0

Guest

Courses New

Q.41 see image …......................................................

Q.41 see image …......................................................

Question Image
Grade:12

1 Answers

Aditya Gupta
2081 Points
4 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:)

Think You Can Provide A Better Answer ?

Other Related Questions on Algebra

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More