Question

1. christine places a ladder against the side of a house so that the top of the ladder makes an angle of $52^{circ}$ with the side of the house. the bottom of the ladder is $1.20$ m from the house.

a) calculate, to the nearest hundredth of a metre, the vertical distance from the top of the ladder to the ground.
b) calculate, to the nearest hundredth of a metre, the length of the ladder.

Explanation:

Part (a)

Step1: Identify the trigonometric relationship

We have a right triangle where the horizontal distance from the house (adjacent side to the angle) is \( 1.20 \, \text{m} \), the vertical distance (opposite side to the angle) is what we need to find, and the angle between the ladder (hypotenuse) and the house (adjacent side) is \( 52^\circ \). We use the tangent function: \( \tan(\theta)=\frac{\text{opposite}}{\text{adjacent}} \)
Let \( y \) be the vertical distance. So \( \tan(52^\circ)=\frac{y}{1.20} \)

Step2: Solve for \( y \)

Multiply both sides by \( 1.20 \): \( y = 1.20\times\tan(52^\circ) \)
Calculate \( \tan(52^\circ)\approx1.2799 \)
Then \( y = 1.20\times1.2799\approx1.5359 \)
Round to the nearest hundredth: \( y\approx1.54 \)

Step1: Identify the trigonometric relationship

We use the cosine function for the right triangle, where \( \cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}} \)
The adjacent side is \( 1.20 \, \text{m} \), the angle \( \theta = 52^\circ \), and the hypotenuse (length of the ladder) is \( L \)
So \( \cos(52^\circ)=\frac{1.20}{L} \)

Step2: Solve for \( L \)

Rearrange the formula: \( L=\frac{1.20}{\cos(52^\circ)} \)
Calculate \( \cos(52^\circ)\approx0.6157 \)
Then \( L=\frac{1.20}{0.6157}\approx1.9489 \)
Round to the nearest hundredth: \( L\approx1.95 \)

Answer:

The vertical distance is approximately \( \boldsymbol{1.54} \) metres.

Part (b)