Given sin A = 3/5, cos B = -24/25, 90° < A < 180°, 180° < B < 270°
Find tan(B - A)
in QII sinA = 3/5, tanA = -3/4
in QIII cosB = -24/25, tanA = 7/24
Now just apply tour sum formula
tan(B-A) = (tanB-tanA)/(1 + tanB tanA)
sinA = 3/5, and A is in II
then cosA = -4/5
leads to tan A = (3/5) / (-4/5) = -3/4
cos B = -24/25 , B in III,
then sinB = -7/25
which lead to tan B = sinB/cosB = (-7/25) / (-24/25) = 7/24
I got the above two by recognizing the 3-4-5 and 7-24-25 right-angled triangles.
You could of course make sketches and use Pythagoras to find them.
tan (B-A) = (tanB - tanA)/(1 + tanAtanB)
= ((7/24) - (-3/4))/(1 + (-3/4)(7/24))
= (25/24) / ( 25/32)
= 4/3
To find tan(B - A), we need to know the values of tan(A) and tan(B).
Given that sin A = 3/5, we can use the Pythagorean identity to find cos A:
cos^2 A = 1 - sin^2 A
cos^2 A = 1 - (3/5)^2
cos^2 A = 1 - 9/25
cos^2 A = 16/25
cos A = √(16/25)
cos A = 4/5
Since 90° < A < 180°, A is in the second quadrant where the cosine is negative. Therefore, cos A = -4/5.
Similarly, given that cos B = -24/25, we can use the Pythagorean identity to find sin B:
sin^2 B = 1 - cos^2 B
sin^2 B = 1 - (-24/25)^2
sin^2 B = 1 - 576/625
sin^2 B = 49/625
sin B = √(49/625)
sin B = 7/25
Since 180° < B < 270°, B is in the third quadrant where both sine and cosine are negative. Therefore, sin B = -7/25.
Now, we can calculate the value of tan(A) and tan(B):
tan A = sin A / cos A
tan A = (3/5) / (-4/5)
tan A = -3/4
tan B = sin B / cos B
tan B = (-7/25) / (-24/25)
tan B = 7/24
Finally, we can find tan(B - A):
tan(B - A) = (tan B - tan A) / (1 + tan A * tan B)
tan(B - A) = (7/24 - (-3/4)) / (1 + (-3/4) * (7/24))
tan(B - A) = (7/24 + 3/4) / (1 - 21/96)
tan(B - A) = (7/24 + 72/96) / (96/96 - 21/96)
tan(B - A) = (7/24 + 72/96) / (75/96)
tan(B - A) = (28/96 + 72/96) / (75/96)
tan(B - A) = 100/96 / 75/96
tan(B - A) = 100/96 * 96/75
tan(B - A) = 100/75
Therefore, tan(B - A) is equal to 100/75.