The 45 yd line represents the hor.
side of a rt. triangle. The 65 yd
line represents the hyp. of the rt.
triangle. The 63 deg. angle lies
between the hyp. and the 45 yd line.
Sin63 = d/65
d = 65 Sine 63 = 59.92 yds
d = r * t
r = d/t = 59.92/48 = 1.21 yds / sec.
From a window in room 7, one can observe the front door of Phillips Hall, which is about 45 yards away. A turn of the head trough an angle of 63 degrees in a counterclockwise direction enables one to see the front door to Jeremiah Smith Hall which is about 65 yds away. If you spot a friend coming out of that door and they proceed directly to Phillips Hall, making the trip in 48 seconds, what would you estimate to be your friend's average speed during the walk?
3 answers
CORRECTION:
d = 65 Sine 63 = 57.92 yds
r = d/t = 57.92/48 = 1.21 yds / sec.
d = 65 Sine 63 = 57.92 yds
r = d/t = 57.92/48 = 1.21 yds / sec.
After further analysis, I noticed that the conditions of the problem did not
create a rt. triangle. So I had to
redo the problem using the Law of Cosines. There is only a small difference in the results, but this
is the right procedure:
(Cos A = b^2 + c^2 - a^2) / 2bc.
(0.4540 = (45)^2 + (65)^2 - a^2) /
2 * 45 * 65,
Cross multiply:
(45)^2 + (65)^2 - a^2 = 0.454*2*45*65,
Solve for a:
6250 - a^2 = 2655.9
a = 59.95 yds. = Dist. traveled
d = r * t
r = d / t
r = 59.95 / 48 = 1.25 yds / sec.
create a rt. triangle. So I had to
redo the problem using the Law of Cosines. There is only a small difference in the results, but this
is the right procedure:
(Cos A = b^2 + c^2 - a^2) / 2bc.
(0.4540 = (45)^2 + (65)^2 - a^2) /
2 * 45 * 65,
Cross multiply:
(45)^2 + (65)^2 - a^2 = 0.454*2*45*65,
Solve for a:
6250 - a^2 = 2655.9
a = 59.95 yds. = Dist. traveled
d = r * t
r = d / t
r = 59.95 / 48 = 1.25 yds / sec.