Geometry Problem

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 619.

## Thursday, June 9, 2011

### Problem 619: Triangle Area, 30 Degrees, Incircle, Incenter, Tangent

Labels:
30 degrees,
area,
incenter,
incircle,
tangency point,
tangent,
triangle

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a=BC=5+2=7

ReplyDelete5=BD=s-b, 2=DC=s-c

∆=√[s(s-a)(s-b)(s-c)]

=√[s(s-a).5.2]

=√10.√[s(s-a)]

tan(A/2)=√[(s-b)(s-c)/s(s-a)]=√10/√[s(s-a)]

Multiplying respective sides:

∆ tan(A/2)=10

∆=10.cot15°=10(2+√3)sq units