+0

# Triangle \$\triangle ABC\$ has a right angle at \$C\$, \$\angle A = 60^\circ\$, and \$AC=10\$. Find the radius of the incircle of \$\triangle ABC\$

0
55
1
+298

Triangle \$\triangle ABC\$ has a right angle at \$C\$, \$\angle A = 60^\circ\$, and \$AC=10\$. Find the radius of the incircle of \$\triangle ABC\$

michaelcai  Oct 2, 2017
Sort:

#1
+77221
+2

See the following image :

The center of the  incircle will be the intersection  of the angle bisectors shown

And the find the x coordinate of the center by solving these two equations :

y  = tan (135) [x -10sqrt(3)]       and  y  = tan(60) [x - 10sqrt (3)] + 10

Set these equations equal   and we have

tan (135) [x -10sqrt(3)]   = tan(60) [x - 10sqrt (3)] + 10       simplify

-1 [x - 10sqrt (3) ]   =  sqrt (3) [ x  - 10sqrt (3) ] + 10

10sqrt (3)  -  x   =  sqrt (3) x  - 30 + 10

x ( sqrt (3) + 1 )  =  10sqrt (3) + 20

x ( sqrt (3)  + 1)   =  10 [ sqrt (3)  + 2 ]

x =   10 [  sqrt (3)  + 2 ] / [ sqrt (3)  + 1 ]            multiply  top /bottom  by  sqrt (3) - 1     and we have

x  = 10 [  sqrt (3)  + 2 ] [ sqrt (3) - 1 ] /  [ 3 - 1]

x =  10  [  sqrt (3)  + 2 ] [ sqrt (3) - 1 ] /  [ 2]

x =  5 [ 3 + sqrt (3)  - 2]

x  =  5 [ 1 + sqrt (3) ]  =  5  + 5sqrt (3)

And the y coordinate of the center is :

And    y =   -1 [ 5 + 5sqrt (3) - 10sqrt (3) ]    =  -1 [  5 - 5sqrt (3) ]  =  5sqrt (3) - 5  =

5 [ sqrt (3)  - 1 ]       and this is the radius of the incircle

CPhill  Oct 2, 2017

### 21 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details