+122 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
+129883 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 !!!
