Lauren 6:21 AM 8/11/01
Subject: solving triangles
Question: I just can't seem to figure out which laws to use on this so it will work out. the other problem I just
can't remember how to solve an isosceles triangle.
1> Solve triangle ABC if B = 58° cm and a = 16 and c = 24. Round to the nearest whole number
2> Find the base of an isosceles triangle if each leg is 12 cm long and each base angle measures 26°. Round to
the nearest tenth.
Answer:
Lauren,
Hey. OK this time I'll just tell you the steps to solving the problem, rather than doing the whole thing; see if
you can follow all the steps. If you're still having trouble after you try doing the steps, feel free to do a
followup question.
PROBLEM 1

Solve triangle ABC if B = 58° cm and a = 16 and c = 24. Round to the nearest whole number.
This one is a little tricky because we can't immediately use the law of sines. We are given three terms: angle
"B", side "a", and side "c". No two of these three terms correspond to an angle and its opposite side  that is,
we don't have an angle and a side that are both the same letter ("A" and "a", or "B" and "b", or "C" and "c"). We
are only given one of each letter: a lowercase "a", an uppercase "B", and a lowercase "c".
Somehow we need to get another letter, so that we'll have a pair of the same letter. Let's try to get angle A.
Then we'll be able to use the law of sines and solve for everything else in the triangle, since (sin A / a)
= (sin B / b) = (sin C / c). How will we get A? Check out this picture:
fIn this picture we have a triangle with vertices K, L, and M, and angles A, B, and C. The length of side ML is a,
the length of side KM is b, and the length of side KL is c. The key to solving this problem is the red line I drew
from vertex M to point O. This red line MO is perpendicular to side KL, and splits the triangle into two right
triangles: triangle KMO and triangle LMO.
Here's the plan to get angle A:
1) Find MO, knowing that sin B = opposite / hypotenuse = MO / a.
2) Find OL, knowing that cos B = adjacent / hypotenuse = OL / a.
3) Find KO, knowing that KO + OL = c.
4) Find A, knowing that tan A = MO / KO. (Take arctan of both sides to get A)
After that, we use law of sines to finish the problem:
5) b = (sin B) / (sin A / a)
5) C = arcsin (c (sin A / a))
PROBLEM 2

Find the base of an isosceles triangle if each leg is 12 cm long and each base angle measures 26°. Round to
the nearest tenth.
Since its an isoceles triangle, LM = KL = 12 cm, and A = C = 26 degrees. (KL and LM the legs, A and C are the
base angles, and KM is the base.) We are given almost enough information to apply the law of sines to get KM
 but first we need angle B. So here's the plan to solve for KM (base):
1) Find angle B (all interior angles of a triangle must equal 180)
2) Use of law of sines (sin B / KM = sin A / LM = sin C / KL) and solve for KM.
Note: You can also solve this problem without using the law of sines. Try solving it by using the red line
LO. A piece of advice in geometry: many, many problems can be solved by simply drawing lines like that
(perpendicular lines that connect one side to a vertex).
Hope that helps!
Wu

Holly 2:25 PM 8/6/01
Subject: algebra
The lenght of a retancle piece of tin is 10cm more than its width. If a
4cm square is cut from each corner and the resulting material is folded to
form a box whose volume is 1500 cm^3, find the orginal length of piece of
tin.
a. 23cm
b. 25cm
c. 33cm
d. 19.4 cm
e. 17cm
what in the world is the solution?
Answer:
Holly,
Hi, thanks for the question. To solve it, you need two equations. Let's look at the sentences from which we can
generate these equations.
(1) The length of a rectangular piece of tin is 10 cm more than its width.
Ok, let's start by assigning names. We will call L the length of our tin rectangle, and W will be the width of our
tin rectangle. Then our first equation is simply:
Equation 1: L = 10 + W
(2) If a 4 cm square is cut from each corner and the resulting material is folded to form a box of volume 1500 cm^3
Now we can make another equation. It helps to draw a picture for this one. First draw a rectangle. Then erase
a small square from each corner. The result should look something like a first aid cross, except with really short
arms. Then you can mentally imagine folding the arms of this cross toward you, so that you get an openlid box.
We know that the volume of a box is the product of its length, width, and height. So what is the length of the box
we just formed? Well, the height of the box is simply the length of one side of the small square we cut out. So
the height is just 4 cm. The length of the box is a little smaller than the length of the original tin rectangle
(before we folded it), and the width of the box is a little smaller than the width of the original tin rectangle.
How much smaller? 8 cm smaller. Because we cut out squares from all four corners, and each square is 4 cm in width,
the length and width of the box are just 8 cm less than the original length and width of the rectangle. Thus we
can write equation 2:
Equation 2:
1500 = height of box * length of box * width of box = 4 * [L  (2 * 4)] * [W  (2 * 4)]
Here's a diagram:
Alright, so now we have two equations and two unknowns (L and W). We can easily solve this system of
equations by substituting one equation into the other. Substituting equation 1 into equation 2:
1500 = 4 * [(10 + W)  (2 * 4)] * [W  (2 * 4)]
= 4 * (W + 2) * (W  8)
= 4 * (W^2  6W  16)
Divide both sides by 4:
375 = W^2  6W  16
Subtract 375 from both sides:
0 = W^2  6W  391
Using quadratic formula:
W = [ b + sqrt(b^2  4ac) ] / (2a)
= [ (6) + sqrt[ (6)^2  4(1)(391) ] / (2*(1))
= [ 6 + sqrt(36 + 1564) ] / 2
= 3 + (1/2)sqrt(1600)
= 3 + (1/2)*40
= 3 + 20
= 23 or 17
Since a width of 17 cm doesn't make any sense, 23 cm must be the correct width.
Thus, if 23 is the width, we can substitute 23 for W in equation 1. Then we discover that the length is
10 cm + 23 cm = 33 cm.
To make sure we have the right answer, let's check our work, and substitute W = 23 and L = 33 into equation 2.
4 * [L  (2 * 4)] * [W  (2 * 4)]
= 4 * [(33)  (2 * 4)] * [(23)  (2 * 4)]
= 4 * (33  8) * (23  8)
= 4 * 25 * 15
= 1500 cm^3.
Thus we have the correct answer, and the length is c., 33 cm.
Hope this helps! Best of luck to you in algebra. The key to algebra word problems is in translating
sentences into equations, and not being afraid to assign variable names to things you don't know.
Wu

