Question

A church spire casts a shadow on the level ground. At the tip of the shadow, the tip of the spire lies 25° upward. Walking 100 m closer, the tip of the spire is at an angle of 55°. What must be the height of the spire in Figure 2? Write your answer with 3 significant figures.

Answer 1

Given:

A church spire casts a shadow on the level ground.

At the tip of the shadow, the tip of the spire lies 25° upward.

Walking 100 m closer, the tip of the spire is at an angle of 55°.

To find:

The height of the spire

Solution:

To solve the above-given problem we will use the following trigonometric ratio of a triangle:

$\boxed{\bold{tan\theta = \frac{Opposite\:side}{Adjacent\:side} }}$

 

Referring to the figure attached below, let's assume,

AB = The height of the church spire

CD = 100 m = the distance walked to come close to the church spire

∠ADC = 25° = the angle of elevation to the tip of the spire

∠ACB = 55° = the angle of elevation to the tip of the spire after walking 100 m towards the spire

Now, using the formula, we will solve further as follows:

In ΔABC, we have

 $tan\ 55 \° = \frac{AB}{BC}$

$\implies AB = BC \times tan\:55\°$  ...... (i)

In ΔABD, we have

$tan\ 25 \° = \frac{AB}{BD}$

$\implies tan\ 25 \° = \frac{AB}{BC + CD}$

substituting the value of AB from (i) and CD = 100 m

$\implies tan\ 25 \° = \frac{BC \times tan\:55\°}{BC + 100}$

$\implies 0.4663 = \frac{BC\: \times\: 1.4281}{BC + 100}$

$\implies 0.4663(BC + 100) = BC\: \times\: 1.4281$

$\implies 0.4663BC + 46.63= 1.4281BC$

$\implies 46.63= 1.4281BC - 0.4663BC$

$\implies 0.9618BC = 46.63$

$\implies BC = \frac{46.63}{0.9618}$

$\implies BC = 48.48$  ...... (ii)

Now, substituting the value of BC from (ii) in (i), we get

$AB = BC \times tan\:55\°$

 $\implies AB = 48.48 \times 1.4281$

 $\implies AB = 69.234$

 we will now round off the above answer to 3 significant figures

 $\implies \bold{AB = 69.2\:m}$

 Thus, the height of the church spire must be 69.2 m.

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