Inside a square ABDE, place a point C so that CDE is an isosceles triangle with angles 15 degrees at D and E. What are the values of the angles of triangle ABC?
Cinelli
Got a credit card? use our Credit Card & Finance Calculators
Thanks to Rhyd6,eyeball08,Wondergirly,bofh,johnstevens77, for Donating to support the site
Triangle
-
- The full Lemon
- Posts: 10813
- Joined: November 4th, 2016, 8:17 pm
- Has thanked: 1471 times
- Been thanked: 3005 times
Re: Triangle
cinelli wrote:Inside a square ABDE, place a point C so that CDE is an isosceles triangle with angles 15 degrees at D and E. What are the values of the angles of triangle ABC?
Cinelli
Are we allowed trig functions? If so, surely far too obvious!
If (for convenience) we take the side of the square as 2 units and X as the midpoint of DE, then DX = XE = 1, and CX = tan(15).
So if Y is the midpoint of AB, then CY = 2 - tan(15).
Thus the angles at A and B are each arc tan(2-tan(15)), and at C is the remainder subtracting those at A and B from 180.
-
- Lemon Quarter
- Posts: 1017
- Joined: December 9th, 2016, 6:44 am
- Has thanked: 233 times
- Been thanked: 308 times
Re: Triangle
UncleEbenezer wrote:cinelli wrote:Inside a square ABDE, place a point C so that CDE is an isosceles triangle with angles 15 degrees at D and E. What are the values of the angles of triangle ABC?
Cinelli
Are we allowed trig functions? If so, surely far too obvious!
If (for convenience) we take the side of the square as 2 units and X as the midpoint of DE, then DX = XE = 1, and CX = tan(15).
So if Y is the midpoint of AB, then CY = 2 - tan(15).
Thus the angles at A and B are each arc tan(2-tan(15)), and at C is the remainder subtracting those at A and B from 180.
If Φ is the angle ABC we can rewrite that as
tan(Φ) + tan(15) =2
You can calculate trig values for 15 exactly using the half angle formula for tan and known values for cos(30) and sin(30)
tan(Θ/2) = (1- cos(Θ)/sin(Θ)
so
tan(15) = (1-cos(30))/sin(30) = (1-sqrt(3)/2)/(1/2) = 2- sqrt(3)
Substituting
tan(Φ) + 2-sqrt(3) =2
tan(Φ) = sqrt(3)
So by inspection Φ = 60
There's probably a trig identity that lets you get there in one step, but that requires memorizing more trig identities.
Once you have that trig identity you can probably disguise it as a calculation that does not use trig.
-
- Lemon Quarter
- Posts: 1017
- Joined: December 9th, 2016, 6:44 am
- Has thanked: 233 times
- Been thanked: 308 times
Re: Triangle
Once we know the answer we can prove it simply.
Construct the equilateral triangle AGB.
Then ∠EAG = 30 and AE = AG.
This is an isoscelese triangle so ∠AEG = (180 -30)/2 = 85
Thus ∠EGD = 15 and points G and C are the same.
Verification simple, derivation requires clairvoyance (or trig).
Please sir may I have another Green's function?
-
- Lemon Slice
- Posts: 553
- Joined: November 9th, 2016, 11:33 am
- Has thanked: 234 times
- Been thanked: 161 times
Re: Triangle
I like these ingenious methods. I have another solution which doesn't involve trig or clairvoyance which I will post in a couple of days (to give time for anyone else to have a go).
Cinelli
Cinelli
-
- Lemon Slice
- Posts: 553
- Joined: November 9th, 2016, 11:33 am
- Has thanked: 234 times
- Been thanked: 161 times
Re: Triangle
Here is my alternative solution.
Construct another isosceles triangle AFE with the angles at A and E 15 degrees. Produce AF to meet CE at G. Then CEF Is equilateral and since angle EAF is 15 degrees and angle AEG is 85 degrees, angle AGE is a right angle. Hence G is the mid point of CE and triangle CAE is isosceles. So AE = AC = AB = BC (by symmetry) and ABC is equilateral.
Cinelli
~
~
~
A................................................E
. . . . ..
. . . . ..
. . F . .
. . . . . .
. . . . . .
. . . G .
. . . . .
. . . . .
. . . . .
. . . . .
. .C .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . ..
. . .
. . .
B................................................D
Construct another isosceles triangle AFE with the angles at A and E 15 degrees. Produce AF to meet CE at G. Then CEF Is equilateral and since angle EAF is 15 degrees and angle AEG is 85 degrees, angle AGE is a right angle. Hence G is the mid point of CE and triangle CAE is isosceles. So AE = AC = AB = BC (by symmetry) and ABC is equilateral.
Cinelli
~
~
~
Return to “Games, Puzzles and Riddles”
Who is online
Users browsing this forum: No registered users and 15 guests