1. What is cos(20 ) csc(70 )? This is a review of the complementary angle theorem that you learned about last time.

Transcription

1 Math 121 (Lesieutre); 8.1, cont.; Novemer 10, What is cos(20 ) csc(70 )? This is a review of the complementary angle theorem that you learned aout last time. cos(20 ) csc(70 ) = cos(20 ) sec(90 70 ) = cos(20 ) sec(20 ) = Solve triangle ABC elow First find the missing side, using the Pythagorean theorem: = = = 25 = 5. Now, let θ e the angle on the left, and ψ e the angle at the top. We see cos θ = 12 1 ( ) 12 θ = cos 1. 1 You can use tan instead, if you prefer, or any other trig function for that matter. For example, tan θ = 5, so θ = ( 12 tan ). The other angle is similar. sin ψ = 12 1 ( ) 12 ψ = sin 1. 1 Using trig identities, you can check that φ + ψ = 90, like you d expect.. A person is standing 50 ft from the ase of a tree. The angle of elevation from the person s feet to the top of the tree is 0 ; find the height of the tree. Here s a triangle representing this situation. 1

2 r 0 50 y The tree is the right edge; the person s feet are the ottom left corner. tan 0 = y tan 0 = y y = 50 feet.. A security camera in a neighorhood ank is mounted on a wall 9 feet aove the floor. What angle of depression should e used if the camera is to e directed to a spot 6 feet aove the floor and 12 feet from the wall? Here s an illustration. θ 6 ft 9 ft 12 ft We want to focus on the triangle at the top. The dashed part of the left edge is 9 6 =. I see tan(θ) = opp adj = 12 = 1, and so ( ) 1 θ = tan Solve triangle ABC elow. (Watch out, it s not a right triangle! This is a preview of the next section.) 2

3 The angles add up to 180, so the third angle is 100. But how to find the sides? The trick is to drop an altitude, as we say in the lingo: 10 a 50 0 c 1 c 2 Now there are two right triangles. We can solve for a: sin 50 = a 10, so a =. Now, sin 0 = a = = sin 0 That gives one side. At last, we need to find oth c 1 and c 2 to get the third side. cos 50 = c 1 10 c 1 = 10 cos 50 cos 0 = c 2 c 2 = cos 0 = sin 0 cos 0 c = c 1 + c 2 = 10 cos 50 + This could e simplified using trig identities, ut let s not. sin 0 cos Suppose you have a compass, and you want to measure the width of a river without crossing it. Can you come up with a method? Here s a hint at one possile method: (Image source: Scout Society)

4 Here s what you do. Identify an oject on the opposite ank: in the picture, you could use the rock. Point the compass directly at it. (If you have a rotating wheel with degrees marked, turn it to record 0 ). Now, walk down the river ank some distance d. Point the compass at the same rock again, and y comparing with your measured north, see how much the angle has changed. That gives you the θ at the opposite ank. Let w denote the width. Looking at the triangle, you have tan θ = d, and so w = d w tan θ = d cot θ. The method I learned in the oy scouts was a little different, and involves no trig. Just walk until θ = 5, counting your steps as you go. Then w = d cot 5 = d. The trig method is etter: maye for some reason you can t walk far enough to get θ = 5, e.g. if the river curves or there s an ostacle on the ank. 7. Consider the rootic arm illustrated elow, which can pivot at the origin and end at the joint. θ 2 a) What are the x and y coordinates of the joint, in terms of and θ 2? This is really a question aout solving the triangle. Let s draw on some more lines to help us out. y 2 θ 2 x 2 y 1 x 1

5 For that, we want to solve the ottom left triangle. It looks like cos = x 1 x 1 = cos sin = y 1 y 1 = sin ) What are the x and y coordinates of the end of the arm, in terms of and θ 2? First, solve the second triangle in gray. Notice that the angle now is actually + θ 2. cos( + θ 2 ) = x 2 x 2 = cos( + θ 2 ) sin( + θ 2 ) = y 2 y 2 = sin( + θ 2 ) The x coordinate is now x 1 + x 2, and the y is y 1 + y 2. So we get x = x 1 + x 2 = cos + cos( + θ 2 ) y = y 1 + y 2 = sin + sin( + θ 2 ) c) ( ) Suppose you want the arm to grasp an oject located at the point (6, 2). What angles and θ 2 could you use to reach this point? (Somewhat surprisingly, this is actually possile to solve exactly (in terms of inverse trig functions), no matter what the point is.) I m not going to work this out. Instead, I refer you to the classic tome A Mathematical Introduction to Rootic Manipulation, availale in full at murray/ooks/mls/pdf/mls9- complete.pdf. You will find on page 116 (of the pdf; 98 according to the numers at the ottom). In the terminology of that ook, we are talking aout inverse kinematics of a planar two-link manipulator. 5