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
ABC , 7 Years ago
Grade 11
Trigonometry> Please answer my question attached in ima...
1 AnswersAsdf124
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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