Find Length of a Triangle Side

A

april19

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I'm trying to help my daughter with her Trig HW and I seem to be stuck in a problem.

There is a triangle with vertices A, B, & C. Vertex A is on top and veritices B & C are on the bottom.
A median is drawn from vertex A to side BC intersecting at point M on side BC.
Length of side AB is 4", side AC is 7", side AM (the median) is 3.5"
What is the length of side BC?

Since AM is a median, it bisects vertex A.
Let x = half of vertex A
y = BM = MC

Using the law of cosines:
y^2=(AB)^2 + (AM)^2 - 2(AB)(AM)cos(x) = 16 + 12.25 - 28cos(x)

y^2=(AC)^2 + (AM)^2 - 2(AC)(AM)cos(x) = 49 + 12.25 - 49cos(x)

Set them equal to each other to solve x.
16 + 12.25 - 28cos(x) = 49 + 12.25 - 49cos(x)
21cos(x) = 33
cos(x) = 1.57

Once I solve the angle x, I should be able to solve side BM and then BC.
My question is the cosine curve oscillates between 1 and -1 so how can cos(x) = 1.57???

Please help. Or am I totally wrong about the approach in solving side BC?
 
I am not sure I should answer at all because I have not studied geometry for fifty years. My recollection (which is probably faulty) is that a median bisects an angle only from the apex of an isocoles (spelling???) triangle. In any case, you are assuming that the median does bisect the angle opposite the bisected line segment, and perhaps you should check that it does. I am 99% sure that a median bisects a line segment. It is not clear to me that the median will necessarily bisect the opposite angle.

I am answering only because it is a school night, your daughter is doing homework, and no one has yet answered. DO NOT RELY ON MY AGED MEMORY, WHICH MAY BE WRONG, but check whether your assumption is correct.
 
Thank you for the reply.

I knew something is not right about it. You are right. The median bisects the vertex only if it's an isosceles triangle.
The problem I have is obvious not an isosceles triangle.
 
Code: 
            A

           (h)


B   (a-b)   H  (b)  M        (a)        C
No need for angles anyway; Pete Pythagoras is your man :wink:
Draw height line AH as shown above; AH is common to 3 right triangles: ABH, AMH and ACH.

Let AH = h, MC = a, MH = b; then BH = a - b

ABH: h^2 = 16 - (a - b)^2
ACH: h^2 = 49 - (a + b)^2
AMH: h^2 = (7/2)^2 - b^2

I'll let you finish it.
 
'm trying to help my daughter with her Trig HW and I seem to be stuck in a problem.

There is a triangle with vertices A, B, & C. Vertex A is on top and veritices B & C are on the bottom.
A median is drawn from vertex A to side BC intersecting at point M on side BC.
Length of side AB is 4", side AC is 7", side AM (the median) is 3.5"
What is the length of side BC?

Since AM is a median, it bisects vertex A. NOT SO---NOT SO

Let x = half of vertex A
y = BM = MC

Using the law of cosines:
y^2=(AB)^2 + (AM)^2 - 2(AB)(AM)cos(x) = 16 + 12.25 - 28cos(x)

y^2=(AC)^2 + (AM)^2 - 2(AC)(AM)cos(x) = 49 + 12.25 - 49cos(x)



The length of a triangle median is defined by (Ma)^2 = (b^2 + c^2)/2 - a^2/4

Letting BC = a, AB = c and AC = b

(3.5^2) = (7^2 + 4^2)/2 - a^2 / 4

4(3.5)^2 = 2(49 + 16) - a^2

Solve for a = BC.
 
Denis,
Thank you for your help.

This is probably so simple that I am not just not seeing it.

I used the first 2 triangles (ABH & ACH) and solved that a=(49-16)/4b=33/4b
Do I substitue this back in the 2 triangles and try to solve b like below?

16 - (33/4b - b)^2 = 49 - (33/4b + b)^2

I tried to solve this but it came out one big mess.

What am I not missing?

Thanks.
 
Thank you for your help.

This is probably so simple that I am not just not seeing it.

I used the first 2 triangles (ABH & ACH) and solved that a=(49-16)/4b=33/4b
Do I substitue this back in the 2 triangles and try to solve b like below?

16 - (33/4b - b)^2 = 49 - (33/4b + b)^2

I tried to solve this but it came out one big mess.

What am I not missing?

The length of a triangle median is defined by (Ma)^2 = (b^2 + c^2)/2 - a^2/4

Letting BC = a, AB = c = 4, AC = b = 7 and AH = 3.5

Then, (3.5^2) = (7^2 + 4^2)/2 - a^2/4

Multiplying through by 4 yields 4(3.5)^2 = 2(49 + 16) - a^2 or

49 = 130 - a^2 or a^2 = 130 - 49 = 81 making a = 9 = BC.
 
april19 said:
I used the first 2 triangles (ABH & ACH) and solved that a = (49 - 16) / 4b = 33 / 4b
Do I substitue this back in the 2 triangles and try to solve b like below?
Click to expand...
Correct! a = 33 / 4b ; or 4ab = 33 ; or 2ab = 33/2 [1]

Next, you need to use triangle AMH with one of the other two; let's use triangle ABH:
ABH: h^2 = 16 - (a - b)^2
AMH: h^2 = (7/2)^2 - b^2
That will lead to 2ab = a^2 - 3.75 ; substitute [1]:
33/2 = a^2 - 3.75 ; leads to a = 4.5 ; so BC = 2a = 9

Using the standard formula (as per TchrWill) gives same results; see TchrWill's last post.
The "3 triangles stuff" I showed was to show the HOW of that standard formula.

It it USUAL to label a triangle ABC, with BC = a, AC = b and AB = c and a < b < c :
Code: 
              C



       U                   V
                               


B                   W                      A
Points U, V and W are midpoints of the sides, hence the 3 medians AU, BV and CW.
In your case, we were solving for side AB, given sides BC and AC, and median CW.

The STANDARD formulas for the 3 medians are:
AU : (Ma)^2 = (b^2 + c^2) / 2 - a^2 / 4 ; Ma = sqrt[2(b^2 + c^2) - a^2] / 2 ; Ma means Median to side a
BV : (Mb)^2 = (a^2 + c^2) / 2 - b^2 / 4 ; Mb = sqrt[2(a^2 + c^2) - b^2] / 2 ; Mb means Median to side b
CW : (Mc)^2 = (a^2 + b^2) / 2 - c^2 / 4 ; Mc = sqrt[2(a^2 + b^2) - c^2] / 2 ; Mc means Median to side c

Hope that was helpful...

TRIVIA:
The smallest case of ALL-INTEGERS is sides 136-170-174, medians being 158-131-127.
Interesting stuff here (under Eulerian triangles): http://hyperion.cc.uregina.ca/~astro/Triangles.html
 
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