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 Tan Settlement (Posted on 2016-01-09)
The point O lies inside the triangle EFG, and ∠OEF = ∠OFG = ∠OGE.
Given that EF = 13, FG = 14, and GE =15, determine tan ∠OEF

 No Solution Yet Submitted by K Sengupta No Rating

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 A lot of Sine work (Solution) | Comment 2 of 3 |
For a problem asking about a tangent value, it took a lot of sines to get there!

Let T be the common measure of angles OEF, OFG, and OGE.  The problem asks us to find tan(T).

Start with triangle EFG.  By Heron's Formula the area is:
s = (13+14+15)/2 = 21
Area = sqrt[21*(21-13)*(21-14)*(21-15)] = 84.

From the sine area formula the area equals:
(1/2)*13*14*sin(EFG) = 91*sin(EFG)
(1/2)*13*15*sin(EFG) = (195/2)*sin(FEG)
(1/2)*14*15*sin(EFG) = 105*sin(EGF)

Equating each sine area to 84 yields:
sin(EFG) = 12/13
sin(FEG) = 56/65
sin(EGF) = 4/5

Looking at the interior angles gives:
angle OGF = angle EGF - T
angle OEG = angle FEG - T
angle OFE = angle GFE - T
angle FOG = pi - angle EGF
angle EOG = pi - angle FEG
angle EOF = pi - angle GFE

Applying the law of sines to triangles EOF, EOG, amd FOG yields:
sin(EOF)/13 = sin(T)/FO
sin(FOG)/14 = sin(T)/GO
sin(EOG)/15 = sin(T)/EO

Rearranging to solve for interior edges:
FO = (169/12)*sin(T)
GO = (35/2)*sin(T)
EO = (975/56)*sin(T)

Then use the sine area formula on all three interior triangles EOF, EOG, and FOG:
area EOF = (1/2)*EO*EF*sin(T) = (12675/112)*(sin(T)^2)
area FOG = (1/2)*FO*FG*sin(T) = (1183/12)*(sin(T)^2)
area EOG = (1/2)*GO*EG*sin(T) = (525/4)*(sin(T)^2)

Add these up and equate to the known total area and solve:
(12675/112)*(sin(T)^2) + (1183/12)*(sin(T)^2) + (525/4)*(sin(T)^2) = 84
(115249/336)*(sin(T)^2) = 84
sin(T) = 168/sqrt(115249)
Then cos(T) = 295/sqrt(115249)

Finally the answer tan(T) = 168/295 = 0.5694915254, which matches Jer's numeric approximation.

The expression for sin(T) turns out to be:
2*area(EFG) / sqrt[(EF*EG)^2 + (EF*FG)^2 + (EG*FG)^2]
Similarily, the expression for cos(T):
(1/2)*(EF^2+EG^2+FG^2) / sqrt[(EF*EG)^2 + (EF*FG)^2 + (EG*FG)^2]
And tan(T) = 4*area(EFG)/(EF^2+EG^2+FG^2)

Edited on January 11, 2016, 1:07 pm
 Posted by Brian Smith on 2016-01-11 12:58:15

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