Question

If tanA=1 tanB=2 tanC=3 then find A+B+C=

Answer 1

Answer:

Answer is :- Value of $(A+B+C) = 0$

Step-by-step explanation:

Given :- $tan(A)=1 , tan (B)=2 , tan (C)=3$

To find :- $A+B+C$ value

Solution :-

Step 1) formula we have used here is ,

$tan ( A+B+C)= [ tan (A) + tan (B) + tan (C) - tan(A)*tan (B)*tan (C) ] / [ 1- tan(A)*tan (B) - tan(B)*tan (C) - tan(C)*tan (A) ]$

Step 2) put the value given in the above equation ,

we get ,

$tan(A+B+C) = [ 1+2+3 - ( 1*2*3) ] / [ 1 - (1*2)-(2*3) - ( 3*1) ]$

Step 3) After solving we get ,

$tan (A+B+C) = [ 6 - 6 ] / [ 1 - 2 - 6 - 3 ]$

Step 4) we got the equation as ,

$tan ( A+B+C) = 0$

Step 5) taking $tan^-1$ on both the sides ,

$tan ^-1 (tan (A+B+C)) = tan ^-1(0)$

we know , that $tan ^-1(tan x) = x \\and tan^-1(0) = 0$

Therefore , answer is $(A+B+C) = 0$

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Answer 2

Answer:

3

Step-by-step explanation:

Given,

tanA=1 tanB=2 tanC=3

To Find,

A+B+C

Theorem: For triangle ABC,

tanAtanBtanC=tanA+tanB+tanC

1(2)tanC=1+2+tanC

2tanC=3+tanC

tanC=3

Hence, the answer is 3.

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