
A boy 1.6 m tall is 20 m away from a tower and observes the angle of elevation of the top of tower to be (i) 45° (ii) 60°
Find the height of the tower in each case.
Answer
486k+ views
Hint: When a boy stands near a tower and observes its top it makes a right triangle with angle 45° and 60°using trigonometric formula we can find height.
Complete step-by-step answer:
Let the height of the tower CD be x
Height of boy (AB) = CE
So, height of DE = CD – CE
= x – 1.6
In the triangle ΔDAE
∠DAE = 45°
From the trigonometric formula
$\tan \theta = \dfrac{{perpendicular}}{{base}}$
∵ Distance between the boy and tower
= 20 cm
So, BC = 20 cm
tan 45 $ = \dfrac{{x - 1.6}}{{20}}$
1 × 20 = x – 1.6 (∵ tan 45° = 1)
x = 21.6
tan 60° $ = \dfrac{{x - 1.6}}{{20}}$
$20\sqrt 3 + 1.6 = x$
x = 36.24
So, the height of tower in case of elevation of 45° = 21.6
And the height of tower in case of elevation of 60° = 36.24.
Note: In this type of problem first try to draw a rough diagram. Mark the angle which is given in question. Using trigonometric formulas we can find the side length and also we can find angle if length is given in the question.
Complete step-by-step answer:
Let the height of the tower CD be x
Height of boy (AB) = CE
So, height of DE = CD – CE
= x – 1.6

In the triangle ΔDAE
∠DAE = 45°
From the trigonometric formula
$\tan \theta = \dfrac{{perpendicular}}{{base}}$
∵ Distance between the boy and tower
= 20 cm
So, BC = 20 cm
tan 45 $ = \dfrac{{x - 1.6}}{{20}}$
1 × 20 = x – 1.6 (∵ tan 45° = 1)
x = 21.6
tan 60° $ = \dfrac{{x - 1.6}}{{20}}$
$20\sqrt 3 + 1.6 = x$
x = 36.24
So, the height of tower in case of elevation of 45° = 21.6
And the height of tower in case of elevation of 60° = 36.24.
Note: In this type of problem first try to draw a rough diagram. Mark the angle which is given in question. Using trigonometric formulas we can find the side length and also we can find angle if length is given in the question.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

The correct geometry and hybridization for XeF4 are class 11 chemistry CBSE

Water softening by Clarks process uses ACalcium bicarbonate class 11 chemistry CBSE

With reference to graphite and diamond which of the class 11 chemistry CBSE

Trending doubts
10 examples of friction in our daily life

Difference Between Prokaryotic Cells and Eukaryotic Cells

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

State and prove Bernoullis theorem class 11 physics CBSE

State the laws of reflection of light

Write down 5 differences between Ntype and Ptype s class 11 physics CBSE
