We begin by drawing a picture. Let \(\displaystyle P\) stand for where the plane reaches a maximum altitude of 31,000 ft. We can also assume that if we draw a line straight down from the plane that it will be perpendicular to the ground. This forms a right triangle.

Now we will use our knowledge of right triangles.  We know the trigonometric identity, \(\displaystyle \sin \left ( \theta \right )=\frac{opposite}{hypotenuse}\).  We can plug 55 in for our \(\displaystyle \theta\) and 31,000 in for the opposite side. Solving for the hypotenuse is solving for the distance from the point of takeoff to the plane when it reaches maximum altitude.

\(\displaystyle \sin \left ( \theta \right )=\frac{opposite}{hypotenuse}\)

\(\displaystyle \sin \left ( 55 \right )=\frac{31,000}h{}\)

\(\displaystyle h=\frac{31,000}{\sin \left ( 55 \right )}\)

\(\displaystyle h=37,844\)

And so the distance from point \(\displaystyle A\) to the plane at maximum altitude is 37,844 ft.

We begin by drawing a picture.  Let point be where your eyes are looking on the billboard and point be where you are standing on the ground.

We must solve for the height of the billboard.  We have information of the adjacent side of the right triangle formed and we wish to know the height, which will be the opposite side from you.  Using trigonometric identities, we will use .  We can plug in 500 ft for our adjacent side and 67 degrees for our .

And the height of the section of the billboard you are looking at is 1,178 ft above the ground.

The important thing to note for this problem is that we are working with similar triangles.  Your shadow forms a right triangle with its three points being your feet, the top of your head, and the top of your shadow’s head.  We have enough information to find the angle of your head’s elevation to your shadow on the ground.  Just like your shadow forms a right triangle to your body, a similar right triangle will also be formed to the sun’s altitude.  We know that this triangle is similar because it shares the angle opposite your body and it shares a right angle, so it’s third angle must also be equal.  Since it shares the angle opposite your body, we must only solve for the angle of your head’s elevation and it will be the same as the sun’s elevation.

We do this by using .  We can plug in 60cm for the opposite since that is your height, and 66cm for the adjacent since that is the length of the shadow.  We are solving for .

And so the sun is at an elevation of 48 degrees.

We don't have this triangle's height to use the formula \(\displaystyle Area=\frac{1}{2}base\cdot height\), so we'll need to use trigonometry to find the area of this triangle. We can use the formula \(\displaystyle R=\frac{a^2\sin B\sin C}{2\sin A}=\frac{b^2\sin A\sin C}{2\sin B}=\frac{c^2\sin A\sin B}{2\sin C}\), where R is the area of a triangle. This formula comes from the fact that \(\displaystyle Area=\frac{1}{2}base\cdot height\) and also incorporates triangle trigonometry and the Law of Sines. Applying the formula, we get:

\(\displaystyle \angle B = 180^\circ-(27^\circ+13^\circ)=140^\circ\)

\(\displaystyle R=\frac{c^2\sin A\sin B}{2\sin C}\)

\(\displaystyle R=\frac{22^2\sin 27^\circ\sin 140^\circ}{2\sin 13^\circ}\)

\(\displaystyle R=313.9cm^2\)

We don't have this triangle's height to use the formula \(\displaystyle Area=\frac{1}{2}base\cdot height\), so we'll need to use trigonometry to find the area of this triangle. We can use the formula \(\displaystyle R=\frac{a^2\sin B\sin C}{2\sin A}=\frac{b^2\sin A\sin C}{2\sin B}=\frac{c^2\sin A\sin B}{2\sin C}\), where R is the area of a triangle. This formula comes from the fact that \(\displaystyle Area=\frac{1}{2}base\cdot height\) and also incorporates triangle trigonometry and the Law of Sines. Applying the formula, we get:

\(\displaystyle \angle C = 180^\circ-(27^\circ+13^\circ)=140^\circ\)

\(\displaystyle R=\frac{c^2\sin A\sin B}{2\sin C}\)

\(\displaystyle R=\frac{22^2\sin 27^\circ\sin 13^\circ}{2\sin 140^\circ}\)

\(\displaystyle R=38.4cm^2\)

Since we are not given the height of this triangle, we'll need to use trigonometry to find its area. If you draw and label the triangle, you'll notice that we are given Side-Side-Angle, which is the ambiguous case, and therefore there may be two different values that satisfy \(\displaystyle \angle B\). Let's start by finding those.

Use the Law of Sines to find

\(\displaystyle \sin\angle B=\frac{b\sin\angle A}{a}=\frac{215\sin17^\circ}{100}=.6286< 1\)

Because this value is less than 1, there are two values that satisfy \(\displaystyle \angle B\). We can now find these using inverse trig and knowing that \(\displaystyle \angle B'=180^\circ-\angle B\). We have:

\(\displaystyle \angle B=\sin^-^1(.6286)=38.9^\circ\) and \(\displaystyle \angle B'=180^\circ-38.9^\circ=141.1^\circ\)

Because we have two values that satisfy \(\displaystyle \angle B\), we'll also have two angles that satisfy \(\displaystyle \angle C\), and two unique triangles with two unique areas.  Each of these is found below.

