Question

mathematics ii
assessment 2: identical twins special right triangles (show your work please)
find the missing side lengths:
1 - )
2 - )
3 - )
4 - )
5 - )
6 - )
a broadcast antenna is on top of a tower. the signal travels from the antenna to your house so you can watch tv. consider the diagram to calculate the height of the tower and the distance the signal travels.
6 - )
this stamp is from the netherlands. suppose the length of each side of the netherlands stamp is 40 millimeters. use the 30 - 60 - 90 triangle theorem to determine the height of the stamp.
7 - )

Explanation:

Step1: Recall 45 - 45 - 90 triangle ratio

In a 45 - 45 - 90 triangle, the ratio of the sides is $1:1:\sqrt{2}$. That is, if the legs are of length $a$, the hypotenuse $c=a\sqrt{2}$.

Step2: Solve for problem 1

In the first 45 - 45 - 90 triangle with leg length 20. Since the legs of a 45 - 45 - 90 triangle are equal, $y = 20$. And the hypotenuse $x=20\sqrt{2}$.

Step3: Solve for problem 2

In the second 45 - 45 - 90 triangle with hypotenuse $m = 18\sqrt{2}$. Let the leg length be $n$. Using the ratio $c=a\sqrt{2}$, we have $18\sqrt{2}=n\sqrt{2}$, so $n = 18$, thus $m = 18\sqrt{2}$ and $n=18$.

Step4: Solve for problem 3

In the third 45 - 45 - 90 triangle with leg length $y = 7\sqrt{2}$. The hypotenuse $s=y\sqrt{2}=7\sqrt{2}\times\sqrt{2}=14$.

Step5: Recall 30 - 60 - 90 triangle ratio

In a 30 - 60 - 90 triangle, the ratio of the sides is $1:\sqrt{3}:2$. If the shorter leg (opposite 30°) is $a$, the longer leg (opposite 60°) is $a\sqrt{3}$ and the hypotenuse is $2a$.

Step6: Solve for problem 4

In the 30 - 60 - 90 triangle, if the shorter leg $x$ (opposite 30°) and the longer leg $y = 17$, then $x=\frac{17}{\sqrt{3}}=\frac{17\sqrt{3}}{3}$ and the hypotenuse $z=\frac{34}{\sqrt{3}}=\frac{34\sqrt{3}}{3}$.

Step7: Solve for problem 5

In the 30 - 60 - 90 triangle with shorter leg $y = 19$, the longer leg $x = 19\sqrt{3}$ and the hypotenuse $z = 38$.

Step8: Solve for problem 6

In the 30 - 60 - 90 triangle with shorter leg $y=\frac{13}{2}$, the longer leg $x=\frac{13\sqrt{3}}{2}$ and the hypotenuse $z = 13$.

Step9: Solve antenna - tower problem

In the 30 - 60 - 90 triangle with base (shorter leg) 500 ft. The height of the tower (longer leg) $h = 500\sqrt{3}$ ft and the distance the signal travels (hypotenuse) $d = 1000$ ft.

Step10: Solve stamp problem

For an equilateral triangle (which can be divided into two 30 - 60 - 90 triangles) with side length 40 mm. The height of the equilateral triangle (longer leg of the 30 - 60 - 90 triangle) $h = 20\sqrt{3}$ mm.

Answer:

1. $x = 20\sqrt{2}$, $y = 20$
2. $m = 18\sqrt{2}$, $n = 18$
3. $s = 14$, $y = 7\sqrt{2}$
4. $x=\frac{17\sqrt{3}}{3}$, $y = 17$
5. $x = 19\sqrt{3}$, $y = 19$
6. $x=\frac{13\sqrt{3}}{2}$, $y=\frac{13}{2}$
7. Tower height: $500\sqrt{3}$ ft, Signal - travel distance: 1000 ft
8. Stamp height: $20\sqrt{3}$ mm