2. calculate the angle of elevation and the slope of the hypotenuse.\na) 11 ft 25 ft\nb) 29 cm 14 cm\n3. a…

Explanation:

Step1: Recall slope and angle - elevation formulas

The slope (m) of a right - triangle is given by (m=\frac{\text{rise}}{\text{run}}), and the angle of elevation (\theta) can be found using (\tan\theta=\frac{\text{rise}}{\text{run}}), so (\theta = \arctan(\frac{\text{rise}}{\text{run}})). For the length of the hypotenuse (c) of a right - triangle with legs (a) and (b), we use the Pythagorean theorem (c=\sqrt{a^{2}+b^{2}}).

Step2: Solve part 2a

The rise is (11) ft and the run is (25) ft. The slope (m=\frac{11}{25}=0.44). The angle of elevation (\theta=\arctan(\frac{11}{25})\approx\arctan(0.44)\approx23.7^{\circ}).

Step3: Solve part 2b

The rise is (29) cm and the run is (14) cm. The slope (m = \frac{29}{14}\approx2.071). The angle of elevation (\theta=\arctan(\frac{29}{14})\approx\arctan(2.071)\approx64.2^{\circ}).

Step4: Solve part 3a

The rise (a = 3) ft and the run (b = 7) ft. Using the Pythagorean theorem (c=\sqrt{3^{2}+7^{2}}=\sqrt{9 + 49}=\sqrt{58}\approx7.62) ft.

Step5: Solve part 3b

The slope (m=\frac{3}{7}\approx0.429). The angle of elevation (\theta=\arctan(\frac{3}{7})\approx\arctan(0.429)\approx23.2^{\circ}).

Answer:

2a) Slope: (0.44), Angle of elevation: approximately (23.7^{\circ}) 2b) Slope: approximately (2.071), Angle of elevation: approximately (64.2^{\circ}) 3a) Length of the surface of the jump: approximately (7.62) ft 3b) Angle of elevation: approximately (23.2^{\circ})