Question

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.

Answer 1

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.

$$sin 60^o = \frac {AC} {AB} = \frac {AB}{4} \\ AB = 4 \times \frac {\sqrt 3}{2} = 2 \times \sqrt {3}$$

$$cos 60^o = \frac {BC}{AB} = \frac {BC}{4} \\ BC = 4 \times \frac {1}{2} = 2$$

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,

$$x + y = 2$$

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

$$y = \sqrt 3 \times x$$

Plug in the $x+ y = 2$

$$x + \sqrt3 x = 2$$

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

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

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

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

Answer:

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