Question

solve please •••••••••••••••​

Answer 1

Answer:

cos(theta) = adj/hyp

cos60° = 10/hyp

hyp = 2×10

= 20m

sin(theta) = opp/hyp

opp = 20×root3/2

= 10 root3

= 17.32 m

heightof the tree = opp+hyp

= 20+17.32

= 37.32 m

Answer 2

Let length of tree before windstorm is BD.

After windstorm the upper part of tree C falls from point C to point A on the ground

. Now, let CD = AC = h2 m AB = 10 m

Broken part makes an angle 60° from the ground.

So, ∠CAB = 60°

From right angled ∆ABC tan 60° = BC/AB ⇒ √3 = h1/10 ⇒ h1 = 10√3 and cos 60° = AB/AC ⇒ 1/2 = 10/h2 ⇒ h2 = 10 × 2 = 20 m

Hence, total length of tree BD = BC + CD = h1 + h1

= 10√3 + 20

= 10 × 1.732 + 20

= 17.32 + 20

= 37.32 m

Hence, height of the tree 37.32 m.