A baseball diamond is a square with sides 224 m The pitchers

A baseball diamond is a square with sides 22.4 m. The pitcher\'s mound is 16.8 m from home plate on the line joining home plate and second base. How far is the pitcher\'s mound from first base? Richard is standing between two buildings in a townhouse development. The building on the left is 9 m away and the angle of elevation to its security spotlight A, is 68 degree. The building on the right is 6 m away and the angle of elevation to its security spotlight, B, is 73 degree. Which spotlight is farther away from Richard, and by how much? A security camera needs to be set so that its angle of view includes the area from a doorway to the edge of a parking lot. The doorway is 16 m from the camera. The edge of the parking lot is 24 m from the camera. The doorway is 28 m from the edge of the parking lot. What angle of view is needed for the camera?

Solution

Since the triangle formed by the pitchers mound, home plate, and first base is not a right triangle, the Pythagorean theorem cannot be used. Instead, the law of cosines gives:

c^2 = a^2 + b^2 - 2abcosC

Note the angle between the line from home to first and the line from home to the pitcher\'s mound is 45 degrees. The line from the pitcher\'s mound to first base is the side of the triangle opposite the 45 degrees angle. Entering the given values into the law of cosines formula:

c^2 = a^2 + b^2 - 2abcosC

= 22.4^2 + 16.8^2 - 2 * 22.4 * 16.8 * cos (45)

=> c = 15.9 m

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let the distances be a and b.

Using trigonometric identities

cos (theta ) = adjacent/hypotenuse

=> cos ( 68 degrees) = 9/a

and

cos (73 degrees) = 6/b

we get a = 24.0

and b = 20.5

So spotlight A is farther away by 3.5 m

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Using the Law of Cosines:

a = 24 m.
b = 28 m.
c = 16 m.

CosB = (a^2 + c^2 - b^2)/2ac
CosB = (576+256-784) / 768 = 0.0625

B = cos^-1(0.0625)
B = 86.4 degrees. = Angle of view needed for the camera.