1. Use the Pythagorean theorem to find the unknown sides. **Problem:** - Triangle ABC: right angle at C, sides AC = 12 m, BC = 5 m, hypotenuse AB = $c$. - Triangle DEF: right angle at E, sides EF = 8 m, hypotenuse DF = 11 m, side DE = $f$. **Formula:** The Pythagorean theorem states: $$c^2 = a^2 + b^2$$ where $c$ is the hypotenuse and $a$, $b$ are the legs of the right triangle. **Step 1: Find $c$ in triangle ABC.** $$c^2 = 12^2 + 5^2 = 144 + 25 = 169$$ $$c = \sqrt{169} = 13$$ So, $c = 13$ m. **Step 2: Find $f$ in triangle DEF.** Here, hypotenuse $DF = 11$, one leg $EF = 8$, and unknown leg $DE = f$. $$11^2 = f^2 + 8^2$$ $$121 = f^2 + 64$$ $$f^2 = 121 - 64 = 57$$ $$f = \sqrt{57} \approx 7.55$$ So, $f \approx 7.55$ m. 2. Determine sine, cosine, and tangent ratios for the given angles. **Triangle ABC, angle $A$: ** - Opposite side to $\angle A$ is $BC = 5$ m. - Adjacent side to $\angle A$ is $AC = 12$ m. - Hypotenuse is $AB = 13$ m. $$\sin A = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{5}{13}$$ $$\cos A = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{12}{13}$$ $$\tan A = \frac{\text{opposite}}{\text{adjacent}} = \frac{5}{12}$$ **Triangle DEF, angle $D$: ** - Opposite side to $\angle D$ is $EF = 8$ m. - Adjacent side to $\angle D$ is $DE = f \approx 7.55$ m. - Hypotenuse is $DF = 11$ m. $$\sin D = \frac{8}{11}$$ $$\cos D = \frac{7.55}{11} \approx 0.686$$ $$\tan D = \frac{8}{7.55} \approx 1.06$$ **Summary:** - $c = 13$ m - $f \approx 7.55$ m - $\sin A = \frac{5}{13}$, $\cos A = \frac{12}{13}$, $\tan A = \frac{5}{12}$ - $\sin D = \frac{8}{11}$, $\cos D \approx 0.686$, $\tan D \approx 1.06$