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Two Triangle Problems


Date: Tue, 11 Jun 1996 21:28:40 -0400 (EDT) 
From: Anonymous
Subject: two triangle problems

Here are two triangle problems I can't solve. 

1. In triangle ABC, angle A is trisected by segments AE and AD 
where points E and D lie on BC. BD = 2, DE = 3, and EC=6. 
Find the length of the shortest side of triangle ABC.

2. If triangle ABC is reflected in its median AM, and AE = 6, 
EC = 12, BD = 10, and AB = k(sqrt3), compute k.

          A
          | \
         /|\ \ B'
        / | \ E
          B---D---M-----C
          | /
          C'

Sorry about the diagram but I hope you can see ABC and AB'C', 
D, and E are intersection points of AC' and BM , and AC and B'M 
respectively. 

Thanks for your help, 

Michael


Date: Thu, 13 Jun 1996 17:26:57 -0400 (EDT) 
From: Dr. Anthony
Subject: Re: two triangle problems

I will look at question (2) tomorrow, but here is question (1) to be 
going on with.

(1) It would be wise to draw an approximate diagram for 
question(1) and refer to it as I go through the maths. It is 
unfortunately a rather long-winded explanation.

In triangle ABD sin(A/3)/2 = sin(B)/AD

In triangle ABE sin(2A/3)/5 = sin(B)/AE

Equate the two values of sin(B) to get

AD*sin(A/3)/2 = AE*sin(2A/3)/5
= 2AE*sin(A/3)cos(A/3)/5

cancel the sin(A/3) on both sides to get 

cos(A/3) = (5/4)(AD/AE)

Now using the angle-bisector theorem on triangle ADC, we have 

AC/AD = 6/3 = 2 

and using same theorem on triangle ABE we have 

AE/AB = 3/2

Multiply these to get (AC/AB)(AE/AD) = 3 or AD/AE = 
(1/3)(AC/AB)...(1) 

Now we need a further relationship, and we get this from further 
applications of the sine rule.

sin(B)/AC = sin(C)/AB so sin(B)/sin(C) = AC/AB ....(2) 

also sin(B)/AE = sin(2A/3)/5 and sin(C)/AD = sin(2A/3)/9 

Divide these and cancel sin(2A/3) to get 

sin(B)/sin(C) = (9/5)(AE/AD) = AC/AB from (2) above. 

Substitute for AC/AB in (1) above

We have AD/AE = (1/3)(9/5)(AE/AD)

(AD/AE)^2 = (3/5)

AD/AE = sqrt(3/5)

Now cos(A/3) = (5/4)sqrt(3/5) = (1/4)sqrt(15) 

At this stage I shall be working with the value of angle A, B, C in 
degrees. 

We have sin(A/3) = sqrt(1 - 15/16) = sqrt(1/16) = 1/4 

So A/3 = 14.4775 degrees and A = 43.4325 degrees. 

We had sin(B)/sin(C) = (9/5)AE/AD = (9/5)sqrt(5/3) = sqrt(27/5) . 
sin(B)/sin(C) = 2.32379

Also B + C = 180 - 43.4325 = 136.5675 degrees 
B = 136.5675-C

Sin(136.5675-C) = 2.32379sinC
sin(136.5675)cos(C) - cos(136.5675)sin(C) = 2.32379sin(C) 
0.6875cos(C) + 0.7262sin(C) = 2.32379sin(C) 
0.6875cos(C) = 1.597605sin(C)

and so tan(C) = 0.43033 and angle C = 23.28374 degrees 

Finally AB/sin(C) = 11/sin(A)

AB = (11*sin(C))/sin(A)
= (11*sin(23.28374))/sin(43.4325)

AB = 6.32456 This is opposite the smallest angle so is 
shortest side

-Doctor Anthony, The Math Forum


Date: Thu, 13 Jun 1996 21:00:02 -0400 (EDT) 
From: Dr. Anthony
Subject: Re: two triangle problems

I sent the answer to question (1) yesterday, and here, as promised, 
is question (2).

Again draw the diagram and refer to it as I work through the 
question. 

Using the angle-bisector theorem with triangle ADC, we see that 
DM:MC is in the ratio 18:6 = 3:1

But M is mid-point of BC, so BD+DM = MC
10+DM = 3DM
10 = 2DM and so DM = 5, MC = 15

We can now find angle C from the triangle EMC since all three 
sides are known.(5,12,15 with 5 opposite angle C) Using the cosine 
formula we get: 

cos(C) = (15^2+12^2-5^2)/(2*15*12)
= 344/360
= 43/45

Now apply the cosine rule to triangle ABC.(BC=30, AC=18) 

AB^2 = 30^2 + 18^2 - 2*30*18*cos(C)
= 900 + 324 - 2*30*18*43/45
= 900 + 324 - 1032
= 192
= 64*3

And taking square roots AB = 8*sqrt(3)

So, in the question k = 8

-Doctor Anthony, The Math Forum

    
Associated Topics:
High School Euclidean/Plane Geometry
High School Geometry
High School Trigonometry

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