Jay Bhurpur 8/3/01 1:36 AM
Subject: curve sketching question...
Hi, Wutang how are you? just curious, what is your ethnic background? and
why do you call yourself WuTang? well anyways this question is a rather
tricky one but is easy, what I would like from you is a confirmation of
its answer. Here is the question word for word.
Sketch the graph of a function that satisfies all of the following
conditions.
I. lim as x approaches negative infinity is f(x) = 0
lim as x approaches positive infinity is f(x) = infinity.
II. f'(x) > 0 if x < 1 or x > 2.
III. f(0) = f'(0) = 1
IV. f''(0) < 0 if x < 0.5
I have already got my graph but I want to see if you will get the same as
I. could you please tell me which way would be best to display the answer
for this question. Please do the question I swear it will only take a
couple of minutes, actually it took me about 5 minutes to complete. The
only reason i am asking you to answer it is because, for me it is worth
marks and I don't have the solutions in front of me to check. I'm just
paranoid, this is the only problem that has gotten me a little rattled(I
mean I answered it alright, but the graph looks like a W and I couldn't
possibly find an equation for it, or one that even looks close to it.)...
Thanks a lot
Answer:
Here's what I came up with, using Photoshop to draw it:
The red line is the main graph, and the brown line is just a relevant tangent line that I drew.
I analyzed the problem in the following manner:
I. As x approaches negative infinity, the curve must go to zero.
As x approaches positive infinity, the curve must go to infinity. (Note that I didn't really draw
the curve as skyrocketing to infinity when x gets bigger than two, but that's only because
I ran out of room and I didn't want to make my image too big  when you draw the graph you
should make it skyrocket to infinity when x gets large)
II. The slope of the curve must be positive for the intervals (1,1) and (2,+infinity).
III. The value of the curve must be 1 for x = 0.
The slope of the curve must be 1 for x = 0.
IV. The curve must be concave down for the interval (0.5, 0.5).
Note: I'm assuming you made a typing error when you said that " f''(0)<0 if /x/<0.5 "
 I think you mean f''(x)<0. My reasons for believing this are because the statement f''(0)<0
doesn't involve x, although right after that statement you say "if x < 0.5". Also, it's not
graphically significant that f''(0)<0. If f'(0) = 0, then it would be significant, because any
point at which the slope is zero and the second derivative is less than zero is a local maximum.
However, we know from part III that the slope is one at x = 0. Thus, the statement "f''(0)<0 if
x<0.5" isn't really relevant.
Putting all this together, I get the above graph. Moving from left to right: the curve gradually diverges away
from zero, and it has a negative slope because the only explicitly mentioned positive slopes are over (1,1) and
(2,+infinity). When we get to 1, the curve has to start ascending, and thus there is a local minimum. So far
the curve has always been concave up. At 0.5 there is a point of inflection, where the concavity switches; from
0.5 to 0.5 the curve is concave down. At x=0, the curve has a yintercept of one and has a slope of one, as shown
by the brown tangent line. Then we have a second point of inflection at 0.5, and the concavity switches to concave
up and stays concave up for the rest of the graph. This concavity transition is probably the hardest part to draw.
Note that at the second pont of inflection, the slope is not zero, although it is perhaps tempting to draw that
part as flat  the conditions specifically say that the slope must be greater than zero over (1, 1). From x = 1
to x = 2, the curve has a negative slope. Then the sign of f'(x) changes from negative to positive at x=2, and
thus we have a local minimum. Finally the graph should approach infinity as x goes to infinity.
So I guess if you got something like a "W", your graph is probably right. I hope mine is right :) Since the
problem doesn't state that you need to find a function that behaves this way, I don't think you need to worry about
it. However I remember that in AP Calculus, I got some graphs like this and I had to come up with functions that
described them approximately. This actually wasn't that hard because they allowed us to make piecewise smooth
curves. That is, we could write down something like: f(x) = x + 2 over (infinity,2) f(x) = x^2 over [2,infinity).
Note that 2 + 2 = 2^2, so there is no abrupt transition at x = 2. So this wasn't hard as long as you were familiar
with some basic curves ... if you ever get a question like that asking you to write an equation for a funky graph,
you might even be allowed to write piecewise discontinuous functions, where abrupt jumps are OK.
Hope that helps!
I call myself WuTang because I'm a huge black dude.
Wu
