Drexel dragonThe Math ForumDonate to the Math Forum



Search All of the Math Forum:

Views expressed in these public forums are not endorsed by Drexel University or The Math Forum.


Math Forum » Discussions » Math Topics » geometry.pre-college.independent

Topic: POW Solution, Feb. 7-11
Replies: 0  

Advanced Search

Back to Topic List Back to Topic List  
Problem of the Week

Posts: 292
Registered: 12/3/04
POW Solution, Feb. 7-11
Posted: Feb 14, 1994 3:09 PM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

****************************************
Problem of the Week, February 7-11

Last week, Jill and Jimmy were attempting to figure out how wide a
river was so that they could begin plans to build a bridge. They
determined that the river was 46 meters wide. Now they have to give
an estimate as to the cost of the bridge, which means they have to
know how long the bridge will be. Here's the catch:

They want to build a bridge, starting 10 meters inland from the edge of
the river, to the top of an already-existing abuttment on the other side
of the river. The abuttment is 16 meters high and 7 meters from the
edge of the river.

How long will the bridge be? Extra: What is the angle of elevation
of the bridge?

****************************************

As the bridge saga continues, we received correct responses from:

Nicole Dunlap Grade 10, Shaler Area High School
rachel laughlin Grade 10, Shaler High School
Ben Ladik, grade 12, and Katie Getsy, grade 9; Steel Valley High School, Pa.
Ryan Ferchak and Bob Gallagher, grade 9, Steel Valley High School
Derek Morrison and Jill Grobelski, grade 9, Steel Valley High School. Pa.
Scott Pisula and Andy Fedoris, grade 9, Steel Valley High School, Pa.
Kristie Tunney and Tonya Kosko, grade 9, Steel Valley Hogh School, Pa.
Susan Quan, grade 7, Masterman School, Philadelphia
Don Mahaney Grade 10, Shaler Area High School
Tim Thiel Grade 10, Shaler Area High School
Jen Eskra Grade 10, Shaler Area High School
Jennifer Strong Grade 10, Shaler Area High School
Mike Orehowsky Grade 10, Shaler High School
Thomas Niebel Grade 10, Shaler High
Dan Schrom Grade 10, Shaler Area High School
Gino Perrotte Grade 10, Shaler Area High School
Alyssa McGrath Grade 10, Shaler Area High School
Matt Bouton Grade 9, Steel Valley High School, Pa.
Tim Schnurr Grade 10, Fairfield HS, Connecticut
Other solvers from Fairfield High School:
Grade 9 Grade 10
Amanda Adams Elissa Colter
Hilary Aleksa Matt Lucas
Lindsey Becker Morgan May
Amy Decrescenzo Natalie Painchaud
Brendan Hogan Ryan Phelan
Molly MacDonald Nghiem Vu
Kathy Medlin
Rebecca Naughton
Agata Raszczyk-Lawska
Phil Rossi
Katy Ruff
Allison Sullivan
Heather Wilcox
Jan Wilson
Sandy Broten Grade 10, Edgerton High School, WI
Jeremy Goede Grade 10, Edgerton High School, WI

**********************************
From: rachel laughlin <laughlin@one.sasd.k12.pa.us>

If the abutment was in the river:
|
? | 16m
|
|
_________|____46m__________________________|______
10m 7m

a^2+b^2=c^2 = 58m

If the abutment was on land:
|
? |
| 16m
|
_________|__________________________________|________|
10m 46m 7m

a^2+b^2=c^2 = 65m

This is all assuming that the bridge has a diagonal slope, lowest at
the bank and highest at the abutment. if the bridge were straight
across, or started at a slope then flattened out, the distance would
change.

Rachel Laughlin Shaler Area High School 10th Grade

**************************
From: Cecil Wenisch <wenischc@one.svsd.k12.pa.us>

Submitted by Ben Ladik, grade 12, and Katie Getsy, grade 9; Steel Valley High
School, Pa.

