Asked in: March 2025 board paper (same type, 80 m and 45° for a church, in June 2026)An observer at a distance of 10 m from a tree looks at the top of the tree; the angle of elevation is 60°. To find the height of the tree complete the activity. (\(\…

Asked in: March 2025 board paper (same type, 80 m and 45° for a church, in June 2026)

An observer at a distance of 10 m from a tree looks at the top of the tree; the angle of elevation is 60°. To find the height of the tree complete the activity. (\(\sqrt3 = 1.73\))

[Figure: AB = h is the tree, B its foot; C is the observer with BC = 10 m; \(\angle BCA = 60^\circ\).]

Activity: \(\tan\theta = \dfrac{\text{____}}{BC}\) ...(I) \(\tan 60^\circ = \text{____}\) ...(II)

\(\dfrac{AB}{BC} = \sqrt3\) (from (I) and (II)) \(AB = BC \times \sqrt3 = 10\sqrt3\)

AB = 10 × 1.73 = ____ \(\therefore\) height of the tree is ____ m.