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
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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