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Private: Learning Math: Measurement
Indirect Measurement and Trigonometry Homework
Session 5, Homework
Problem H1
At a distance of 160 m from a tower, you look up at an angle of 23 degrees and see the top of the tower:
What is the height of the tower?
Problem H2
Compute the height of this extremely steep road at point C for the drawing below:
Problem H3
Draw a side view of the flight path for a glider whose glide angle is 5 degrees. What is the glide ratio?
Problem H4
One glider has a glide ratio of 1:40, while another has a glide angle of 3 degrees. Which glider flies farther? Explain why.
Problem H5
Suppose that a glider has a glide ratio of 1:40. It is flying over a village at an altitude of 230 m, and it’s 9 km from an airstrip. Can it reach the airstrip? Explain.
Problem H6
An electricity line pole makes an angle of 75 degrees with the road surface, as shown below:
How much does the road rise over a horizontal distance of 100 m?
Problem H7
- Your friend places a mirror 30 ft. from the base of a tall tree. Then she steps back from the mirror until she sees the top of the tree in the mirror’s center. What can be said about the angle formed from the treetop to the mirror to the base of the tree, and the angle formed from her head to the mirror to the base of her feet? What do you know about the other angles in the triangles formed below?
- How might this information be used to determine the height of the tree?
- You know that your friend is 6 ft. tall and that the mirror is 30 ft. from the base of the tree. After your friend moves back 4 ft. from the mirror, she can see the treetop’s reflection. How tall is the tree?
Take It Further
Problem H8
Pretend that you are standing at the equator at noon one day, and the Sun’s rays are directly overhead (casting no shadow). Meanwhile, your friend, who is located 787 km away, calls you and tells you that at that very moment the Sun is casting a shadow, and that he had measured and calculated that the Sun’s rays are coming in at 7.2 degrees. Knowing that there are 360 degrees around the Earth from its center point, use this information to estimate the Earth’s circumference. Compare this estimate of the Earth’s circumference to today’s known value of 40,075.16 km.
Draw some pictures. The 7.2-degree angle will be opposite the length of 787 km. Fifty of these 7.2-degree angles give a complete circle.
Notes
Note 10
This problem has a great deal of history behind it. Eratosthenes is credited with using this approach to calculate the circumference of the Earth. He concluded that the only explanation for why no shadows fell at Syene at midday on June 21 but did fall at Alexandria was because of the curvature of the Earth. He then devised a method for finding the circumference of the Earth that is based on constant-ratio calculations involving proportions. On the day of the summer solstice, he measured the direction of the Sun’s rays as they struck an obelisk in Alexandria and an angle between them. This angle (we’ll call it ∠A) can be compared to 360 degrees, and this ratio can in turn be used in the following proportion:
∠A = Distance Between Alexandria and Syene
360° Circumference of Earth
He measured ∠A to be 8 degrees and the distance from Alexandria to Syene as 4,800 Greek stadia (a stadia corresponds to approximately 606.75 ft.). He then set up a proportion similar to the one above. Eratosthenes’s approximate measurement for the circumference of the Earth was very close to today’s modern value.
Solutions
Problem H1
Using a calculator, we see that tan 23° = 0.42. Therefore, 0.42 = h/160, so h = 160 • 0.42, or 67.2 m.
Problem H2
Using a calculator, we see that tan 12° = 0.21. Therefore, 0.21 = h/10, so h = 10 • 0.21, or 2.1 m.
Problem H3
The tangent of a 5-degree angle is 0.0875. This is a glide ratio of about 1:11.4, so the glider flies 11.4 m for every meter it drops.
Problem H4
The tangent of a 3-degree angle is 0.0524. This is a glide ratio of about 1:19, which is much less than 1:40, so the glider with ratio 1:40 flies more than twice as far.
Problem H5
The distance the glider can travel is 230 • 40 = 9,200 m, or 9.2 km. So yes, the glider can reach the airstrip.
Problem H6
The angle of 75 degrees means that the road rises at a 15-degree angle. Tan 15o is about 0.27, which equals h/100; therefore, h is about 27 m.
Problem H7
- The two angles are equal. They are known as the angle of incidence and the angle of reflection, and from physics we know they are equal. Also, since we know they are both right triangles, the angle at the top of the tree is equal to the angle at your friend’s head.
- Since we’ve determined that the angles in the triangles are the same, they are similar triangles. We can set up the following proportion to help us find the height of the tree:
- Using this proportion, we get the following:
x/6 = 30/4
The height of the tree is 45 ft.
Problem H8
Since 50 of these 7.2-degree angles give a complete circle and we know the length between where the Sun is overhead and where it is at an angle, we can use this to approximate the circumference of the Earth:
360 ÷ 7.2 = 50
50 • 787 = 39350 km
It’s pretty close! Note 10