In order to start this problem, you must first know the distance from our
start and end points, which is 63 meters. Next, we took the height, 16 m, of
the abuttment and used it as one of our legs and 63 m as the other. We uaed
the Pythagorean Theorem to find the length of the span, which was 65 m.

63 squared + 16 squared = x squared
4225 = x squared
the square root of 4225 = x
65 = x

We also scaled down the meters to millimeters on our drawing, then
measured,and still got 65.

**************************
From: Cecil Wenisch <wenischc@one.svsd.k12.pa.us>

Subnitted by Ryan Ferchak and Bob Gallagher, grade 9, Steel Valley High
School, Pa.
]


16

[------10-----------------46----------------7-----]

Notice, the sum of the bases of the bridge is 63 meters. The abuttment is 16
meters high. From that knowledge, you can form the two legs of the triangle.
Connect the new area for the abuttment to the existing one to find the
hypotenuse. At this point the hypotenuse is unknown, but by using the
Pythagorean Theorem you can find the hypotenuse's length. (a squared + b
squared = c squared) Take 63 and square it; you should get 3969. Take 16 and
square it; you should get 256. Take the two and add them. You now have 4225.
Take the square root of that to get the measure of c. c = 65

**************************
From: Cecil Wenisch <wenischc@one.svsd.k12.pa.us>

Submitted by Derek Morrison and Jill Grobelski, grade 9, Steel Valley High
School. Pa.

First, we added 46 meters, the width of the river, to 10 meters, the distance
inland from the edge of the river, to 7 meters, the distance from the other
edge of the river to the abuttment. The sum of the three was 63.

Second, we knew that the abuttment was 16 meters high and 7 meters from the
edge
of the river.

We now could form a triangle by using 16 meters as the shorter leg and 63
meters as the longer leg. Now all that we needed was to find the hypotenuse
or length of the bridge.
By using the Pythagorean Theorem, leg one squared + leg two squared =
hypotenuse
squared, we could determine that the hypotenuse or length of the bridge is 65
meters.
16 squared + 63 squared = h squared
256 + 3963 = 4225
the square root of 4225 = 65
Therefore, the length of the bridge is 65 meters.

**************************
From: Cecil Wenisch <wenischc@one.svsd.k12.pa.us>

Submitted by Scott Pisula and Andy Fedoris, grade 9, Steel Valley High School,
Pa.

We made a drawing and figured that we have both legs of a triangle. So then we
used a squared + b squared = c squared to find our answer.
63 squared + 16 squared = c squared
3963 + 256 = 4225
c = 65

**************************
From: Cecil Wenisch <wenischc@one.svsd.k12.pa.us>

Submitted by Kristie Tunney and Tonya Kosko, grade 9, Steel Valley Hogh
School, Pa.
]


16

[------10-----------------46----------------7-----]

------------------- 63 ----------------------------


The length of the bridge will be 65 m long. This is proved by the Pythagorean
Theorem, which is leg squared + leg squared = hypotenuse squared.

*****************************
From: myron@forum.swarthmore.edu (Myron Goldman)

Solution from Susan Quan, grade 7, Masterman School, Philadelphia

Assuming the bridge follows a straight line from a point 10 ft. from the river
to the front edge of an abuttment 16 ft. high, the bridge would be 65 ft. long.
The bridge is the hypotenuse of a right triangle with legs 63 ft. and 16 ft. so
using the Pythagorean Theorem gives the length of the bridge.

