Arc Length and Area

Consider a central angle q of 137 degrees on a circle of radius 20 centimeters. (a) Find the length of s of the arc that is cut off. (b) Calculate the area A of the sector that is cut off. (c) Determine the area A0 of the shaded region.


Solution: We must measure the angle in radians. q (p /180o) = 0.7611p radians = 2.391= t.

  1. s = rt = 20 (2.391) = 47.82 cm.
  2. A = (1/2)r2t = (1/2) 202(2.391) = 478.2 cm2.
  3. A0 = A - area of the triangle = A - (r cos(q /2) r sin(q /2)) = 478.2 - 136.4 = 342 cm2.

Linear Velocity on a Spinning Circle

Jack is sitting at home in Wilmington on the 34th parallel. Assume the earth has a depth to its center of 3960 miles. How fast in miles per hour is he moving due to the spinning of the earth on its axis?


Solution: Linear velocity is given by v = rv , where r is the radius of the circle and v is the angular velocity in radians per unit time. We need to determine r and v . The hypotenuse, 3960 miles, times cos q gives us r. We find v by fact that the earth completes one revolution (2p radians) every 24 hours.

v = rv = 3960 cos 34 (2p / 24) = 859 mph.

 

Inverse Trigonometric Functions and Triangles


Consider a right triangle with hypotenuse of length 1. Call the length of one leg x, then the other leg has length , by the Pythagorean theorem. The two acute angles are q = sin-1x and b = cos-1x. From this concept we can derive three basic identities:

  1. cos(sin-1x) = .
  2. sin(cos-1x) = .
  3. sin-1x + cos-1x = p / 2.

In each of the following triangles write q explicitly in terms of x.


Solution: First triangle: q = sin-1(x / 7). Second triangle: q = tan-1(3.5 / x) - tan-1(1 / x).