53. A 60-foot drawbridge is 24 feet above water level when closed. When opened the bridge makes an angle of 33 degrees with the horizontal.
a. How high is the tip P of the open bridge above the water?
b. When the bridge is open, what is distance from P to Q?
53. A 60-foot drawbridge is 24 feet above water level when closed. When opened the bridge makes an angle of 33 degrees with the horizontal.
a. How high is the tip P of the open bridge above the water?
b. When the bridge is open, what is distance from P to Q?
Let the open bridge form a trapezoid ABPQ, with bases AB and PQ
AB = 60
AP = QB = 30
height is h such that h/30 = sin 33° = .5446
h = 16.3 ft
PQ = 60 - 2*30*cos33° = 9.7 ft
Hey Im late by 10 years but whatever
This one is pretty simple, lets draw it out:
P Q
/
/
/
/
/————————
A B
WaitI didnt finish the last one :((((
BUt anyways you know that the angle at the bottom left A is 33 degrees and the actual draw bridge itself is 60 degrees so if we use a bit of trig we can find:
sin 33 = h/60 since sin = opposite of hypoteneuse
with some basic algebra we find h = 60 * sin 33 which is about 32.67, when the bridge is closed its 24 feet above water so add the 32.67 and 24 and viola you get your answer ~ 56.7 feet above water
Does the drawbridge open at the middle, with both ends tilting upward? Where are P and Q?
Omg I meant sin = opposite over hypotenuese
the abreviation is soh cah toa if you want that, anyways happy studying
Oh yeah for part b.
Assumably the drawbridge all the way down is the same as it is up, so the most bottom line of you diagram
__________(end of triangle)________(bottom length same position of p and q)
that entire length is 60ft so first we find the base of the triangle portion, which is tan 33 = 32.67/x
which gets to about 50.3 the remaining portion is 60 - 50.3 which results to around 9.7 ft
sorry for splitting up my answers so much, but goodbye now and have a nice day or night