if tan a = 7/24 and sin b = 15/17 and angles ...

# Explanation: ## Step1: Find $\cos B$ Since $\sin^{2}B+\cos^{2}B = 1$ and $\sin B=\frac{15}{17}$, then $\cos B=\sqrt{1 - \sin^{2}B}=\sqrt{1 - (\frac{15}{17})^{2}}=\sqrt{\frac{289 - 225}{289}}=\sqrt{\frac{64}{289}}=\frac{8}{17}$ (because $B$ is in Quadrant I). ## Step2: Find $\tan B$ $\tan B=\frac{\sin B}{\cos B}=\frac{\frac{15}{17}}{\frac{8}{17}}=\frac{15}{8}$. ## Step3: Use the formula for $\tan(A - B)$ The formula for $\tan(A - B)=\frac{\tan A-\tan B}{1 + \tan A\tan B}$. Given $\tan A=\frac{7}{24}$ and $\tan B=\frac{15}{8}$, we substitute these values: \[ \begin{align*} \tan(A - B)&=\frac{\frac{7}{24}-\frac{15}{8}}{1+\frac{7}{24}\times\frac{15}{8}}\\ &=\frac{\frac{7 - 45}{24}}{1+\frac{105}{192}}\\ &=\frac{-\frac{38}{24}}{\frac{192 + 105}{192}}\\ &=\frac{-\frac{38}{24}}{\frac{297}{192}}\\ &=-\frac{38}{24}\times\frac{192}{297}\\ &=-\frac{38\times8}{297}\\ &=-\frac{304}{297} \end{align*} \] # Answer: $-\frac{304}{297}$