In \(\Delta\)ABC, if \(r_1=r_2+r_3+r\) then a) A=\(45^0\) b) B=\(90^0\) c) C=\(90^0\) d) A=\(90^0\)
r1,r2,r3,r ?
r1, r2, r3 are ex-radius opposite to angles A, B and C. r is in-radius.
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i can solve it with some formulas but i dont know if u are familiar with this formulas or not?
we know that \(\large r_{1}=p \tan \frac{A}{2} \) , \(\large r_{2}=p \tan \frac{B}{2} \) , \(\large r_{3}=p \tan \frac{C}{2} \) , \(\large r=p \tan \frac{A}{2} \tan \frac{B}{2} \tan \frac{C}{2}\) where p is half of Perimeter of Triangle
r1=r2+r3+r so \(\large \tan \frac{A}{2}=\tan \frac{B}{2}+\tan \frac{C}{2}+\tan \frac{A}{2}\tan \frac{B}{2} \tan \frac{C}{2} \) then u have \( \large \tan \frac{A}{2}=\frac{\tan \frac{B}{2}+\tan \frac{C}{2}}{1-\tan \frac{B}{2}\tan \frac{C}{2}} =\tan (\frac{B}{2}+\frac{C}{2})=\tan (\frac{\pi}{2}-\frac{A}{2})=\cot \frac{A}{2} \) so A=90
Thanks a lot! yeah, I was not familiar with those formulas but now i am! Thanks a lot @mukushla
welcome my friend
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