Drawing Figures to Solve Problems

When we want to solve the problems of heights and distances, we should consider the following:

(i) All the objects such as towers, trees, buildings, ships, mountains etc. shall be considered as linear for mathematical convenience.

(ii) The angle of elevation or angle of depression is considered with reference to the horizontal line.

(iii) The height of the observer is neglected, if it is not given in the problem.

When we try to find heights and distances at an angle of elevation or depression, we need to visualize geometrically. To find heights and distances, we need to draw figures and with the help of these figures we can solve the problems. Let us see some examples.

1. A tower stands vertically on the ground. From a point on the ground, which is 15 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 60°. Find the height of the tower.

Instructions

Solution: First let us draw a simple diagram to represent the problem. Here AB represents the tower, CB is the distance of the point from the tower and ∠ACB is the angle of elevation.

We need to determine the height of the tower, i.e., AB. Also, ACB is a triangle, right-angledacute-angled at B.

To solve the problem, we choose the trigonometric ratio tan (or cot), as the ratio involves ABAD and BCAC.

Now, tan 60° = ABBCBCAB

Putting values,

33 = AB15AC15

AB = 15315

Hence, the height of the tower is 153 m.

2. An electrician has to repair an electric fault on a pole of height 5 m. She needs to reach a point 1.3 m below the top of the pole to undertake the repair work. What should be the length of the ladder that she should use which, when inclined at an angle of 60° to the horizontal, would enable her to reach the required position? Also, how far from the foot of the pole should she place the foot of the ladder? (take 3 = 1.73)

Instructions

Solution: The electrician is required to reach the point B on the pole AD.

So, BD = –

= 5 – 1.3 = m.

Here, BC represents the ladder. We need to find its length of the of the right triangle BDC.

Now, which trigonometic ratio should we consider?

It should be sin 60°cos 60°.

So, BDBCBCBD = sin 60°

/BC = 3213

Therefore, BC = 3.7×23 = m (upto two decimal places)

Thus, the length of the ladder should be 4.28 m

Now, DCBD = cot 60°tan 60° = 133

DC = 3.73 = m (upto two decimal places)

Therefore, she should place the foot of the ladder at a distance of 2.14 m from the pole.

3. An observer 1.5 m tall is 28.5 m away from a chimney. The angle of elevation of the top of the chimney from her eyes is 45°. What is the height of the chimney?

Instructions

Solution : Here, AB is the chimney, CD the observer and ∠ ADE the angle of elevation. In this case, ADE is a triangle, -angled at E and we are required to find the height of the chimney.

We have AB = + BE = + and DE = CB = m

To determine AE, we choose a trigonometric ratio, which involves both AE and DE.

So, tan 45°sin 45° = AEDE

= AE28.5

Therefore, AE = m

So the height of the chimney (AB) = AE + BE = + m = m.

Thus, height of the chimney is 30 m.