1. **Problem statement:** Given the equation $$13 \cos \alpha - 5 = 0$$ and the condition $$\tan \alpha < 0$$, find the value of $$13 \sin \alpha + 25 \tan^2 \alpha$$ without using a calculator and with a suitable diagram. 2. **Step 1: Solve for $$\cos \alpha$$** From the equation: $$13 \cos \alpha - 5 = 0$$ Add 5 to both sides: $$13 \cos \alpha = 5$$ Divide both sides by 13: $$\cos \alpha = \frac{5}{13}$$ 3. **Step 2: Determine the quadrant of $$\alpha$$** Since $$\tan \alpha < 0$$ and $$\cos \alpha = \frac{5}{13} > 0$$, cosine is positive and tangent is negative. Cosine is positive in Quadrants I and IV. Tangent is negative in Quadrants II and IV. Therefore, $$\alpha$$ lies in Quadrant IV. 4. **Step 3: Find $$\sin \alpha$$** Use the Pythagorean identity: $$\sin^2 \alpha + \cos^2 \alpha = 1$$ Substitute $$\cos \alpha = \frac{5}{13}$$: $$\sin^2 \alpha + \left(\frac{5}{13}\right)^2 = 1$$ $$\sin^2 \alpha + \frac{25}{169} = 1$$ $$\sin^2 \alpha = 1 - \frac{25}{169} = \frac{169}{169} - \frac{25}{169} = \frac{144}{169}$$ $$\sin \alpha = \pm \frac{12}{13}$$ Since $$\alpha$$ is in Quadrant IV, where sine is negative, we have: $$\sin \alpha = -\frac{12}{13}$$ 5. **Step 4: Find $$\tan \alpha$$** Recall: $$\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{-\frac{12}{13}}{\frac{5}{13}} = -\frac{12}{5}$$ 6. **Step 5: Calculate $$13 \sin \alpha + 25 \tan^2 \alpha$$** Calculate each term: $$13 \sin \alpha = 13 \times \left(-\frac{12}{13}\right) = -12$$ Calculate $$\tan^2 \alpha$$: $$\left(-\frac{12}{5}\right)^2 = \frac{144}{25}$$ Multiply by 25: $$25 \times \frac{144}{25} = 144$$ Sum the two terms: $$-12 + 144 = 132$$ **Final answer:** $$\boxed{132}$$