tan A = 4/7 , tan B = 1/8 , tan C= 1/7 find A + B + C
Answers
Step-by-step explanation:
Given :-
tan A = 4/7 ,
tan B = 1/8 ,
tan C= 1/7
To find:-
Find A + B + C ?
Solution :-
Method -1:-
Given that :-
tan A = 4/7 ,
tan B = 1/8 ,
tan C= 1/7
We know that
Tan (A+ B ) = (Tan A + Tan B )/(1-Tan A Tan B)
On Substituting these values in the above formula
=> Tan (A+B) = [(4/7)+(1/8)]/[1-(4/7)(1/8)]
LCM of 7 and 8 = 56
=> Tan (A+B) = [(32+7)/56]/[1-(4/56)]
=> Tan (A+B) =(39/56)/[(56-4)/56]
=> Tan (A+B) = (39/56)/(52/56)
=> Tan(A+B) = 39/52
=> Tan (A+B) = 3/4---------(1)
and
Now finding Tan(A+B+C)
=> Tan [(A+B)+C]
=> [Tan (A+B) +Tan C]/[1-Tan(A+B)Tan C]
On Substituting these values in the above formula
=> [(3/4)+(1/7)]/[1-(3/4)(1/7)]
LCM of 4 and 7 = 28
=> [(21+4)/28]/[1-(3/28)]
=> (25/28)/[(28-3)/28]
=> (25/28)/(25/28)
=> 1
=> Tan (A+B+C) = 1
=> Tan (A+B+C) = Tan 45°
=> A+B+C = 45°
Method -2:-
Given that :-
tan A = 4/7 ,
tan B = 1/8 ,
tan C= 1/7
We know that
Tan(A+B+C)=[(TanA+TanB+TanC-(TanA TanBTanC)]/[(1-TanATan B-TanB Tan C -TanCTanA)]
On Substituting these values in the above formula
=>[(4/7)+(1/8)+(1/7)-(4/7)(1/8)(1/7)]/[1-(4/7)(1/8)-(1/8)(1/7)-(1/7)(4/7)]
=> [(5/7)+(1/8)-(4/392)][1-(4/56)-(1/56)-(4/49)]
LCM of 7 ,8 = 56
=> [{(40+7)/56}-(1/98)]/ [ 1-((4+1)/56)-(4/49)]
=> [(47/56)-(1/98)]/[1-(5/56)-(4/49)]
LCM of 56 and 98 = 392
LCM of 56 and 49 = 392
=> (329-4)/392]/[392-35-32)/392]
=> (325/392)/(392-67/392)
=>(325/392)/(325/392)
=> 1
=> Tan (A+B+C) = 1
=> Tan (A+B+C) = Tan 45°
=> A+B+C = 45°
Answer:-
The value of A+B+C for the given problem is 45°
Used formulae:-
-» Tan(A+B ) =(Tan A+Tan B)/(1-Tan ATan B)
-»Tan(A+B+C)=(TanA+TanB+TanC-(TanATanBTanC)/(1-TanATan B-TanB Tan C -TanCTanA)