To solve the given problem, let's follow these steps in detail:
### Step 1: Determine [tex]\(\tan A\)[/tex] and [tex]\(\sin B\)[/tex]
We are given:
[tex]\[ 8 \tan A = -15 \][/tex]
Dividing both sides by 8, we get:
[tex]\[ \tan A = -\frac{15}{8} \][/tex]
Thus, [tex]\(\tan A = -1.875\)[/tex].
Also given:
[tex]\[ 25 \sin B = -7 \][/tex]
Dividing both sides by 25, we get:
[tex]\[ \sin B = -\frac{7}{25} \][/tex]
Thus, [tex]\(\sin B = -0.28\)[/tex].
### Step 2: Determine [tex]\(\cos A\)[/tex] and [tex]\(\sin A\)[/tex]
Using the identity for [tex]\(\tan\)[/tex]:
[tex]\[ \tan^2 A + 1 = \sec^2 A \implies \sec^2 A = 1 + \left(-\frac{15}{8}\right)^2 \][/tex]
[tex]\[ \sec^2 A = 1 + \frac{225}{64} \][/tex]
[tex]\[ \sec^2 A = \frac{64}{64} + \frac{225}{64} \][/tex]
[tex]\[ \sec^2 A = \frac{289}{64} \][/tex]
Thus,
[tex]\[ \cos^2 A = \frac{1}{\sec^2 A} = \frac{64}{289} \][/tex]
[tex]\[ \cos A = \pm \sqrt{\frac{64}{289}} = \pm \frac{8}{17} \][/tex]
Since [tex]\(\tan A\)[/tex] is negative and [tex]\(A\)[/tex] is not in the fourth quadrant, [tex]\(A\)[/tex] is in the second quadrant where cosine is negative:
[tex]\[ \cos A = -\frac{8}{17} \][/tex]
Thus, [tex]\(\cos A = -0.47058823529411764\)[/tex].
Now using the Pythagorean identity for sine and cosine:
[tex]\[ \sin^2 A + \cos^2 A = 1 \][/tex]
[tex]\[ \sin^2 A = 1 - \cos^2 A \][/tex]
[tex]\[ \sin^2 A = 1 - \left(-\frac{8}{17}\right)^2 \][/tex]
[tex]\[ \sin^2 A = 1 - \frac{64}{289} \][/tex]
[tex]\[ \sin^2 A = \frac{289}{289} - \frac{64}{289} \][/tex]
[tex]\[ \sin^2 A = \frac{225}{289} \][/tex]
[tex]\[ \sin A = \sqrt{\frac{225}{289}} = \frac{15}{17} \][/tex]
Since [tex]\(A\)[/tex] is in the second quadrant where sine is positive:
[tex]\[ \sin A = \frac{15}{17} \][/tex]
Thus, [tex]\(\sin A = 0.8823529411764706\)[/tex].
### Step 3: Determine [tex]\(\cos B\)[/tex]
Using the identity:
[tex]\[ \sin^2 B + \cos^2 B = 1 \][/tex]
[tex]\[ \cos^2 B = 1 - \sin^2 B \][/tex]
[tex]\[ \cos^2 B = 1 - \left(-\frac{7}{25}\right)^2 \][/tex]
[tex]\[ \cos^2 B = 1 - \frac{49}{625} \][/tex]
[tex]\[ \cos^2 B = \frac{625}{625} - \frac{49}{625} \][/tex]
[tex]\[ \cos^2 B = \frac{576}{625} \][/tex]
[tex]\[ \cos B = \pm \sqrt{\frac{576}{625}} = \pm \frac{24}{25} \][/tex]
Since [tex]\(\sin B\)[/tex] is negative and [tex]\(B\)[/tex] is not in the fourth quadrant, [tex]\(B\)[/tex] is in the third quadrant where cosine is also negative:
[tex]\[ \cos B = -\frac{24}{25} \][/tex]
Thus, [tex]\(\cos B = -0.96\)[/tex].
### Step 4: Compute [tex]\(\sin A \cos B + \cos A \sin B\)[/tex]
Finally, substitute the values:
[tex]\[ \sin A \cos B + \cos A \sin B \][/tex]
[tex]\[ = \left(\frac{15}{17}\right) \left(-\frac{24}{25}\right) + \left(-\frac{8}{17}\right) \left(-\frac{7}{25}\right) \][/tex]
[tex]\[ = -\frac{360}{425} + \frac{56}{425} \][/tex]
[tex]\[ = -\frac{360 - 56}{425} \][/tex]
[tex]\[ = -\frac{304}{425} \][/tex]
Thus,
[tex]\[ \sin A \cos B + \cos A \sin B = -\frac{304}{425} \][/tex]
Which evaluates to approximately [tex]\(-0.7152941176470589\)[/tex].
This completes the solution.
