A dense metal sphere is dropped from a 10-meter tower, and at the exact same time an identical metal sphere is thrown horizontal
2 answers:
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Answer:
Length of third side = 3.97m
Explanation:
First of all, let we draw the triangle from the given information assuming our first corner is A. second corner is B and third corner is C.
From A-B we have distance = 5.5m = Say it c
From B-C we have distance = 4.3m = Say it a
From A-C we have distance = ? = Say it b which we have to find out.
Using the Law of Cosines: The square of the unknown side equals to the sum of squares of other 2 sides and subtracting 2*(Product of other sides)*(Cos(Angle opposite to the unknown side)
For our case it is:
b² = a² + c² - 2acCos(B) - Say it equation 1
From the attached triangle you may see that, a & c are our known sides and B is the angle opposite to the side b.
There values are:
a = 4.3m; c = 5.5m ; Angle B = 0.8 rad = 0.8 * 57.3 = 45.84 degrees where 1 rad = 57.3 degrees
Now by putting the respectve values in equation 1 we have:
b² = (4.3)² + (5.5)² - 2*4.3*5.5*Cos(45.84)
b² = 18.49 + 30.25 - 32.95
b² = 15.79
b = √15.79
b = 3.97m
Thus the length of third side is 3.97m.
PS: The picture of triangle is being attached for yours understanding.
