Draw a right triangle with a height of 30,000 ft. and a base of 40,000 ft. The hypotenuse, or distance travelled, is then 50,000ft using the Pythagorean Theorem. Then dividing distance by speed will give us time, which is 200 seconds, or 3 minutes and 20 seconds.

We can use the Pythagorean Theorem to solve for x.

9² + x² = 15²

81 + x² = 225

x² = 144

x = 12

\(\displaystyle Area= \frac{1}{2}\times base\times height\)

\(\displaystyle 42=\frac{1}{2}\times base\times 12\)

\(\displaystyle 42=6\times base\)

\(\displaystyle base=7\)

Use the Pythagorean Theorem. Let a = 8 and c = 10 (because it is the hypotenuse)

\(\displaystyle \small a^2+x^2=c^2\)

\(\displaystyle \small 8^2+x^2=10^2\)

\(\displaystyle \small 64+x^2=100\)

\(\displaystyle \small x^2=100-64=36\)

\(\displaystyle \small x=6\)

This is a \(\displaystyle 5-12-13\) triangle. We can find the value of the other leg by using the Pythagorean Theorem.

Remembering that

\(\displaystyle sin=\frac{opposite}{hypotenuse}\).

Thus,

\(\displaystyle a^2+12^2=13^2\)

\(\displaystyle a^2+144=169\)

\(\displaystyle a^2=25\)

\(\displaystyle a=5\).

If \(\displaystyle sin(x)=\frac{12}{13}\), you know the adjacent side is \(\displaystyle 5\).

Thus, making

\(\displaystyle tan(x)=\frac{opposite}{adjacent}=\frac{12}{5}\) because tangent is opposite/adjacent.

Given two angles of unequal measure in a triangle, the side opposite the greater angle is longer than the side opposite the other angle. Since \(\displaystyle m \angle A > m \angle B\), it follows that \(\displaystyle BC > AC\). Also, in a right triangle, the hypotenuse must be the longest side; in \(\displaystyle \bigtriangleup ABC\), since \(\displaystyle \angle C\) is the right side, this hypotenuse is \(\displaystyle \overline{AB}\). It follows that \(\displaystyle AB > BC > AC\), and that (I) is the only statement that can possibly be true.

AB is the leg adjacent to Angle A and BC is the leg opposite Angle A.

Since we have a \(\displaystyle 30^{\circ}-60^{\circ}-90^{\circ}\) triangle, the opposites sides of those angles will be in the ratio \(\displaystyle x-x\sqrt{3}-2x\).

Here, we know the side opposite the sixty degree angle. Thus, we can set that value equal to \(\displaystyle x\sqrt{3}\).

\(\displaystyle 4=x\sqrt{3}\)

\(\displaystyle \frac{4}{\sqrt{3}}=\frac{x\sqrt{3}}{\sqrt{3}}\)

\(\displaystyle \frac{4}{\sqrt{3}}\cdot \frac{\sqrt{3}}{\sqrt{3}}=x\)

\(\displaystyle x=\frac{4\sqrt{3}}{3}\)

which also means

\(\displaystyle \overline{BC}=\frac{4\sqrt{3}}{3}\)

The ladder leaning against the wall forms a right triangle. The hypotenuse of the triangle is 5 ft., the length of the ladder. 

Because sin x= opposite/hypotenuse, sine of the angle is equal to the length of the side opposite the angle divided by the hypotenuse. In this case, the length of the side opposite the angle is h, the height of the end of the ladder that is touching the wall. Thus,

\(\displaystyle \sin x^{\circ} = \frac{h}{5}\)

Because we are told that 

\(\displaystyle \sin x^{\circ}=\frac{3}{5}\)

we know that h=3. Therefore, 3 feet is the height of the ladder. If the ladder is falling at a rate of 6 feet per second, we can find the number of seconds it will take the ladder to hit the ground with the equation

\(\displaystyle h=6s\)

where h represents the height the ladder is falling from, and s represents the number of seconds it takes the ladder to fall. We can now solve for s: 

\(\displaystyle s=\frac{h}{6}\)

\(\displaystyle s=\frac{3}{6}=\frac{1}{2}=0.5\)

It takes the ladder 0.5 seconds to fall to the ground. 

The figure shows a right triangle. The acute angles of a right triangle have measures whose sum is \(\displaystyle 90^{\circ }\), so

\(\displaystyle m \angle A+ m \angle C = 90^{\circ }\)

Substituting \(\displaystyle 45^{\circ }\) for \(\displaystyle m \angle C\):

\(\displaystyle m \angle A+ 45^{\circ } = 90^{\circ }\)

\(\displaystyle m \angle A= 45^{\circ }\)

This makes \(\displaystyle \bigtriangleup ABC\) a 45-45-90 triangle. By the 45-45-90 Triangle Theorem, the length of leg \(\displaystyle \overline{BC}\) is equal to that of hypotenuse \(\displaystyle \overline{AC}\), the length of which is 20. Therefore,

\(\displaystyle BC = \frac{AC}{\sqrt{2}} = \frac{20}{\sqrt{2}}\)

Rationalize the denominator by multiplying both halves of the fraction by \(\displaystyle \sqrt{2}\):

\(\displaystyle BC = \frac{20 \cdot \sqrt{2}}{\sqrt{2} \cdot \sqrt{2}} = \frac{20 \sqrt{2}}{2} = 10 \sqrt{2}\)

Because DC and AB are parallel, this means that angles CDB and ABD are equal. When two parallel lines are cut by a transversal line, alternate interior angles (such as CDB and ABD) are congruent.

Now, we can show that triangles ABD and BDC are similar. Both ABD and BDC are right triangles. This means that they have one angle that is the same—their right angle. Also, we just established that angles CDB and ABD are congruent. By the angle-angle similarity theorem, if two triangles have two angles that are congruent, they are similar. Thus triangles ABD and BDC are similar triangles.

We can use the similarity between triangles ABD and BDC to find the lengths of BC and CD. The length of BC is proportional to the length of AD, and the length of CD is proportional to the length of DB, because these sides correspond.

We don’t know the length of DB, but we can find it using the Pythagorean Theorem. Let a, b, and c represent the lengths of AD, AB, and BD respectively. According to the Pythagorean Theorem:

 a² + b² = c²

15² + 20² = c²

625 = c²

c = 25

The length of BD is 25.

We now have what we need to find the perimeter of the quadrilateral.

Perimeter = sum of the lengths of AB, BC, CD, and DA.

Perimeter = 20 + 18.75 + 31.25 + 15 = 85

The answer is 85.