\(\displaystyle \angle C=180^\circ-(\angle A+\angle B)=180^\circ-(17^\circ+38.9^\circ)=124.1^\circ\)

 \(\displaystyle \angle C'=180^\circ-(\angle A+\angle B')=180^\circ-(17^\circ+141.1^\circ)=21.9^\circ\)

Next, we can use the formula \(\displaystyle R=\frac{a^2\sin B\sin C}{2\sin A}=\frac{b^2\sin A\sin C}{2\sin B}=\frac{c^2\sin A\sin B}{2\sin C}\), where R is the area of a triangle. This formula comes from the fact that \(\displaystyle Area=\frac{1}{2}base\cdot height\) and also incorporates triangle trigonometry and the Law of Sines. Applying the formula, we get:

\(\displaystyle R=\frac{a^2\sin B\sin C}{2\sin A}\)

\(\displaystyle R=\frac{100^2\sin 38.9^\circ\sin 124.1^\circ}{2\sin 17^\circ}\)

\(\displaystyle R=8892.64m^2\)

 Now, find \(\displaystyle R'\) for our second triangle. 

\(\displaystyle R'=\frac{a^2\sin B\sin C}{2\sin A}\)

\(\displaystyle R'=\frac{100^2\sin 141.1^\circ\sin 21.9^\circ}{2\sin 17^\circ}\)

\(\displaystyle R'=4005.56m^2\)

Since we are not given the height of this triangle, we'll need to use trigonometry to find its area. If you draw and label the triangle, you'll notice that we are given Side-Side-Angle, which is the ambiguous case, and therefore there may be two different values that satisfy \(\displaystyle \angle B\). Let's start by finding those.

Use the Law of Sines to find

\(\displaystyle \sin\angle B=\frac{b\sin\angle A}{a}=\frac{95\sin40^\circ}{125}=.4885\)

\(\displaystyle \angle B=sin^-^1(.4885)=29.2^\circ\)

While this is the ambiguous case, there is only one solution if the side opposite the given angle (a=125 ft) is equal to or greater than the other given side (b=95 ft). Because \(\displaystyle 125>95\), there is only one value of \(\displaystyle \angle B\) and only one solution to this triangle's area. Therefore, we can continue onward to find this triangle's area. 

\(\displaystyle \angle C=180^\circ-(\angle A+\angle B)=180^\circ-(40^\circ+29.2^\circ)=110.8^\circ\)

Next, we can use the formula \(\displaystyle R=\frac{a^2\sin B\sin C}{2\sin A}=\frac{b^2\sin A\sin C}{2\sin B}=\frac{c^2\sin A\sin B}{2\sin C}\), where R is the area of a triangle. This formula comes from the fact that \(\displaystyle Area=\frac{1}{2}base\cdot height\) and also incorporates triangle trigonometry and the Law of Sines. Since we know a and b both, we could use either of the first two formulas to find the area. You'll get the same answer either way; feel free to test the other one to see this shown. Applying the formula, we get:

\(\displaystyle R=\frac{a^2\sin B\sin C}{2\sin A}\)

\(\displaystyle R=\frac{125^2\sin 29.2^\circ\sin 110.8^\circ}{2\sin 40^\circ}\)

\(\displaystyle R=5543.04ft^2\)

Using the formula for area of a triangle equal to \(\displaystyle \frac{1}{2}base\cdot height\), drawing \(\displaystyle \bigtriangleup ABC\) and labelling its sides, angles, and height h, then using triangle trigonometry and substitution, we can derive the formulae \(\displaystyle R=\frac{1}{2}ab\sin C=\frac{1}{2}ac\sin B=\frac{1}{2}bc\sin A\), where R is equal to area. This can be used to find the area of a triangle when we know two of its sides and the included angle. Plugging in, we get:

\(\displaystyle R=\frac{1}{2}bc\sin A=\frac{1}{2}\cdot 31\cdot 12\sin 44^\circ=129.2\)

Therefore the area of this triangle is 129.2 square inches.

Using the formula for area of a triangle equal to \(\displaystyle \frac{1}{2}base\cdot height\), drawing \(\displaystyle \bigtriangleup ABC\) and labelling its sides, angles, and height h, then using triangle trigonometry and substitution, we can derive the formulae \(\displaystyle R=\frac{1}{2}ab\sin C=\frac{1}{2}ac\sin B=\frac{1}{2}bc\sin A\), which can be used to find the area of a triangle when we know two of its sides and the included angle. Plugging in, we get:

\(\displaystyle R=\frac{1}{2}ac\sin B=\frac{1}{2}\cdot 11.24\cdot 16.98\cdot \sin88.2^\circ=95.38\)

Therefore, the area of the triangle is 95.38 square centimeters.

Because we do not know the height of this triangle, we need to lean into trigonometry. Luckily, Heron's Formula will help us solve this problem. You can use it when you know three sides of a triangle and want to find the area. Please note that in order to find a triangle's area, you must make sure that the triangle exists by satisfying the Triangle Inequality, which states that the sum of any two sides of the triangle must always be larger than the remaining side. This is true for the lengths 4, 8, and 9 given in this problem. Heron's formula can be described as:

\(\displaystyle R=\sqrt{s(s-a)(s-b)(s-c)}\) where R is the area of the triangle and s is the semiperimeter equivalent to \(\displaystyle s=\frac{1}{2}(a+b+c)\). Using these formulas and plugging in, we get:

\(\displaystyle s=\frac{1}{2}(a+b+c)\)\(\displaystyle =\frac{1}{2}(4+8+9)=\frac{21}{2}=10.5\)

\(\displaystyle R=\sqrt{s(s-a)(s-b)(s-c)}\)

\(\displaystyle =\sqrt{10.5(10.5-4)(10.5-8)(10.5-9)}\)

\(\displaystyle =\sqrt{10.5(6.5)(2.5)(1.5)}\)

\(\displaystyle =\sqrt{255.9375}\)

\(\displaystyle =15.998\)

Therefore the area of \(\displaystyle \bigtriangleup ABC\) is 15.998 meters.