Geometry Forum - Problem of the Week
Solutions - Alley Hoop, June 6-10, 1994
Sonia Timberlake
____________________
My name is Sonia Timberlake. I'm a 7th grader at Odle Middle
school in Bellevue, but I take a morning math course at Sammamish
High School, which is where I was given this problem by my math
teacher on Tuesday morning (yesterday). I came up with the
approximate answer of10.77, and here is how:
A: /:B This is the diagram I made. The capital letters are
:l / : names of points- the point in the middle- where the hoop
: l / : would be hung- is point H. Lines HF and HC are
: l / : perpendicular to AE and BD, respectively. Line AD
: l / : represents the 30-foot ladder, and line BE the 20-foot.
F:____l_____:C Using pythagoras in the triangle ADE, you can determine
: /l : the length of AE; 20 root 2. Using Pythagoras in the
: / 1l : triangle EBD, you can determine the length of BD; 10
: / 1 l : root 3
: / 1 l :
E:/____1___l:D Now you see that triangle EBD's lengths are those of a
30-60-90 triangle, with angle BED being 60 degrees,
G so triangle EGH is also 30-60-90. GH is named x.
Because of the laws of a 30-60-90 triangle, we know that EG has the
length x divided by root 3, and EH has the length 2x divided by root 3.
Then, since EFHG is a rectangle, FH also has length x divided by root
3, and EF is also x. Now we know that length AF is 20 root 2 minus a.
Now, we go back and find the measure of angle EAD. The sine
of this angle is equal to 1/3, so using a calculator, you can find that
the angle is approximately 19.5 degrees.
Then, in the triangle AFH, you can say that the tangent of 19.5=
FH divided by AF. We already know the lengths of these two sides in
terms of a, and we can find out the tangent of 19.5, and solve the
equation, coming up with the answer a=10.77
Paul Curcio
____________________
The height of the basket would be about 10.74225 feet. If you write equations for the lines of the two ladders, the thirty-foot ladder's equation is y = ((sqrt800)/10)x, and the twenty-foot ladder's is y = ((sqrt300)/-10)x - (sqrt300)/10. By putting these equations into a graphing calculator and callating the intersect, the y of the intersection is 10.74225.
Susan Quan
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I found the equation for each line and then found the intersection. The equation of the 30 ft. ladder with coordinates (10,0) and (0,20*2) is 2*2 x + y = 20*2; the equation of the 20 ft. ladder with coordinates at (0,0) and (10,10*3) is y = 10*3. By substitution, y = 16*3 - 12*2 or about 10.74 ft. The basketball hoop is about 10.74 ft. above the ground.
Karen Brown
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The 20 ft. ladder has endpoints at (0,10*3) and the 30 ft. ladder has endpoints at (0,0) and (10,20*2). The equation of the 20 ft. ladder is *3 x + y = 10*3 and the equation of the 30 ft. ladder is 2*2 x - y = 0. Solving by elimination gives x = -6 + 4*6 and since y = 2*2 x it follows that y = 2*2(-6+4*6) or -12*2 +16*3. So y is about 10.74 ft.and this is the height of the hoop.
Meera Shah
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Label corner of the 20 foot ladder and the 10 foot alley, A. Label corner of the 30 foot ladder and the 10 foot alley, B. Label corner where the 20 foot ladder meets the upper wall, C. Label corner where the 30 foot ladder meets the wall, D. Label the point where the ladders meet, E. Triangle ABC is a 30-60-90 triangle since the hypotenuse is double the shorter leg. Therefore, angle CAB=60, angle C=30, and CB=10*sqrt(3). cos(angle DBA)=10/30; therefore, angle DBA=70.53 degrees. Angle AEB=180-70.53-60= 49.47 degrees. sin(49.47)/10 = sin(60)/BE. BE=11.36 feet. sin(70.53) = x/11.36 where x is the height of the basket. x = 10.74 feet.
Ryan Lansing
____________________
Step 1. Drop a perpendicular from the intersection of the
ladders.
Step 2. Label the line segment from the bottom left corner
to the perpendicular line as x. Label the line segment from
the perpendicular to the right corner as 10 - x.
Step 3. Label the angle made by the 20 foot ladder and the
ground as alpha or A. Label the angle made by the 30 foot
ladder and the ground as beta or B.
Step 4. cos A = 10/20 A = 60 degrees
cos B = 10/30 B = 70.5 degrees
Step 5. Label the perpendicular as y.
Step 6. tan 60 = y/x tan 70.5 = y/(10 - x)
x tan 60 = y (10 - x) tan 70.5 = y
x tan 60 = (10 - x) tan 70.5
x tan 60 + x tan 70.5 = 10 tan 70.5
x (tan 60 + tan 70.5) = 10 tan 70.5
x = 6.2 feet
Step 7. tan 60 = y/6.2 y = 10.74
The basket is 10.74 feet high.
Roger Head
____________________
This is an excellent problem, and easily solved by trigonometry...and
very useful in showing where trig is derived from...or could be
derived from. The answer puts the hoop above 10 feet by my
calculations...10.75 feet or 10 feet 9 inches. Maybe that's what the
hoop should be raised to.
Solution: Using the diagram (which I will not attempt to draw!), drop
a vertical line from the two ladders' intersection to the ground. Call
this line "h", the distance we're looking for. If we look closely, we
can see several "similar" triangles. Two of them are formed by the
line h, the ladders, and the ground. The two large ones are formed
by the ladders, walls and ground. The properties of similar triangles
tell us that the ratios of corresponding sides are equal. Looking
closely at the similar triangles (large and small), the sides
corresponding to h are the wall sides of the large triangles. We need
to know these sides. Using the Pythagorean Theorem, the 30 foot
ladder triangle has a 28.28 ft. side and the 20 foot ladder triangle
has a 17.32 ft. side.
Now we're ready to set up our ratios and solve. One more little
detail...the horizontal sides of the smaller triangles are "x" and "10-x",
since we only know the total length of the two (10 feet).
The large triangle with the 30 foot ladder and its small similar
triangle:
28.28/h = 10/x
The large triangle with the 20 foot ladder and its small similar
triangle:
17.32/h = 10/10 - x
Solving 28.28 x = 10 h and 17.32 (10 - x) = 10 h
28.28 x = 17.32 (10 - x) or 45.6 x = 173.2 or x = 3.8
Solving for h 10 h = 28.28 (3.8) and h = 10.75 ft
Checking with Trigonometry:
Let A be the angle between the horizontal and the 30 ft ladder and B
be the angle between the horizontal and the 20 ft ladder. Using the
Cosine functions...
Cos A = 10/30 and Cos B = 10/20
Then A=70.55degrees and B=60degrees
Now using the same vertical line as above from the intersection on
the two ladders and using the tangent functions....and x and 10 - x as
the length of the horizontal sides...
Tan A = h/x and Tan B = h/(10 - x)
and Tan 70.55 = 2.83 and Tan 60 = 1.73
h/x = 2.83 and h/(10 -x) = 1.73
solving we get x = 3.8 and h = 10.75 a good check on our answer
above.
The excellent point that can be made in working this problem both
with similar right triangles and the trig functions is their relationship
to right triangles and the Pythagorean Theorem. Thanks.
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