 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  By all means (Posted on 2011-06-16) In isosceles triangle AB=BC=n.

What value of AC warrants the largest area of the triangle ABC?
Solve by:
a) Plane geometry.
b) Trigonometry.
c) Calculus.
d) Any other way is welcome.

 See The Solution Submitted by Ady TZIDON No Rating Comments: ( Back to comment list | You must be logged in to post comments.) Another trigonometry way (this is too fun) | Comment 4 of 5 | Call angle BAC=a and AC = x

Then the area by the formula A=.5absin(C) is
A = .5xnsin(a)

By the law of cosines
n�= n� + x� - 2nxcos(a)
x� = 2nxcos(a)
x = 2ncos(a)

substitute this into the area formula
A = .5*2ncos(a)nsin(a)
= n�cos(a)sin(a)

Use the trig identity sin(2θ) = 2sin(θ)cos(θ) to rewrite this as
A = .5n�sin(2a)

For fixed n this has maximum value when 2a=90�
A = .5n�

Since a=45� substitute this into the first area equation above
.5xnsin(45�) = .5n�
xn(√(2)/2) = n�
x = n√2
 Posted by Jer on 2011-06-16 15:35:29 Please log in:

 Search: Search body:
Forums (0)