Originally posted by smaiaI got about 19.85 degrees, but i want to find the nice simple way
You will be amazed how beautiful and simple the solution is.
No need to use calculators or perform complex computations.
It can be resolved mentally if you discover a few hidden things.
(there are omitted degrees symbols....)
let x = AB = BC = AC
BD = DC = x/2
using simple trig, BED = 50 deg, EBD = 60 deg, BDE = 70 deg
ED/sin(60) = (x/2)/sin(50) (sine rule)
ED = x*sin(60)/sin(50) = sqroot(3)*x/(2sin(50))
sqr(EC) = sqr(ED) + sqr(DC) - 2*ED*DC*cos(110) (cosine rule, 110 deg from basic trig)
EC = sqroot(3sqr(x)/(4sqr(sin(50))) + sqr(x)/4 - 2 * sqroot(3)*sqr(x)/(4sin(50)) * cos(110)) = 1.39x
DC/sin(DEC) = EC/sin(110)
sin(DEC) = x/2 * sin(110)/EC
DEC = 19.849404914... degrees
very messy....
correct me if im wrong please
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Originally posted by sb3700can you resolve that mentally with no need for complex calculations, as the poster of this problem said the solution is simple ( manageable in the mind). If you can do all of that in your head, your the man. if i had to guess, your solution is not of the simple variety, just a thought😵
I got about 19.85 degrees, but i want to find the nice simple way
(there are omitted degrees symbols....)
let x = AB = BC = AC
BD = DC = x/2
using simple trig, BED = 50 deg, EBD = 60 deg, BDE = 70 deg
ED/sin(60) = (x/2)/sin(50) (sine rule)
ED = x*sin(60)/sin(50) = sqroot(3)*x/(2sin(50))
sqr(EC) = sqr(ED) + sqr(DC) - 2*ED*DC*cos(110) (cosine rul ...[text shortened]... * sin(110)/EC
DEC = 19.849404914... degrees
very messy....
correct me if im wrong please
Originally posted by sb3700Nice try, but sorry, this is not the correct answer. You don't need to develop such amazingly complicated calculations to solve this problem.
I got about 19.85 degrees, but i want to find the nice simple way
(there are omitted degrees symbols....)
let x = AB = BC = AC
BD = DC = x/2
using simple trig, BED = 50 deg, EBD = 60 deg, BDE = 70 deg
ED/sin(60) = (x/2)/sin(50) (sine rule)
ED = x*sin(60)/sin(50) = sqroot(3)*x/(2sin(50))
sqr(EC) = sqr(ED) + sqr(DC) - 2*ED*DC*cos(110) (cosine rul ...[text shortened]... * sin(110)/EC
DEC = 19.849404914... degrees
very messy....
correct me if im wrong please
I assure you: all you need is
1- the sum of the internal angles of a triangle is 180
2- The properties of similar triangles
3- the opposite angles of two intersecting lines are equal and other axioms of euclidian geometry
Good luck.
1 edit
Originally posted by smaiaAgain, irrelevant. It's about similifying your expressions.
True, but assume you are not allowed to use a calculator.
I got (first attempt - will check it later):
DEC = arctan[sin(50)sin(70)/(sin(50)cos(70) + sin(60))]
No calculators in sight. Clearly not the 'simple' method you're looking for, though.
2 edits
Originally posted by mtthwI came up with:
Again, irrelevant. It's about similifying your expressions.
I got (first attempt - will check it later):
DEC = arctan[sin(50)sin(70)/(sin(50)cos(70) + sin(60))]
No calculators in sight. Clearly not the 'simple' method you're looking for, though.
Angle CED = arccot { [ 2 (sin 60) - (sin 110) (cos 50) ] / [ (sin 110) (sin 50) ] }
which is equiivent with yours and smasia's first post
appx. = 32.544
if three of us agree then maybe his solution is flawed