please answer Q6 & Q7...ITS REALLY URGENT!!!
Attachments:
Answers
Answered by
1
{6} Let h be the height of the hill and x m be the distance between the foot of the hill and foot of the tower.
In rt.△ABC,
Cos 60° = x/hx = h cot 60°........(i)
In rt.△DBC,
cot 30° = x/50x = 50
cot 30°........(ii)Equating (i) and (ii)
h cot 60° = 50 cot 30°h
= 50 cot 30°/cot 60°h
= 50 × √3/1√3
= 50 × 3 = 150 m
Therefore the height of the hill is 150 m.
Hope it helps u.
In rt.△ABC,
Cos 60° = x/hx = h cot 60°........(i)
In rt.△DBC,
cot 30° = x/50x = 50
cot 30°........(ii)Equating (i) and (ii)
h cot 60° = 50 cot 30°h
= 50 cot 30°/cot 60°h
= 50 × √3/1√3
= 50 × 3 = 150 m
Therefore the height of the hill is 150 m.
Hope it helps u.
shraddha33204:
is this q. from goyal
no its not
okay take your time
i am so sorry i am not able to solve it
its okay...thank you for trying
but your sure of the first one right?
ya
i am sure it's correct
is this question from some applications from trigonometry of class 10
Answered by
0
by Sujeet yaduvanshi ✌
Attachments:
thanks
Similar questions