16^2 + 63^2 = x^2 where x is the length of the bridge

256 + 3969 = x^2

4225 = x^2

x = 65

*********************
From: m2 <m2@one.sasd.k12.pa.us>

The length of the bridge can be determined by using the
Pythagorean theorem. The first side, which we will call A, has a
length is 63 meters.(The river is 46 meters wide, plus the 10 meters
inland on one side and the 7 meters inland on the other.) The second
side, which we will call B, is 16 meters long, and that corresponds
to the 16 meter height of the abutment. Note:Side A is perpendicular
to side B because of the first minimum theorem. The third side, C,
will be the length of the bridge.
The numerical value for C is found by taking both sides A and
B, squaring them, and adding the new values together. The result,
4225, is C squared. Finally, taking the square root of 4225 you get a
result of 65 meters; which is the length of the bridge.
To find the angle of elevation, the trigonometric functions
of sine, cosine, and tangent must be used:

The sine of angle 1 = 16/65 = .264
The cosine of angle 1 = 63/65 = .969
The tangent of angle 1 = 16/63 = .254

* Since I have not done any of this in geometry class I was
forced to look ahead in the book a few chapters.

Taking a calculator, and using any of the three values above,
all that is needed to find the measure of the angle is to push the
key that will give the reciprocal value of sine, cosine, or tangent.
After doing this for each of the values, I found the measure of the
angle of elevation to be 14.25 degrees.

Don Mahaney

**********************
From: m1 <m1@one.sasd.k12.pa.us>

|
|
| 16meters
|
-----------------------------------------------

10meters + 46meters + 7 meters

( I am assuming that the bridge is flat and not arched in the middle
like a normal bridge)

As shown by the picture a right triangle is created from the
abuttment, the existing lenght, and the bridge which turns out to be
the hypotenuse. The height of the abuttment is given as 16 meters
and is a leg of the triangle created. the other leg of the triangle
is found by adding the width of the river which is 46 meters, the
distance the abuttment is away from the river which is 7 meters, and
the distance the bridge will be from the edge of the river which is
10 meters. The distance calculated from the addition is 63 meters.
To find the length of the bridge(which is the hypotenuse) I used the
Pythagoreon Theorem.(a^2+b^2=c^2) Let a=16 meters and Let b=63
meters. I used the following calculations to find c:
16^2+63^2= c^2
c= (16^2+63^2)^.5
c= (4225)^.5
c= 65 meters The length of the bridge(or the
hypotenuse of the triangle) is found to be 65 meters long.
To find the angle of elevation (which is the angle whose
sides in the triangle are the hypotenuse and the 63 meter leg) can be
found by using a simple trigometric function. To find the angle of
elevation for the bridge I took the inverse tangent(arc tangent) of
the two legs of the triangle:
a= 16 meters b= 63 meters
angle of elevation= arc tan (16meters/63 meters)
angle of elevation= 14.25 degrees ( the 14.25 degree angle of
elevation is larger than the normal angle of elevation for a bridge
would be in real life.)
final answer:
length of bridge= 65 meters
angle of elevation= 14.25 degrees

tim thiel sophmore at shaler area

*************************
From: m5 <m5@one.sasd.k12.pa.us>

To figure out the length of the bridge, Jill and Jimmy must use
the Pythagorean theorem. This theorem is (a*a)+(b*b)=c*c. a is 16m,
b is 63m (add 46m+10m+7m). After squaring these variables, the
problem is 256+4096=c*c. c*c ends up being 4225, so c=65. Therefore
the length of the bridge is 65 meters. But Jimmy and Jill should buy
an extra meter or so of construction material in case of
constructional error.

Jen Eskra grade 10 Shaler Area High School

*************************
From: strong jenn <strong@one.sasd.k12.pa.us>

My name is Jennifer Strong and I am in a tenth grade honors
geometry class at Shaler Area High School. Here is my answer for
POW:
(n^2 = n squared. r* = r degrees)
The length of the bridge can be very easily determined using
the Pythagorean Theorem. However, it must first be assumed that the
abutment meets the river bank at a 90 degree angle. Second, all
lines (the abutment, the bridge, and the bank to river to bank) must
be assumed straight also.
Since a^2 + b^2 = c^2, substituting the abutment for a (16 m)
and the bank-river-bank for b (10+46+7=63 m), one gets 16^2 + 63^2 =
256 + 3939 = 4225 = c^2.
c then equals the square root of 4225. Therefore, the length of the
bridge, c , is 65 meters.

