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# help with old problem please ;-;

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hi there!!!

i'm been having some trouble with this precalc problem so i turned to the internet for help but all i found was this:

https://web2.0calc.com/questions/trigonometry_20584

the solution to the problem there is incorrect so i was hoping someone could walk me through how to do this problem? i'm really struggling smh. thank you so so so much in advance!!! :D

Sep 6, 2023

#1
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tan (A + B) =   [ tan A + tan B ]  / [ 1 - tan A tanB ] =   [ 5/2 + 1/2] / [ 1 - (5/2)(1/2) ] = 3 / (-1/4)  =  -12

We can solve for tan G thusly :

tan (B + G)  = [ tan B + tan G ] / [ 1 - tanB tanG ]

3/2 =   [   1/2 + tan G ] / [ 1 - (1/2) tan G ]

(3/2)  [ 1 - (1/2) tan G ]   =  1/2 + tan G

(3/2) - (3/4)tanG = 1/2+ tanG

(3/2) - (1/2)  = tan G + (3/4)tan G

1 =   ( 1 + 3/4) tan G

1 =  (7/4) tan G

tan G  =  4/7

tan ( A + B + G)  =   [ (tan A + tan B + tan G)  - tanA tanB tan G ]  /  [ 1 - ( tanAtanB + tanAtanG + tanBtanG) ]

[ (5/2 + 1/2 + 4/7 )  - (5/2)(1/2)(4/7) ] / [ 1 - ( 5/4 + 20/14 + 4/14) ] =

[  (3 + 4/7 ) - ( 20/ 28) ] / [ 1 - ( 5/4 + 12/7 ) ]  =

[   25/7 - 5/7 ]  / [  -55/28 ]  =

[ 20/7 ] / [ -55/28 ] =

-(20 * 28) / ( 7 *55)  =

-20 * 4  / 55  =

-80  / 55  =

-16 / 11

I'm not too sure about the formula for the triple sum, but I think that it's correct !!!   Sep 6, 2023
#2
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yeah!! that's so helpful!! thank you so much i get it now :D

heartSTORM907  Sep 6, 2023