Q.2 (A)(3) An observer at a distance of 10 m from tree looks at the top of the tree, the angle of elevation is 60º. To find the height of tree complete the activity. (\(\sqrt3\) = 1.73)[Figure: right-angled Δ ABC, right angle at B; AB is the tree, BC…

Q.2 (A)(3) An observer at a distance of 10 m from tree looks at the top of the tree, the angle of elevation is 60º. To find the height of tree complete the activity. (\(\sqrt3\) = 1.73)

[Figure: right-angled Δ ABC, right angle at B; AB is the tree, BC = 10 m, ∠BCA = 60º.]

Activity :

In the figure given above, AB = h = height of tree, BC = 10 m, distance of the observer from the tree.
Angle of elevation (θ) = ∠BCA = 60º
tan θ = ____ / BC .................. (I)
tan 60º = ____ .................. (II)
\(\dfrac{AB}{BC} = \sqrt3\) (From (I) and (II))
AB = BC × \(\sqrt3\) = \(10\sqrt3\)
AB = 10 × 1.73 = ____
∴ height of the tree is ____ m.