1. Problem 5: Use the Pythagorean Theorem to find the length of side $b$ (hypotenuse $AC$) in triangle $ABC$ where $AB=6$ cm and $BC=8$ cm. 2. The Pythagorean Theorem states: $$a^2 + b^2 = c^2$$ where $c$ is the hypotenuse. 3. Here, $AB=6$ cm and $BC=8$ cm are the legs, and $AC=b$ is the hypotenuse. 4. Substitute values: $$6^2 + 8^2 = b^2$$ 5. Calculate squares: $$36 + 64 = b^2$$ 6. Sum: $$100 = b^2$$ 7. Take square root: $$b = \sqrt{100} = 10$$ cm. --- 1. Problem 6: Find measure of angle $\angle C$ in triangle $ABC$ with $AB=13$ cm, $BC=5$ cm, and $AC$ unknown. 2. Use cosine rule or trigonometry. Since $\angle C$ is opposite side $AB=13$ cm, and adjacent sides are $BC=5$ cm and $AC$ unknown, we use cosine: 3. Use cosine definition: $$\cos(\angle C) = \frac{\text{adjacent}}{\text{hypotenuse}}$$ 4. Here, adjacent side to $\angle C$ is $BC=5$ cm, hypotenuse is $AB=13$ cm. 5. Calculate cosine: $$\cos(\angle C) = \frac{5}{13}$$ 6. Find angle: $$\angle C = \cos^{-1}\left(\frac{5}{13}\right) \approx 67.38^\circ$$. --- 1. Problem 7: Find side $x$ in a right triangle with angle $17^\circ$, opposite side $9.9$ m, adjacent side $x$. 2. Use tangent function: $$\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}$$ 3. Substitute values: $$\tan(17^\circ) = \frac{9.9}{x}$$ 4. Solve for $x$: $$x = \frac{9.9}{\tan(17^\circ)}$$ 5. Calculate $\tan(17^\circ) \approx 0.3057$. 6. Compute $x$: $$x = \frac{9.9}{0.3057} \approx 32.39$$ m. Final answers: - Problem 5: $b=10$ cm - Problem 6: $\angle C \approx 67.38^\circ$ - Problem 7: $x \approx 32.39$ m