 Teacher resources and professional development across the curriculum

Teacher professional development and classroom resources across the curriculum             A B C Homework Solutions for Session 5 Homework

See solutions for Problems: H1 | H2 | H3 | H4 | H5 | H6 | H7 | H8    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

 a. 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. b. 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: c. 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     