Knowing this, there are several ways to find the angle of
elevation. These methods include:
tan r* = a/b, cos r* = b/c, or sin r*=a/c
Keeping the variables the same, tan r* = a/b seems the most
applicable because we are given a and b in the original problem.

a/b = 16/63 = tan r*
Then taking tan^-1 of each side, we are left with r* equals
approximately 14.25*.

cos r*= b/c or sin r* = a/c would both be
solved in a similar fashion.

*************************
From: m4 <m4@one.sasd.k12.pa.us>

Assume that the river is straight. The equation to find the
length of the bridge is a squared times b squared = c squared. Which
is sixty-three squared times sixteen squared = c squared. a is the
length of the river (46m) plus the ten meters from the start of the
bridge to the river and the seven meters to the abutment. b is the
sixteen meters is the hight of the abutment. c is the length of the
bridge which is the hypotenuse of the right triangle formed by the
abutment and the ground length. When you work out the equation it
gives you an answer of 65 meters.

3969 + 256 = c squared. when you take the square-root of the
left side you get an answer of 65 meters.
To find the angle of elevation of the bridge you have to use a
tangent. To come about with this answer I took a/b and pressed the
tan-1 key on my calculator and got an answer of 14.25 degrees

Mike Orehowsky, Grade 10 Shaler High School

**************************
From: m3 <m3@one.sasd.k12.pa.us>

The length of the bridge is sixty-five meters, assuming the
river is straight, the abuttment is at a right angle to the shore,
and the bridge starts at ground level(on the shore opposite the
abuttment). Then, the length across the river, up the abuttment, and
the bridge connecting them is a right triangle. This allows the
Pythagorean Theorem to be used to find the length of the bridge.
Substituting sixteen for a(the height of the abuttment), and
sixty-three for b(the distance across the river, inshore ten meters,
and inshore to the abuttment) into the equation of: a squared+ b
squared= c squared. This gives me the equation: 256+3969= c squared.
Therefore, c squared is 4225, and its square root is 65 meters, the
length of the bridge.

To find the angle of elevation of the bridge, I was forced to
use a tangent. I found the tangent by dividing 16(side a) by 63(side
b). This gave me a value of .2539. Then by pressing the tan-1 key
on my calculator, it gave me the angle of elevation which was 14.25
degrees.

Thomas Niebel
Grade 10 Shaler High

*******************************
From Dan Schrom <m5@sasd.k12.pa.us>

First, to find the length of the bridge, you must use the
Pythagorean Theorem. The PT says that A^2 + B^2 = C^2. In other
words, the sum of the squares of the two legs in a right triangle
equal the the square of the hypotnuse. Although you must say that
the abuttment forms a 90 degree angle. This holds true because

63^2 + 16^2 = 65^2. After you find the length of the
hypotnuse(length of bridge), you can then find the angle of elevation
by using sin.
sin = a = a/c = 16/65 = 14.25 degrees.
!
65 !
!
________/_____________________________/____!

10 + 43 + 7

Therefore, the length of the bridge equals 65meters and the angle of
elevation is 14.25 degrees.

*****************************
From: m4 <m4@one.sasd.k12.pa.us>

Gino Perrotte
I am a tenth grade honors geometry student. This is the first time
that I have used the NEXT computer system.

