Trigonometry again?

Question

Some people say this can be accomplished by using laws of cosines. I'd like to see the answer using any means possible.

Given a triangle (abc) with all angles (ABC) acute. A line is drawn from each angle to perpendicular of the opposite side. (You end up with 6 overlapping right triangles). If the distances for the perpendicular lines from each angle to each side opposite are given, can you find the perimeter (a+b+c)?

Answer 1

One can take advantage of the fact given x, y, z, the altitudes of the triangle, a triangle with sides 1/x, 1/y, 1/z will be proportional to the original. Let A be the original triangle, B the second, and let s be the semiperimeter of A, and t of B. Then by Heron's formula, we know:

1/2 * 1/x * H = Sqrt( t(t-1/x)(t-1/y)(t-1/z) )

where H is the altitude of B. We then also have the following ratio, through proportionality:

x / H = s / t

or (x t) / H = s

Using H from above, we have:

(x t) / (2x * Sqrt( t(t-1/x)(t-1/y)(t-1/z) )) = s

or t / Sqrt( t(t-1/x)(t-1/y)(t-1/z) ) = 2s = perimeter of A, which is the answer sought, where

t = 1/2(1/x + 1/y + 1/z), as defined above.

As a check, for an equilateral triangle, where x, y, z = 1,

t = 3/2, and thus

2s = (3/2) / (Sqrt(3/16)) = 2 Sqrt(3), which is the correct answer.

Answer 2

no cosines; like this omly; check me though!
S=sqrt{p(p-a)(p-b)(p-c)} – heron’s formula for area of triangle, where p=(a+b+c)/2; on the other hand S=a*h1/2 = b*h2/2 = c*h3/2;
Or p = S/h1+S/h2+S/h3; thus heron S^2 = S^4 (1/h1+1/h2+1/h3) (1/h1+1/h2) (1/h1+1/h3) (1/h2+1/h3), hence S^2 = 1 / {(1/h1+1/h2+1/h3) (1/h1+1/h2) (1/h1+1/h3) (1/h2+1/h3)} or S=1/sqrt(g) and a+b+c = 2S (1/h1+1/h2+1/h3) / sqrt(g) = 2 sqrt(1/h1+1/h2+1/h3) / sqrt((1/h1+1/h2) (1/h1+1/h3) (1/h2+1/h3));
xmas! once more

Answer 3

hiya! hear We no longer equals the equation to 0 in spite of the indisputable fact that it quite is in accordance to question. we would desire to continually discover the fee of x for which it quite is 0. as an occasion as u have given cos2x-sinx=0 . it quite is a situation. we would desire to continually resolve it. Its answer would be cos2x=sinx sin(ninety-2x)=sinx ninety-2x=x 3x=ninety x=30