can you solve any one of this 10th class questions please ☺️
Attachments:
Answers
Answered by
6
Answer:
Given tower be AB .
when Sun's altitude is 60°
ACB = 60°
& Length of shadow is = BC .
Shadow is 50 metre when angle change from 60° to 30°
CD = 50m .
we need to find height of tower i.e. AB
Since tower is vertically to ground .
hence , angle ABC = 90°
In right Angled Triangle ABC ,
tan c = sides opposite to angle C /
sides adjacent to angle C .
= tan 60° =AB / CB
= √3 =AB / CB
= CB = AB / √3. ....................(I)
In right angle triangle ABD
tan 30° = AB / DB .
1 /√3 = AB / DB .
DB = √3AB.
DC + CB = √3AB.
50 + CB = √3AB.
CB= √3AB - 50 . .................... (ii)
From (I ) & ( II)
= AB /√3 = √3 AB- AD .
= AB = √3( √3AB ) -50√3 .
= AB= 3AB - 50 √ 3 .
= 50√3 = 3AB -AB .
= 50√3 = 2AB .
= 2AB = 50√3 .
=AB = 50√3 / 2 .
= AB = 25√3 .
Hence , Height of the tower = AB = 20√3 .
Similar questions