Question

1. a telecommunication mast and a pillar stand on the same horizontal level. from the mid - point, o, between the foot of the pillar, s, and the foot of the mast, t, the angle of elevation of the top of the mast, g, is 70°. from the point, o, the angle of elevation of the top of the pillar, h, is 28°. if the pillar is 10 m high:

(a) represent the information in a diagram;
(b) calculate, correct to the nearest whole number:
(i) the height of the mast;
(ii) |hg|.

Explanation:

Step1: Find the distance from O to the top of the pillar

Let the distance from $O$ to the base of the pillar and mast be $x$. For the pillar of height $h_{1} = 10$ m and angle of elevation $\theta_{1}=28^{\circ}$, we use the tangent - function $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$. So, $\tan28^{\circ}=\frac{10}{x}$, and $x = \frac{10}{\tan28^{\circ}}$.

Step2: Find the height of the mast

For the mast, the angle of elevation $\theta_{2} = 70^{\circ}$ and the adjacent side to the angle of elevation is also $x$. Let the height of the mast be $h_{2}$. Using the tangent - function $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$, we have $\tan70^{\circ}=\frac{h_{2}}{x}$. Since $x=\frac{10}{\tan28^{\circ}}$, then $h_{2}=x\tan70^{\circ}=\frac{10\tan70^{\circ}}{\tan28^{\circ}}$.
We know that $\tan70^{\circ}\approx2.7475$ and $\tan28^{\circ}\approx0.5317$. So, $h_{2}=\frac{10\times2.7475}{0.5317}\approx52$ m.

Step3: Use the Pythagorean theorem to find $|HG|$

The vertical distance between the top of the mast and the top of the pillar is $h = h_{2}-h_{1}=52 - 10 = 42$ m, and the horizontal distance between the mast and the pillar is $2x$. But we can also use the right - triangle formed by the line segment $HG$ and the vertical and horizontal displacements.
We know that the vertical difference in heights is $h = 42$ m and the horizontal distance between the mast and the pillar is $2x$. Since $x=\frac{10}{\tan28^{\circ}}\approx18.8$ m, the horizontal distance between the mast and the pillar is $2x\approx37.6$ m.
By the Pythagorean theorem $|HG|=\sqrt{(h_{2}-h_{1})^{2}+(2x)^{2}}$. Another way is to consider the right - triangle with vertical side $h_{2}-h_{1}$ and horizontal side $0$ (in the context of the relative position of the tops of the two structures).
We use the fact that the vertical difference between the tops of the two structures is $h = h_{2}-h_{1}=42$ m and the horizontal distance between their bases is $2x$. But we can also use the right - triangle formed by the line joining the tops of the two structures.
Since the vertical difference between the tops of the two structures is $h = h_{2}-h_{1}=42$ m and the horizontal distance between them is considered in the right - triangle formed by the line segment $HG$.
We know that the vertical difference between the tops of the two structures is $42$ m and the horizontal distance between the bases of the two structures (which is relevant for the right - triangle with hypotenuse $HG$) is $2x$. But if we consider the right - triangle with vertical side $h_{2}-h_{1}$ and assume the horizontal displacement between the tops is $0$ (in the 2 - D vertical - horizontal plane), we can use the vertical difference directly.
$|HG|=\sqrt{(52 - 10)^{2}+0^{2}}=42$ m (assuming no horizontal displacement between the tops in the plane of interest). In a more general sense, using the fact that the vertical difference between the tops of the two structures is $h=h_{2}-h_{1}$ and the horizontal distance between the bases of the two structures is $2x$.
$|HG|=\sqrt{(52 - 10)^{2}+(2\times\frac{10}{\tan28^{\circ}})^{2}}\approx\sqrt{42^{2}+(2\times18.8)^{2}}=\sqrt{1764 + 1411.52}=\sqrt{3175.52}\approx56$ m.

Answer:

(i) $52$ m
(ii) $56$ m