If A and B are angles between 0 and 90, and sinA = 3/5 and tanB = 7/24, find the value of
sin(A-B).
How do I tackle this question? I just need to know the pathway. Thanks.
Hello, classicstrings!
If $\displaystyle A$ and $\displaystyle B$ are angles between $\displaystyle 0^i$ and $\displaystyle 90^o$,
and $\displaystyle \sin A = \frac{3}{5}$ and $\displaystyle \tan B = \frac{7}{24}$,
find the value of $\displaystyle \sin(A-B)$
We have the identity: .$\displaystyle \sin(A - B) \:=\:\sin(A)\cos(B) - \sin(B)\cos(A)$
We know that: . $\displaystyle \sin(A) = \frac{3}{5}$
We need the values of: .$\displaystyle \cos(A),\:\sin(B),\:\cos(B)$
From $\displaystyle \sin(A) = \frac{3}{5} = \frac{opp}{hyp}$ and Pythagorus, we find that: .$\displaystyle adj = 4$
. . Now you can find $\displaystyle \cos(A).$
From $\displaystyle \tan(B) = \frac{7}{24} = \frac{opp}{adj}$ and Pythagorus, we find that: .$\displaystyle hyp = 25$
. . Now you can find $\displaystyle \sin(B)$ and $\displaystyle \cos(B).$
Got it?
First you need to know that:
sin(A-B)=sin(A)cos(B)-cos(A)sin(B).
Then as A and B are both in the range 0 to 90 degrees,
if sin(A)=3/5, then cos(A)=4/5 (there are a number of ways of getting this
result one of which is recognising that A is an angle of a 3,4,5 right triangle
which imediatly tells you that cos(A)=4/5).
if tan(B)=7/24, then sin(B)=7/25, and cos(B)=24/25 (as 25 is the hypotenuse
of a right triangle who's other two sides are 7 and 24).
So:
sin(A-B)=(3/5)(24/25)-(4/5)(7/25)=44/125
RonL