Q.41 see image …......................................................
Harsh vijay , 5 Years ago
Grade 12
Algebra> Q.41 see image …............................
1 AnswersAditya Gupta
Last Activity: 5 Years ago
my ans is coming A= 2arcsin(2/3) so that secantA= 9
let point of intersection of altitudes be P. let A/2= x. then angle PCB= x too!
so tanx= 2r/a, where r is inradius and a= BC/2.
further, area= r*s where s is semi perimeter.
so ½*2a*h= r*(2a+2l)/2 where h is height of triangle from A and l=AB=AC
now obviously a/h= tanx and r/a= tanx/2 and l/h= 1/cosx
so solving the eqns, you will obtain sinx= 2/3
now secA= 1/cos2x= 1/(1 – 2sin^2x)
= 9
kindly approve:)
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