Length of the shadow of a 15 meters high pole is 5\sqrt{3} meters at 7 o’clock in the morning. Then, what is the angle of elevation of the Sun rays with the ground at the time?

Step 1: Note down the given information. According to that make an imaginary figure.
NOTE: Pole height AB= 15 meters
Pole shadow length BC = 5 \sqrt{3}
Angle of elevation of sun rays = ?
Step 2: Find the Appropriate trigonometric ratios to calculate angle of elevation.
From the right triangle.
\tan \theta = \frac{opposite}{adjacent}
\tan \theta = \frac{AB}{BC} = \frac{15}{5 \sqrt{3}}
\tan \theta = \frac{3}{ \sqrt{3}} = \frac{(\sqrt{3})^2}{\sqrt{3}}
\tan \theta = \sqrt{3}
Step 3: Search for the equivalent trigonometric ratio angle values.
EXAMPLE: \tan \theta = \sqrt{3}
\tan \theta = \tan 60\degree (Since, \tan 60\degree = \sqrt{3})
\theta = 60 \degree
Therefore the angle of elevation = 60\degree