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.
a) FALSE (would be TRUE if it said LN(56)-LN(7)=LN(8))
b) TRUE (because LN(x+y)-LN(x+y)=LN[(x+y)/(x+y)]=LN(1)=0)
Answer:
Step-by-step explanation:
Triangle IJK is a right angle triangle.
From the given right angle triangle
JK represents the hypotenuse of the right angle triangle.
With ∠K as the reference angle,
IK represents the adjacent side of the right angle triangle.
IJ represents the opposite side of the right angle triangle.
To determine Cos K, we would apply trigonometric ratio
Cos θ = adjacent side/hypotenuse. Therefore,
Cos K = 36/85
Cos K = 0.4235
Rounding up to the nearest hundredth,
Cos K = 0.42
6 1/12 would be the answer
Rewriting our equation with parts separated
1/3+5+3/4
Solving the fraction parts
1/3+3/4=?
Find the LCD of 1/3 and 3/4 and rewrite to solve with the equivalent fractions.
LCD = 12
4/12+9/12=13/12
Simplifying the fraction part, 13/12,
13/12=11/12
Combining the whole and fraction parts
5+1+1/12=6 1/12
