applications of trigonometry - A straight highway leads to the foot of a tower. Ramaiah standing at the top of the tower observes a car at an angle of depression 30\degree.

Krishna (last edited 4 years ago) (Expert, Educator @Qalaxia)

Step 1: Analyse the given question. based up on the given information make a suitable figure.

FIGURE: Heights of the tower CD = h meters

Let the distance traveled by the car in 6 seconds = AB = x meters

The remaining distance to be traveled by the car BC = d meters

Distance AC = AB + BC = (x + d) meters

GIVEN: \angle PDA = \angle DAB = 30\degree

\angle PDB = \angle DBC = 60\degree ( \because Alternating interior angles)

Step 2: "Find the common height of the two triangles by using the trigonometric ratios definitions.

EXAMPLE: We know opposite side, we have to find the adjacent side so

take \tan \theta = \frac{opposite}{adjacent}

From \triangle ACD

\tan 30\degree = \frac{CD}{AC}

\frac{1}{\sqrt{3}} = \frac{h}{x + d} ( \because \tan 30\degree = \frac{1}{\sqrt{3}})

h = \frac{x + d}{\sqrt{3}}.................................(1)

From \triangle BCD

\tan 60\degree = \frac{CD}{BC}

\sqrt{3} = \frac{h}{d} ( \because \tan 60\degree = \sqrt{3} )

h = \sqrt{3}d ................................(2)

Step 3: Express d in terms of x by solving equation (1) & (2)

h = \frac{x + d}{\sqrt{3}}.........................(1)

h = \sqrt{3}d ......................(2)

Equating R.H.S

\sqrt{3}d = \frac{x + d}{\sqrt{3}}

d = \frac{x + d}{(\sqrt{3})^2}

x + d = 3d

x = 2d

d = \frac{x}{2} ................................(3)

Step 4: Find the time taken by the car to reach the foot of the tower from B point.

EXPLANATION: Time taken to travel ‘x’ meters = 6 seconds.

Time taken to travel the distance of ‘d’ meters = \frac{x}{2} [ \because equation (3)]

= \frac{6}{2}

= 3 seconds.