SOLUTION: Prove that in any triangle that has the sides proportionate with 4, 5 and 6, the least angle measure is half of the biggest angle measure. With the Cosinus Theorem, I found that(A

Algebra ->  Triangles -> SOLUTION: Prove that in any triangle that has the sides proportionate with 4, 5 and 6, the least angle measure is half of the biggest angle measure. With the Cosinus Theorem, I found that(A      Log On


   

Question 1026992: Prove that in any triangle that has the sides proportionate with 4, 5 and 6, the least angle measure is half of the biggest angle measure.
With the Cosinus Theorem, I found that(AB=4k , AC=5k, BC=6k)
cos A = 2/16
cos B = 9/16
cos C = 12/16
Help me, I dont know what to do further!
Found 2 solutions by ikleyn, Theo:
Answer by ikleyn(52788) About Me  (Show Source):
You can put this solution on YOUR website!
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Prove that in any triangle that has the sides proportionate with 4, 5 and 6, the least angle measure is half of the biggest angle measure.
With the Cosinus Theorem, I found that(AB=4k , AC=5k, BC=6k)
cos A = 2/16
cos B = 9/16
cos C = 12/16
Help me, I dont know what to do further!
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 

1.  I checked your calculations for cosines and confirm that they are correct.

2. Now calculate cos%28A%2F2%29.   (Notice that in your notations A is the largest angle, since it is opposite to the longest side BC.)

   Use the formula for the cosines of the half argument of Trigonometry   

   cos%28A%2F2%29 = sqrt%28%281%2Bcos%28A%29%29%2F2%29.         (See the lesson Trigonometric functions of half argument in this site).

   You will get  cos%28A%2F2%29 = sqrt%28%281+%2B+%282%2F16%29%29%2F2%29 = sqrt%2818%2F%2816%2A2%29%29 = sqrt%289%2F16%29 = 3%2F4 = 12%2F16 = cos%28C%29.

3.  Hence, the angle  A%2F2  is congruent to the angle  C.   (They both are acute.)

4.  Proved.

I didn't know this fact before. It is new for me. Thanks !


Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
there is only one set of angles that will be in a triangle that has sides in the ratio of 4 to 5 to 6.

assuming that side 4 is opposite angle A and side 5 is opposite angle B and side 6 is opposite angle C, then you get the law of sines ratio of:

4/sin(A) = 5/sin(B) = 6/sin(C)

the smallest angle in this triangle will be angle A because it is opposite side 4.

the largest angle in this triangle will be angle C because it is opposite side 6.

you want to prove that the smallest angle is 1/2 the size of the biggest angle.

this also means that the biggest angle is 2 times the size of the smallest angle.

assuming this is true, then you can solve for the angles as follows:

the law of sines ratio is shown below again for ease of reference.

4/sin(A) = 5/sin(B) = 6/sin(C)

since A is the smallest angle and C is the largest, then use those ratios as shown below:

4/sin(A) = 6/sin(c)

cross multiply to get:

4 * sin(C) = 6 * sin(A)

divide both sides of this equation by sin(A0 and divide both sides of this equation by 4 to get:

sin(C) / sin(A) = 6/4

since C is 2 * A, replace C with 2 * A in this equation to get:

sin(2A) / sin(A) = 6/4

lookup the trig identities for double angles to get:

sin(2A) = 2sin(A)cos(A).

replace sin(2A) with its equivalent identity to get:

2sin(A)cos(A) / sin(A) = 6/4

the sin(A) in the numerator and denominator cancel out and you get:

2cos(A) = 6/4

divide both sides of this equation by 2 to get:

cos(A) = 6/8

solve for A to get:

A = 41.40962211 degrees.

since angle C is twice angle A, then angle C must be equal to 82.81924422 degrees.

since sum of the angle of a triangle = 180 degrees, then angle B must be equal to 180 minus angle A minus angle C = 180 - 41.40962211 - 82.81924422.

this makes B = 55.77113367 degrees.

if these angles are correct, then the law of sines must be observed and the ratio between the length of the sides divided by their opposite angles must be the same.

the law of sines states, once again:

4/sin(A) = 5/sin(B) = 6/sin(C)

we also know that:

angle A = 41.20962211 degrees
angle B = 55.77113367 degrees
angle C = 82.81924422 degrees

4/sin(A) becomes 4/sin(41.20962211) = 6.047451568
5/sin(B) becomes 5/sin(55.77113367) = 6.047431568
6/sin(C) becomes 6/sin(82.81924422) = 6.047431568

the ratio is same, so the law confirms that the angles and sides are correct.

since the sides are in the ratio of 4 to 5 to 6, and since the biggest angle is twice the smallest angle, than the hypothesis is proven.

you should be able to make any multiple of 4 to 5 to 6 and get the same solution.

for example:

if the smallest side is 8, then you should get:

8/sin(A) = 10/sin(B) = 12/sin(C)

i checked them out and the ratio is the same for each.
it's 12.09486314.

since i've never done any like this before, i don't know any other way to prove it except by finding the angles and then determining that the law of sines works with those angle.

once you know the angle, you could also probably use the law of cosines as well, but that's not necessary since you already used the law of sines and both laws support each other, as they should.

i did need to use the trig identities to help solve this as well.

best i can do.

i don't know any better.