A vertical tower subtends a right angle on the top of 10 m high flagstaff. If the distance between them is 20m then find the height of tower.
Answers
Step-by-step explanation:
Let AB be the flagstaff and CD be the vertical tower.
AB=10m=EC
BC=20m=AE
In triangle ABC we have,
tanx=AB/BC
tanx=10/20
1/tanx=20/10
10/tanx=20.............................1
In triangle ADE we have
tan(90-x)=DE/AE
cotx=DE/20
1/cotx=20/DE
DE/cotx=20.........................2
by eq1 and. eq 2 we get
10/tanx=DE/cotx
10/(AB/BC)=DE/(BC/AB)
10/(10/20)=DE/(20/10)
10×20/10=DE×10/20
DE=40m
Therefore height of vertical tower = DE+EC=40+10=50m
SOLUTION
Let AB = Tower
& CD = Flagstaff
Given that,
Angle ACB= 90°
BC= EC= 20m
EB = CD = 10m
Now, in ∆BCD, by Pythagoras Theorem
BC^2 = CD^2 +BD^2
=) 10^2 + 20^2
=)100 + 400
=) 500
In right ∆ABC, again Pythagoras Theorem
=) AB^2=AC^2 + BC^2 =AE^2+ EC^2+BC^2
=) AE^2 +BD^2 + BC^2
=) (AB-BE)^2 + (20)^2 + 500
=) (AB- 10)^2 +400 +500
=) AB^2 - 20AB +100 +900
=) AB= 1000/20
=) AB= 50 Metres. (required Height)
hope it helps ✔️
