Question:

Step 1: Rewrite and Group Equations The given equations are: $$2\cos \alpha + 6\cos \beta + 7\cos \gamma + 9\cos \delta = 0 \quad (1)$$ $$2\sin \alpha - 6\sin \beta + 7\sin \gamma - 9\sin \delta = 0 \quad (2)$$ Rearrange the terms to group coefficients $2$ with $9$ and $6$ with $7$: From equation (1): $$2\cos \alpha + 9\cos \delta = -(6\cos \beta + 7\cos \gamma)$$ From equation (2): $$2\sin \alpha - 9\sin \delta = 6\sin \beta - 7\sin \gamma$$ Step 2: Square and Add the Rearranged Equations Square both rearranged equations: $$(2\cos \alpha + 9\cos \delta)^2 = (-(6\cos \beta + 7\cos \gamma))^2 = (6\cos \beta + 7\cos \gamma)^2$$ $$(2\sin \alpha - 9\sin \delta)^2 = (6\sin \beta - 7\sin \gamma)^2$$ Add the squared equations: $$(2\cos \alpha + 9\cos \delta)^2 + (2\sin \alpha - 9\sin \delta)^2 = (6\cos \beta + 7\cos \gamma)^2 + (6\sin \beta - 7\sin \gamma)^2$$ Step 3: Expand and Simplify Expand the left-hand side (LHS): $$\text{LHS} = (4\cos^2 \alpha + 81\cos^2 \delta + 36\cos \alpha \cos \delta) + (4\sin^2 \alpha + 81\sin^2 \delta - 36\sin \alpha \sin \delta)$$ $$\text{LHS} = 4(\cos^2 \alpha + \sin^2 \alpha) + 81(\cos^2 \delta + \sin^2 \delta) + 36(\cos \alpha \cos \delta - \sin \alpha \sin \delta)$$ Using the identity $\cos^2 x + \sin^2 x = 1$ and the angle addition formula $\cos(A+B) = \cos A \cos B - \sin A \sin B$: $$\text{LHS} = 4(1) + 81(1) + 36\cos(\alpha + \delta)$$ $$\text{LHS} = 85 + 36\cos(\alpha + \delta)$$ Expand the right-hand side (RHS): $$\text{RHS} = (36\cos^2 \beta + 49\cos^2 \gamma + 84\cos \beta \cos \gamma) + (36\sin^2 \beta + 49\sin^2 \gamma - 84\sin \beta \sin \gamma)$$ $$\text{RHS} = 36(\cos^2 \beta + \sin^2 \beta) + 49(\cos^2 \gamma + \sin^2 \gamma) + 84(\cos \beta \cos \gamma - \sin \beta \sin \gamma)$$ Using the identity $\cos^2 x + \sin^2 x = 1$ and the angle addition formula $\cos(A+B) = \cos A \cos B - \sin A \sin B$: $$\text{RHS} = 36(1) + 49(1) + 84\cos(\beta + \gamma)$$ $$\text{RHS} = 85 + 84\cos(\beta + \gamma)$$ Step 4: Equate and Determine the Ratio Equating the LHS and RHS: $$85 + 36\cos(\alpha + \delta) = 85 + 84\cos(\beta + \gamma)$$ $$36\cos(\alpha + \delta) = 84\cos(\beta + \gamma)$$ Since $\beta$ and $\gamma$ are not complementary, $\beta+\gamma \neq \frac{\pi}{2}$. Thus, $\cos(\beta+\gamma) \neq 0$, and we can divide by $\cos(\beta+\gamma)$: $$\frac{\cos(\alpha + \delta)}{\cos(\beta + \gamma)} = \frac{84}{36}$$ Simplify the fraction: $$\frac{84}{36} = \frac{12 \times 7}{12 \times 3} = \frac{7}{3}$$ Step 5: Find (m + n) Given that $\frac{\cos(\alpha + \delta)}{\cos(\beta + \gamma)} = \frac{m}{n}$, we have $m=7$ and $n=3$. These are relatively prime positive numbers. The value of $(m+n)$ is: $$m+n = 7+3 = 10$$