Question

From the top of a cliff 25 metres high, the angle of elevation of the top of a tower found to be equal to the angle of depression of the foot of the tower, then the height of the tower is.

• A. 25 metres
• B. $25 \sqrt {2}$ metres
• C. $25 \sqrt {2}$ metres
• D. 50 metres

Answer 1

Answer: (D)

50 metres

Answer 2

AB is the cliff of height 25 m and PQ is the tower. Let $\Delta$PBL = $\Delta$QBI. = $\theta$ From rt. $\Delta d \, \Delta$QAB, $\frac{25}{x}$ = tan $\theta \, \therefore \, x$ = 25 cot $\theta$ From $rt \, \Delta d \, \Delta$BPL, $\frac{PL}{x}$ = tan $\theta$ $\therefore$ PL = $x$ tan $\theta$ = 25.cot $\theta$ tan $\theta$ = 25 $\therefore$ Height of the tower -Q P = QL + I.P = AB + PL = 25 + 25 = 50 m