Finding tan(A+B) where sinA=7/25, sinB=5/13

[lemon]

1. If sinA=7/25 and sinB=5/13, where A is acute and B is obtuse, find the exact value of tan(A+B)



2. Tan(A+B)=tanA+tanB/1-tanAtanB



3. TanA=7/24
TanB=5/-12
Tan(A+B)=7/24+5/-12/1-7/24x5/-12=-36/323 or -0.1 (1d.p.)

Could somebody please check this for me, please?

 

[tiny-tim]

If sinA=7/25 and sinB=5/13, where A is acute and B is obtuse, find the exact value of tan(A+B)
…
Tan(A+B)=7/24+5/-12/1-7/24x5/-12=-36/323 or -0.1 (1d.p.)

Hi lemon!

Very good , except …

read the question … it asks for the exact value, which I assume means leave it as a fraction (exactly as the original data were given).

 

[Mark44]

Removed the extra bold tags...

1. If sinA=7/25 and sinB=5/13, where A is acute and B is obtuse, find the exact value of tan(A+B)



2. Tan(A+B)=tanA+tanB/1-tanAtanB

Please use parentheses. You have everything jammed together, so it's difficult to tell what's in the numerator and what's in the denominator. This should be written as

tan(A + B) = (tan A + tan B)/(1 - tan A * tan B)

3. TanA=7/24
TanB=5/-12
Tan(A+B)=7/24+5/-12/1-7/24x5/-12=-36/323 or -0.1 (1d.p.)

Again, please use parentheses. It would be clearer as

tan(A + B) = (7/24 - 5/12)/(1 - (7/24)(-5/12))

Could somebody please check this for me, please?

Your answer of -36/323 $\approx$ -0.111455 $\approx$ -0.1 is correct.

 

[lemon]

Understood. Thanks to you both