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Let stadium 1 be the one on the left and stadium 2 the one on the right.
Angle above stadium 1 is 72.9° and the angle above stadium 2 is 34.1° using the angle property of alternate angles(because both the ground and the dotted line are parallel).
For the next part we need to use the trigonometric function of tangent.
As tan x = opposite / adjacent,
Tan 72.9°=1500/ adjacent ( the ground from O to stadium 1)
Therefore the adjacent is 1500/tan 72.9°= 461.46 m( to 5 s.f.)
Same for the next angle,
Tan 34.1°=1500/ adjacent ( the ground from O to stadium 2)
Therefore, the adjacent is 1500/tan 34.1° = 2215.49 m (to 5 s.f.)
Thus, the distance between both stadiums is 2215.49-461.46= 1754.03 m
Correcting the answer to whole number gives you 1754 m which is the option C.
Use algebra for such problems
let, Angle COD = x
Angle KOD = y
Angle KPC = z
Given ,. x - y = 61° ( Equation 1 )
x - z = 53° ( Equation 2 )
Subtract 1st equation from 2nd and you'll get :
z - y = 8° ( Equation 3 )
Now since , x + y + z = 180° ( Equation 4 )
Add Equation 3 to Equation 4 and you'll get
x + 2z = 188° ( Equation 5)
From Equation 2 we know that , x -z = 53°
or, x = 53° + z ( Equation 6 )
Put this value of 'x' in Equation 5 and solve for z. you'll get :
(53° + z) + 2z = 188°
or
3z = 188 - 53 = 135°
solving for z we get
z = 45°
put this value of z in Equation 5
x + ( 2 x 45° ) = 188°
or
x = 188° - 90° = 98°
hence , Angle COD = 98°
