Given: FGKL is a trapezoid, m∠F=90°, m∠K=120°, FK=LK=a Find: The length of midsegment.
1 answer:
Answer:
(3/4)a
Step-by-step explanation:
The angle at K is 120°, so the angle at L is its supplement: 60°. That makes triangle FKL an equilateral triangle with a base of FL = a. The vertex at K is centered over the base, so is a/2 from G.
The midsegement length is the average of GK and FL, so is ...
midsegment = (GK +FL)/2 = (a/2 +a)/2
midsegment = (3/4)a
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Draw a diagram to illustrate the problem as shown in the figure below.
Let h = the height of the hill.
At position A, the angle of elevation is 40°, and the horizontal distance to the foot of the hill is x.
By definition,
tan(40°) = h/x
h = x tan40 = 0.8391x (1)
At position B, Joe is (x - 450) ft from the foot of the hill. His angle of elevation is 40 + 18 = 58°.
By definition,
tan(58°) = h/(x - 450)
h = (x - 450) tan(58°) = 1.6003(x-450)
h = 1.6003x - 720.135 (2)
Equate (1) and (2).
1.6003x - 720.135 = 0.8391x
0.7612x = 720.135
x = 946.0523
From (1), obtain
h = 0.8391*946.0523 = 793.8 ft
Answer: The height of the hill is approximately 794 ft (nearest integer)
Let h = the height of the hill.
At position A, the angle of elevation is 40°, and the horizontal distance to the foot of the hill is x.
By definition,
tan(40°) = h/x
h = x tan40 = 0.8391x (1)
At position B, Joe is (x - 450) ft from the foot of the hill. His angle of elevation is 40 + 18 = 58°.
By definition,
tan(58°) = h/(x - 450)
h = (x - 450) tan(58°) = 1.6003(x-450)
h = 1.6003x - 720.135 (2)
Equate (1) and (2).
1.6003x - 720.135 = 0.8391x
0.7612x = 720.135
x = 946.0523
From (1), obtain
h = 0.8391*946.0523 = 793.8 ft
Answer: The height of the hill is approximately 794 ft (nearest integer)