A vertical tower subtends a right angle on the top of 10m high flagstaff.if distance between them is 20m,find the height of the tower! ans given is 50m...but i am not able 2 draw a figure n i have understood it ! plz help!
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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
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
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