Pls solve it and explain me pls it's very important for me
Attachments:

Answers
Answered by
0
I hope it will be answer of your question
Attachments:

Answered by
0
firstly we find the perpendicular of the Brocken tree
tan 30 =perpendicular/base
tan 30=perpendicular/10
1/√3=perpendicular/10
on cross multiplying,we have
√3×perpendicular=10
perpendicular=10/√3
=3.34√3
=5.78m
to find hypotens ,we have
cos30°=base/hypotenuse
√3/2=10/hypotenuse
on cross multiplying,we have
hypotenuse×√3=2×10
hypotenuse=20/√3
hypotenuse=6.67√3
=11.53m
height of the tree=hypotenuse+perpendicular
height of the tree=11.53+5.78
height of the tree=17.31
tan 30 =perpendicular/base
tan 30=perpendicular/10
1/√3=perpendicular/10
on cross multiplying,we have
√3×perpendicular=10
perpendicular=10/√3
=3.34√3
=5.78m
to find hypotens ,we have
cos30°=base/hypotenuse
√3/2=10/hypotenuse
on cross multiplying,we have
hypotenuse×√3=2×10
hypotenuse=20/√3
hypotenuse=6.67√3
=11.53m
height of the tree=hypotenuse+perpendicular
height of the tree=11.53+5.78
height of the tree=17.31
Attachments:

Similar questions