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Please answer my question attached in image. The answer given in book is C. SecA:secB:secC

Please answer my question attached in image.
The answer given in book is 
C. SecA:secB:secC

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Grade:11

1 Answers

Asdf124
15 Points
4 years ago
Let there be a triangle ABC such that angle A angle B and angle C be acute .
 Let the altitude from A touch BC at D such that it is perpendicular to BC
Let altitude from B touch AD at E such that it is perpendicular to AD 
Let altitude from C touch AB at F such that it is perpendicular to AB
 
In triangle BFC
angle FCB = 90-B
 
In triangle HDC 
angle HCD =90-B
Therefore angle DHC =B
Since sec DHC= secB
 
Therefore SecB=HC/HD (eq1)
 
In triangle AFC
angleFCA=90-A 
 
In triangle HEC
angle HCE=90-90-A 
 
Therefore angle EHC=A
 
Therefore secEHC =secA
 
Therefore sec A =HC/HE (eq 2) 
 
From eq 1 and 2 
HC =secA×HE=sec×HD 
 
Therefore HD/HE =secA/secB
 
So HD = secA and HE=secB
 
In triangle ABD
angle BAD=90-B
 
In triangle HAF 
angleHAF=90-B 
 
Therefore angle FHA=B
 
SsecFHA =secB
sec B=HA/HF (eq1 3)
 
In triangle ADC 
angle DAC=90-C
In triangle HAE
angle HAE =90-C
 
Therefore angle AHE =C
secAHE=secB
 
secC =HA/HE (eq 4)
 
From eq 3 and 4
HA = HE×secC = HF×secB
 
HE/HF = secB/secC
 
So HE=secB and HF=secC
 
Therefore HD:HE:HF= secA:secB:secC
 
 

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