Premium Online Home Tutors
3 Tutor System
Starting just at 265/hour
Answer :
Let BC be the building of height 20 m and CD be the tower of height x metres. Let A be a point on the ground at a distance of y metres away from the foot B of the building.
In \( ∆ \ ABC \), we have
\( \frac{BC}{AB} \ = \ tan45º \ \)
\( => \ \frac{20}{y} \ = \ 1 \)
\( => \ y \ = \ 20\)
In \( ∆ \ ABD \), we have
\( \frac{BD}{AB} \ = \ tan60º \ \)
\( => \ \frac{20+x}{20} \ = \ \sqrt{3} \)
\( => 20+x \ = \ 20 \sqrt{3} \ \)
\( => \ x \ = \ 20( \sqrt{3} -1) \) \(=> x \ = \ 20(1.732 - 1) \ \)
\( => \ x \ = \ 14.64 \)
\(\therefore \) the height of the tower is 14.64 metres.