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.

Printable View

- Feb 4th 2007, 07:06 AMclassicstringsCompound Angle Forumla
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. - Feb 4th 2007, 07:37 AMSoroban
Hello, classicstrings!

Quote:

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?

- Feb 4th 2007, 07:47 AMCaptainBlack
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