1. Given a triangle with a vertical left side of length 9, a horizontal base of length 10, and the left angle of 50°. 2. Use the Law of Cosines or trigonometric ratios to find the diagonal length $c$: $$c = \sqrt{9^2 + 10^2 - 2 \times 9 \times 10 \times \cos(50^\circ)}$$ 3. Calculate: $$c = \sqrt{81 + 100 - 180 \times \cos(50^\circ)}$$ 4. Approximate $\cos(50^\circ) \approx 0.6428$: $$c = \sqrt{181 - 180 \times 0.6428} = \sqrt{181 - 115.7} = \sqrt{65.3} \approx 8.08$$ 1. Triangle with base 9, left slanted side 9, and angle at top vertex B. 2. Since two sides are equal (9 and 9), it is isosceles. 3. The base angles are equal; the top angle is given. 4. No further calculation requested. 1. Triangle with vertical left side 5, horizontal base 10, diagonal 6, angle at A marked. 2. Use Law of Cosines to find angle at A: $$\cos(A) = \frac{5^2 + 10^2 - 6^2}{2 \times 5 \times 10} = \frac{25 + 100 - 36}{100} = \frac{89}{100} = 0.89$$ 3. Calculate angle: $$A = \cos^{-1}(0.89) \approx 27.13^\circ$$