+0

# Last Sum & Difference question

0
219
6
+260

So this is as far as i have got for this particular problem. I am scared to submit it because i think im forgeting some rule or conjuction. Anyway here is what i got.

sin A 4/5

sin B 3/5      I got these from the pythagorean theorem.

For tan A i did

4/5 / 3/5 = 4/5 x 5/3 = tan A 4/3

For tan B i did

3/5 / 25/5 = tan B 35/10

Im am confused about inserting the values in the formula. Im getting too big of answers, so someone please show me how you do this problem. This is the last one and them im going to sleep for 6 hours.

Veteran  Mar 18, 2017
Sort:

#1
+17711
0

I get a different value for sin(B).

If  cos(B)  =  ( 2sqrt(5) ) / 5, I draw the reference right triangle to have an adjacent side (x-side) of 2sqrt(5) and hypotenuse of 5.

Then, using the Pythagorean Theorem, I get the opposite side (y-side) to have a value of sqrt(5),

giving a value for sin(B) to be sqrt(5) / 5.

geno3141  Mar 18, 2017
#3
+260
0

Veteran  Mar 18, 2017
#2
0

Yes, Like gene3141 I get:

Sin(B) =1/sqrt(5), which is the same as: Sqrt(5) / 5

Guest Mar 18, 2017
#4
+260
0

11/2 is the answer. You dont even know what kind of values i ended up trying to convert for the formula. All because of a little error. thanks guys

Veteran  Mar 18, 2017
#5
0

Are you using this formula: tan(A + B) = sin(A + B) / cos(A + B).

Guest Mar 18, 2017
#6
+82522
0

sin A  = 4/5

sin B   =   sqrt (r^2 - x^2) / r  =  sqrt (5 - 4) / sqrt (5)    =   1/sqrt (5)

tan A    =   [ 4/5[ / [3/5] =   4/3

tan B    =  [1/sqrt(5) / [ 2/sqrt (5)]  =  1/2

tan (A + B)   =   [ tanA + tanB] / [1 - tanA*tanB]  =   [ (4/3) + (1/2)] / [ 1 - (4/3)(1/2) ] =

[11/6] / [ 1 - 4/6]  =    [11/6] / [ 2/6]  =   11/2

CPhill  Mar 18, 2017

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