Answer to Question #100152 in Geometry for Alden

Question #100152
A right triangle contains a 60degree angle. If the measure of the hypotenuse is 4, find the distance from the point of intersection of the 2 legs of the triangle to the point of intersection of the angle bisectors.
1
Expert's answer
2019-12-09T11:46:26-0500

We need to find the distance from point of intersection of the 2 legs of the triangle to the point of intersection of the Angular bisectors.






sin60o=ACAB=AB4AB=4×32=2×3sin 60^o = \frac {AC} {AB} = \frac {AB}{4} \\ AB = 4 \times \frac {\sqrt 3}{2} = 2 \times \sqrt {3}


cos60o=BCAB=BC4BC=4×12=2cos 60^o = \frac {BC}{AB} = \frac {BC}{4} \\ BC = 4 \times \frac {1}{2} = 2

The Point where the angular bisectors intersect is called "Incentre".

Denoted by "I".


Draw a line from the Incentre to the side of length 2 units.

This Line divides that side BC into two parts.

Let denote those parts by x and y.


This line makes two triangles, such that

one triangle is 30 - 60 - 90 triangle and the other one is 45-45 -90 triangle.


Let x be the length of the part in the 45-45-90 triangle and

y be the length of the part in the 30-60-90 triangle.


So,

x+y=2x + y = 2

The side opposite to 60o60^o is 3\sqrt 3 times the side opposite 30o30^o,



y=3×xy = \sqrt 3 \times x

Plug in the x+y=2x+ y = 2

x+3x=2x + \sqrt3 x = 2


x(3+1)=2x ( \sqrt {3} +1 ) = 2


x=23+1=31x = \frac {2}{ \sqrt {3} +1} = \sqrt {3} - 1

So, the Hypotenuse in the triangle 45-45-90 is 2\sqrt {2} times the length of a leg.


So, the Hypotenuse is 2(31)=62\sqrt 2 (\sqrt {3} - 1) = \sqrt 6 - \sqrt 2


Answer:

The distance from the point of intersection of the legs to Incentre is 62\sqrt 6 - \sqrt 2 .

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