To find out how long the bridge will be, I used the pythagorean
theorem. To use this theorem you must assume that the abuttment is
perpendicular to the ground. The legs of the right triangle will be
the river and the abuttment. The length of the abuttment is given to
be 16 meters. The length of the river is given to be 46 meters.
Then you have to add 10 meters and 7 meters to the river length
because that is how far in they want the brigde to start. Then I
applied the Pythagorean theorem, which is c^2= a^2 + b^2. Doing the
math:
C^2= 16^2 + 63^2; c^2= 4225; take the square root of both
sides and you get c= 65. The hypotenuse of the triangle is 65 meters
long, and the hypotenuse is the bridge. Therefore the bridge is 65
meters long.
To find the angle of elevation, the equation a/c sin^-1 is used. In
the triangle the abuttment is side a, the river is side b, and the
bridge is side c. Substituting the numbers into the equation you
get:
(16/65) sin^-1= 14.25 degrees
The angle of elevation is 14.25 degrees.
|
| 16m
|

__________ _______|
10m ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 7m
46m

********************************
From: m2 <m2@one.sasd.k12.pa.us>

My name is Alyssa McGrath and this is my solution to the problem of
the week. I attend Shaler Area High School in Pittsburgh, PA. I am
in 10th grade and I'm in Honors Geometry.

For the 1st part of the problem I found the length of the bridge. To
find it I added 10m+46m+7m. The sum is 63m (the base of the
triangle). The hight is 16m as stated in the problem. I assumed the
angle here would be 90 degrees, so I used the Pythagorean Theorem to
find the length of the hypotenuse (final side). The theorem gives
you the equation: a^2+b^2=c^2

When the values were filled in it became:
16^2+63^2=c^2
4225=c^2
65=c
This means the hypotenuse is 65 meters.
To find the angle of elevation I looked in my book to find the
formulas for:
Sin = a/c =0.246
Cos = b/c =0.969
Tan = a/b =0.254
I then looked on the table of Trigonometric Ratios. When the three
numbers matched up the chart said the angle of elevation is 14
degrees.
c !

! a

!

_________ _______!
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
b
I couldn't draw the hypotenuse but it forms the angle of elevation.

***************************
From: Cecil Wenisch <wenischc@one.svsd.k12.pa.us>

Submitted by Matt Bouton, grade 9, Steel Valley High School, Pa.

[


16


[-----7------------46-------------10--------]

10+46+7=63

16 squared + 63 squared = 65 squared

65 = bridge length

*************************
From: PDALEY@fair1.fairfield.edu

Looking at the bridge from down river, you can see it forms a right
triangle. You can find the length of the bridge by finding the lengths
of the two legs then using the Pythagorean Theorem. You can use this
because one of the angles in the triangle is a right angle. The two legs
are the overall distances at the base, which is 7 + 46 + 10, and the
abuttment, which is 16 meters high. Therefore, 16^2 + 63^2 = the square
of the bridge length. Next, 256 + 3969 = x^2, 4225 = x^2, and x = 65.
So the bridge is 65 meters long.
Tim Schnurr, Fairfield HS, grade 10.

************************************
From: akuemmel@students.wisc.edu (Andrew Kuemmel)

Solution by Sandy Broten, Grade 10, Edgerton High School, WI

The bridge forms a right triangle. We know two dimentions of the triangle
(height is 16 meters and span is 63 meters) so knowing the Pythagoream
Theorem you can find the length of the third side.
2 2 2
16 + 63 = x
256 + 3969 = 4225
the square root of 4225 = 65, so the bridge will be 65 m long.

**************************************
From: akuemmel@students.wisc.edu (Andrew Kuemmel)

Solution by Jeremy Goede, Grade 10, Edgerton High School, WI

The river is 46 m wide, add 17 m for where you start the bridge and where
the abuttment is which equals 63 m. This is a right triangle, so sou
square 63 and 16 and add them to find the square of the length of the
bridge. 63^2 + 16^2 = 4225 and the square root of 4225 is 65. The bridge
is 65 m long.

To find the angle of elevation,

Let x be the angle of elevation.

tan x = opposite leg / adjacent leg = 16/63 = .253968254

x = tan^-1 (.253968254)

x = 14.25

The angle of elevation is 14.25 degrees.





Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© Drexel University 1994-2014. All Rights Reserved.
The Math Forum is a research and educational enterprise of the Drexel University School of Education.