Question

UI
12. From the top of a 7 m high building, the angle of elevation of the top of a cable tower is
60° and the angle of depression of its foot is 45º. Determine the height of the tower,
13. As observed from the top of a 75 m high lighthouse from the sea-level.
level the​

Answer 1

$\sf\blue{Question:}$

$\sf{From \ the \ top \ of \ 7 \ m \ high \ building,}$

$\sf{the \ angle \ of \ elevation \ of \ the \ top \ of \ a}$

$\sf{cable \ tower \ is \ 60° \ and \ the \ angle \ of}$

$\sf{depression \ of \ it's \ foot \ is \ 45°. \ Determine}$

$\sf{the \ height \ of \ the \ tower.}$

____________________________________

$\sf\red{\underline{\underline{Answer:}}}$

$\sf{The \ height \ of \ cable \ tower \ is \ 19.11 \ m}$

$\sf\orange{Given:}$

$\sf{\implies{From \ the \ top \ of \ a \ 7 \ m \ high}}$

$\sf{building, \ the \ angle \ of \ elevation \ of \ the \ top}$

$\sf{of \ a \ cable \ tower \ is \ 60°}$

$\sf{\implies{Angle \ of \ depression \ of \ it's }}$

$\sf{foot \ is \ 45°}$

$\sf\pink{To \ find:}$

$\sf{The \ height \ of \ the \ cable \ tower.}$

$\sf\green{\underline{\underline{Solution:}}}$

$\sf{Let \ AB \ represent \ building \ and \ CE \ represent}$

$\sf{the \ tower.}$

$\sf{\therefore{AB=7 \ m}}$

$\sf{Angle \ of \ Depression \ is \ 45°}$

$\sf{\angle \ DAC=\angle \ ACB}$

$\sf{... Alternate \ angles}$

$\sf{\therefore{\angle \ DAC=45°}}$

$\sf{In \ right \ angled \ triangle \ ABC \ \angle \ ABC=90°,}$

$\sf{tan45°=\frac{AB}{BC}}$

$\sf{\therefore{1=\frac{7}{BC}}}$

$\sf{\therefore{BC=7 \ m}}$

$\sf{In \ quadrilateral \ ABCD,}$

$\sf{\angle \ ABC=90°...given}$

$\sf{\angle \ BCD=90°...given}$

$\sf{\angle \ ADC=90°...AD \ is \ perpendicular \ to \ EC}$

$\sf{\angle \ DAB=90°... remaining \ angle}$

$\sf{Also, \ adjacent \ sides \ AB \ and \ BC \ are \ congruent.}$

$\sf{\therefore{The \ quadrilateral \ ABCD \ is \ square.}}$

$\sf{\therefore{AB=BC=CD=AD=7 \ m}}$

$\sf{...all \ sides \ of \ square \ are \ congruent.}$

$\sf{In \ \triangle \ AED, \angle \ ADE=90°}$

$\sf{Angle \ of \ elevation=60°}$

$\sf{\therefore{\angle \ EAD=60°}}$

$\sf{tan \ 60°=\frac{ED}{AD}}$

$\sf{\sqrt3=\frac{ED}{7}}$

$\sf{\therefore{ED=7\sqrt3 \ m}}$

$\sf{Height \ of \ cable \ tower=CD+ED}$

$\sf{=7+7\sqrt3}$

$\sf{=7(1+\sqrt3)}$

$\sf{=7(1+1.73)}$

$\sf{=7\times2.73}$

$\sf{=19.11 \ m}$

$\sf\purple{\therefore{\tt{The \ height \ of \ cable \ tower \ is \ 19.11 \ m}}}$