IB Security 2019 Question 9
Question 9
The angle of elevation of the top of a tower from a point on ground which is 30 mtrs. away from the foot of the tower is 30°. Height of the tower is:
Solution
Here, AB is the height of the tower.
BC is the distance between the foot of the tower and the point.
$$tan 30^\circ = \dfrac{AB}{30}$$
=> $$\dfrac{1}{\sqrt{3}}= \dfrac{AB}{30}$$
=> $$AB = \dfrac{30}{\sqrt{3}}$$
=> $$AB = \dfrac{30}{\sqrt{3}}\times\dfrac{\sqrt{3}}{\sqrt{3}}$$
$$= 10\sqrt{3} m$$
Here, AB is the height of the tower.
BC is the distance between the foot of the tower and the point.
$$tan 30^\circ = \dfrac{AB}{30}$$
=> $$\dfrac{1}{\sqrt{3}}= \dfrac{AB}{30}$$
=> $$AB = \dfrac{30}{\sqrt{3}}$$
=> $$AB = \dfrac{30}{\sqrt{3}}\times\dfrac{\sqrt{3}}{\sqrt{3}}$$
$$= 10\sqrt{3} m$$
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