An observer at a distance of 10 m from tree looks at the top of the tree, the angle of elevation is 60$^\circ$. To find the height of tree complete the activity. ($\sqrt{3} = 1.73$) Activity : In the figure given above, AB = h = height of tree, BC = 10 m, distance of the observer from the tree. Angle of elevation ($\theta$) = $\angle$BCA = 60$^\circ$ tan $\theta$ = $\frac{\boxed{\phantom{AB}}}{BC}$ $\dots$ (I) tan 60$^\circ$ = $\boxed{\phantom{\sqrt{3}}}$ $\dots$ (II) $\frac{AB}{BC} = \sqrt{3}$ $\dots$ (From (I) and (II)) AB = BC $\times$ $\sqrt{3}$ = 10$\sqrt{3}$ AB = 10 $\times$ 1.73 = $\boxed{\phantom{17.3}}$ $\therefore$ height of the tree is $\boxed{\phantom{17.3}}$ m.

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An observer at a distance of 10 m from tree looks at the top of the tree, the angle of elevation is 60$^\circ$. To find the height of tree complete the activity. ($\sqrt{3} = 1.73$)  Activity :   In the figure given above, AB = h = height of tree, BC = 10 m, distance of the observer from the tree.  Angle of elevation ($\theta$) = $\angle$BCA = 60$^\circ$  tan $\theta$ = $\frac{\boxed{\phantom{AB}}}{BC}$ $\dots$ (I) tan 60$^\circ$ = $\boxed{\phantom{\sqrt{3}}}$ $\dots$ (II) $\frac{AB}{BC} = \sqrt{3}$ $\dots$ (From (I) and (II)) AB = BC $\times$ $\sqrt{3}$ = 10$\sqrt{3}$ AB = 10 $\times$ 1.73 = $\boxed{\phantom{17.3}}$ $\therefore$ height of the tree is $\boxed{\phantom{17.3}}$ m.

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Step 1: Understanding the Concept:   This problem uses trigonometry to find the height of an object given the distance from the object and the angle of elevation. The situation forms a right-angled triangle where the height is the opposite side and the distance is the adjacent side to the angle of elevation.    Step 2: Key Formula or Approach:   In a right-angled triangle, the tangent of an angle is the ratio of the length of the opposite side to the length of the adjacent side.  $$ \tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} $$ Step 3: Detailed Explanation:   Here is the completed activity with the blanks filled in.  In the figure given above, AB = h = height of tree, BC = 10 m, distance of the observer from the tree.  Angle of elevation ($\theta$) = $\angle$BCA = 60$^\circ$  In right-angled $\triangle$ABC,  tan $\theta$ = $\frac{\boxed{AB}}{BC}$ $\dots$ (I) We know that, tan 60$^\circ$ = $\boxed{\sqrt{3}}$ $\dots$ (II) $\therefore$ $\frac{AB}{BC} = \sqrt{3}$ $\dots$ (From (I) and (II)) AB = BC $\times$ $\sqrt{3}$ = 10$\sqrt{3}$ AB = 10 $\times$ 1.73 = $\boxed{17.3}$ $\therefore$ height of the tree is $\boxed{17.3}$ m. Step 4: Final Answer:   The height of the tree is 17.3 m.

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