Question

1. a safe angle for a ladder is about 75° from the ground. if you have a 3.8 metre ladder, how far from a wall should you place the base of the ladder? give your answer to 3 significant figures. 13) a safe angle for a ladder is about 75° from the ground. if you have a 3.5 metre ladder, how high can it reach up a wall? round your answer to 3 significant figures. 14) holly is looking up at a helicopter. the direct distance from holly to the helicopter is 16 km. the vertical distance from holly to the helicopter is 11 km. calculate the angle of elevation from holly to the helicopter, giving your answer to 1 decimal place. 15) the angle of elevation from rose to a helicopter is 46°. the horizontal distance from rose to the helicopter is 4 km. calculate the direct distance from rose to the helicopter, giving your answer to 3 significant figures.

Explanation:

12)

1. Identify the trigonometric relation:

The ladder forms a right - triangle with the wall and the ground. The length of the ladder is the hypotenuse \(c = 3.8\) m and the angle between the ladder and the ground \(\theta=75^{\circ}\). We want to find the distance \(x\) from the base of the ladder to the wall. We use the cosine function \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\), so \(x = c\cos\theta\).

1. Calculate the value:

\(\cos75^{\circ}=\cos(45^{\circ} + 30^{\circ})=\cos45^{\circ}\cos30^{\circ}-\sin45^{\circ}\sin30^{\circ}=\frac{\sqrt{2}}{2}\times\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}\times\frac{1}{2}=\frac{\sqrt{6}-\sqrt{2}}{4}\approx0.259\). Then \(x = 3.8\times\cos75^{\circ}\approx3.8\times0.259 = 0.984\) m.

13)

1. Identify the trigonometric relation:

The length of the ladder is the hypotenuse \(c = 3.5\) m and the angle between the ladder and the ground \(\theta = 75^{\circ}\). We want to find the height \(h\) the ladder reaches on the wall. We use the sine function \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\), so \(h = c\sin\theta\).

1. Calculate the value:

\(\sin75^{\circ}=\sin(45^{\circ}+ 30^{\circ})=\sin45^{\circ}\cos30^{\circ}+\cos45^{\circ}\sin30^{\circ}=\frac{\sqrt{2}}{2}\times\frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}\times\frac{1}{2}=\frac{\sqrt{6}+\sqrt{2}}{4}\approx0.966\). Then \(h = 3.5\times\sin75^{\circ}\approx3.5\times0.966 = 3.38\) m.

14)

1. Identify the trigonometric relation:

The direct distance from Holly to the helicopter is the hypotenuse \(c = 16\) km and the vertical distance is the opposite side \(y = 11\) km. We use the sine function \(\sin\theta=\frac{y}{c}\), and then \(\theta=\sin^{- 1}(\frac{y}{c})\).

1. Calculate the value:

\(\theta=\sin^{-1}(\frac{11}{16})\). Using a calculator, \(\sin^{-1}(\frac{11}{16})\approx42.8^{\circ}\).

15)

1. Identify the trigonometric relation:

The horizontal distance \(x\) is related to the direct distance \(d\) (hypotenuse) by the cosine function \(\cos\theta=\frac{x}{d}\), and we want to find \(d\). So \(d=\frac{x}{\cos\theta}\). Here \(\theta = 46^{\circ}\) and \(x = 4\) km.

1. Calculate the value:

\(\cos46^{\circ}\approx0.695\), then \(d=\frac{4}{\cos46^{\circ}}\approx\frac{4}{0.695}\approx5.76\) km.

Answer:

1. \(0.984\) m
2. \(3.38\) m
3. \(42.8^{\circ}\)
4. \(5.76\) km