Step 1: Let the height of the pillar be \( h \) m.
Let the horizontal distance between the tower and pillar be \( x \) m.
Step 2: Use trigonometric ratios for the angles of depression.
For the top of the pillar (\(45^\circ\)): \[ \tan 45^\circ = \dfrac{50 - h}{x} \Rightarrow 1 = \dfrac{50 - h}{x} \Rightarrow x = 50 - h \] For the bottom of the pillar (\(60^\circ\)): \[ \tan 60^\circ = \dfrac{50}{x} \Rightarrow \sqrt{3} = \dfrac{50}{x} \Rightarrow x = \dfrac{50}{\sqrt{3}} \] Step 3: Equate both expressions for \(x\).
\[ 50 - h = \dfrac{50}{\sqrt{3}} \Rightarrow h = 50 - \dfrac{50}{\sqrt{3}} \] Step 4: Simplify.
\[ h = 50\left(1 - \dfrac{1}{\sqrt{3}}\right) = 50\left(\dfrac{\sqrt{3} - 1}{\sqrt{3}}\right) = \dfrac{50(\sqrt{3} - 1)}{1.732} \] \[ h \approx 50(0.577) = 28.85 \, \text{m} \] Step 5: Conclusion.
Hence, the height of the pillar is approximately 28.87 m.
The shadow of a tower on level ground is $30\ \text{m}$ longer when the sun's altitude is $30^\circ$ than when it is $60^\circ$. Find the height of the tower. (Use $\sqrt{3}=1.732$.)
Find the unknown frequency if 24 is the median of the following frequency distribution:
\[\begin{array}{|c|c|c|c|c|c|} \hline \text{Class-interval} & 0-10 & 10-20 & 20-30 & 30-40 & 40-50 \\ \hline \text{Frequency} & 5 & 25 & 25 & \text{$p$} & 7 \\ \hline \end{array}\]