The number of hours of daylight, H, on day t of any given year (on January 1, tequals1) in a particular city can be modeled
by the function H(t)equals13 plus 8.2 sine [StartFraction 2 pi Over 365 EndFraction (t minus 77 )]. a. March 18, the 77th day of the year, is the spring equinox. Find the number of hours of daylight in the city on this day. b. June 18, the 169th day of the year, is the summer solstice, the day with the maximum number of hours of daylight. Find the number of hours of daylight in the city on this day. c. December 18, the 352nd day of the year, is the winter solstice, the day with the minimum number of hours of daylight. Find the number of hours of daylight in the city on this day.
1 answer:
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Answer:
Height of mound = 794 ft
Step-by-step explanation:
To illustrate the angle of elevation and distance, i have drawn it and attached below.
Now, from my diagram;
h = the height of the mound
At his first point of his trip to the foot of the mound, the angle of elevation is 40°, while the horizontal distance to the foot of the mound is "X"
So, by triangle definition,
tan(40°) = h/x
And so;
h = x tan40
h = 0.8391x - - - - (eq 1)
At his second point of the trip to the foot of the mound, Joe is now,
"(x - 450) ft" from the foot of the mound.
Thus, his angle of elevation is 40 + 18 = 58°.
So, by triangle definition,
tan(58°) = h/(x - 450)
h = (x - 450)•(tan(58°))
h = 1.6003(x - 450)
h = 1.6003x - 720.135 - - - - -(eq2)
To get the height(h) of the mound, let's equate (eq1) to (eq2).
0.8391x = 1.6003x - 720.135
1.6003x - 0.8391x = 720.135
0.7612x = 720.135
x = 720.135/0.7612
x = 946.0523 ft
Let's put this value for x in eq (1);
h = 0.8391 x 946.0523 = 793.83 ft ≈ 794ft
