
Now we can apply the law of sines to find [latex]r{.}[/latex] We have [latex]\dfrac{r}{sin 161.5°} = \dfrac{20}{sin 2.6°}[/latex] Solve for [latex]r{.}[/latex]
[latex]r = \sin 161.5° \cdot \dfrac{20}{sin 2.6°} \approx 139.9[/latex]
So [latex]r[/latex] is about 139.9 yards. Finally, using the right triangle [latex]ADC[/latex] and the definition of sine, we can write
[latex]\dfrac{h}{r} = \sin 15.9°[/latex]. Solve for [latex]h[/latex].
[latex]h = r \cdot \sin 15.9° \approx 38.33[/latex]

The castle is about 38.33 yards tall.

We let [latex]x[/latex] represent the distance to the asteroid. The asteroid and the two observers make an isosceles triangle with base 1000 km and equal sides of length [latex]x{,}[/latex] as shown below.
The base angles of the triangle are both [latex]\dfrac{180° - 0.001°}{2} = 89.999°{.}[/latex] By the law of sines,
[latex]\dfrac{x}{sin 89.999°} = \dfrac{1000}{sin 0.001°}[/latex] Solve for x.
[latex]x = \sin 89.999° \cdot \dfrac{1000}{sin 0.001°}[/latex]
[latex]x \approx 57,000,000[/latex]

The asteroid is about 57 million kilometers from Earth (roughly one-third of the distance to the Sun).

In the figure below, the star Wolf 359 is located at point [latex]O{.}[/latex] Our Sun is located at point [latex]C{,}[/latex] halfway between the two observation points at [latex]A[/latex] and [latex]B{.}[/latex] Thus, the distance between the observation points is twice the distance from the Earth to the Sun, or 2 AU.
The altitude from [latex]O[/latex] to side [latex]\overline{AB}[/latex] forms a right triangle [latex]\triangle ACO[/latex] and bisects the angle at [latex]O{.}[/latex] Thus
[latex]\angle AOC = \dfrac{1}{2}(0.85^{\prime} = 0.425^{\prime}[/latex]
We use the definition of tangent to find
[latex]\tan 0.425^{\prime \prime} = \dfrac{opposite}{adjacent}[/latex]
[latex]\tan (0.425° / 3600) = \dfrac{1}{x}[/latex]. Solve for x.
[latex]x = \dfrac{1}{\tan 0.425° / 3600} \approx 485,000[/latex]

The star is approximately [latex]485,000[/latex] AU from Earth, or nearly half a million times the distance from Earth to the Sun.