Show that \( |\angle ACB| = 116.5^\circ \), correct to 1 decimal place - Leaving Cert Mathematics - Question 7 - 2021

To find the area of triangle ABC, we can use the formula:

$\text{Area} = \frac{1}{2} a b \sin C$ Using sides a = 28, b = 4, and ( C = 116.5^\circ ):

$\text{Area} = \frac{1}{2} (28)(4) \sin(116.5^\circ)$

Calculating ( \sin(116.5^\circ) \approx 0.8372 ):

$\text{Area} = \frac{1}{2} (28)(4)(0.8372) \approx 50.11 \, \text{km}^2$ Rounding gives us 50.1 km² (to 1 decimal place).

To find the shortest distance from point C to line segment AB, we can use the area found previously:

Using the formula for area, we derive:

$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$

Let the base be ( AB ). From earlier calculations, we know:

$\text{Area} = 50.1 \, \text{km}^2$ Now we calculate the base AB using the sine formula:

$AB = \sqrt{28^2 + 4^2 - 2(28)(4) \cdot \cos(116.5^\circ)} \approx \sqrt{900.01} \approx 30.0$

Setting area formula with height:

$50.1 = \frac{1}{2} (30.0) \cdot h$

Solving for height h:

$h = \frac{50.1 \times 2}{30.0} \approx 3.34 \, \text{km}$ So the shortest distance is approximately 3.3 km (to 1 decimal place).

To find the height (|AT|), we can use the tangent function from point B:

Given: ( \tan(0.05^\circ) = \frac{|AT|}{AB} )

Where ( AB \approx 30.0 ):

$\tan(0.05^\circ) = \frac{|AT|}{30.0}$

This gives us:

$|AT| = 30.0 \cdot \tan(0.05^\circ) \approx 30.0 \cdot 0.00087266 \approx 0.02627 \, ext{km} \approx 26 \, ext{m}$

Thus the vertical height of the tower is approximately 26 m.