Question

1. two office towers are 150 m apart. from the top of the shorter building, the angle of depression to the base of the taller building is \\(15^\circ\\). the angle of elevation to the top of the taller building, from the top of the shorter building is \\(20^\circ\\). determine the height of each tower to the nearest metre. (6 marks)

1. a rectangular box has a width of 15 in, a depth of 27.5 in, and a height of 10 in. draw an isometric drawing of the box. assume the diagonal distance between each dot is 2.5 in. (5 marks)

Explanation:

Identify given values and geometric relations

Using the Right Triangle Trigonometry knowledge point, we model the two office towers. Let the shorter tower have height \(h_1\) and the taller tower have height \(h_2\). The horizontal distance between them is \(d = 150\text{ m}\).

From the top of the shorter building:

• The angle of depression to the base of the taller building is \(15^\circ\). This forms a right triangle where the adjacent side is \(d = 150\text{ m}\) and the opposite side is the height of the shorter building \(h_1\).
• The angle of elevation to the top of the taller building is \(20^\circ\). This forms a right triangle where the adjacent side is \(d = 150\text{ m}\) and the opposite side is the height difference \(y\) between the taller and shorter building, so \(h_2 = h_1 + y\).

Calculate the height of the shorter tower

Using the Right Triangle Trigonometry knowledge point:

$$ \tan(15^\circ) = \frac{h_1}{150} $$

$$ h_1 = 150 \cdot \tan(15^\circ) \approx 150 \cdot 0.26795 = 40.19\text{ m} $$

Rounding to the nearest metre:

$$ h_1 \approx 40\text{ m} $$

Calculate the height of the taller tower

Using the Right Triangle Trigonometry knowledge point, we find the height difference \(y\):

$$ \tan(20^\circ) = \frac{y}{150} $$

$$ y = 150 \cdot \tan(20^\circ) \approx 150 \cdot 0.36397 = 54.59\text{ m} $$

The total height of the taller tower is:

$$ h_2 = h_1 + y = 40.19 + 54.59 = 94.78\text{ m} $$

Rounding to the nearest metre:

$$ h_2 \approx 95\text{ m} $$

Determine isometric grid dimensions for the box

Using the Isometric Projection knowledge point, we convert the box dimensions to grid units. The diagonal distance between adjacent dots is \(2.5\text{ in}\).

• Width: \(\frac{15\text{ in}}{2.5\text{ in/unit}} = 6\text{ units}\) (6 diagonal spaces along the \(30^\circ\) isometric axis)
• Depth: \(\frac{27.5\text{ in}}{2.5\text{ in/unit}} = 11\text{ units}\) (11 diagonal spaces along the \(150^\circ\) isometric axis)
• Height: \(\frac{10\text{ in}}{2.5\text{ in/unit}} = 4\text{ units}\) (4 vertical spaces)

Answer:

Question 8

• Height of the shorter tower: \(40\text{ m}\)
• Height of the taller tower: \(95\text{ m}\)

Question 9

To draw the isometric projection of the rectangular box:

• Width (left-down diagonal axis): Count \(6\) units (diagonal spaces between dots).
• Depth (right-down diagonal axis): Count \(11\) units (diagonal spaces between dots).
• Height (vertical axis): Count \(4\) units (vertical spaces between dots).