Tree top’s angle of elevation is 30° from a point on ground, 300m away the tree. When the tree grew up its angle of elevation became 60° from the same point. How much did the tree grow?

Answer: Option B

Solution

Answer: Option B

Solution:

$$\begin{array}{rl}& \text{Original tree height}=\text{h}=\text{MQ}\\ & \text{New}\text{tree}\text{height}=PQ\\ & \text{in}\Delta MQN,\tan 30^{\circ }=\frac{1}{\sqrt{3}}=\frac{MQ}{NQ}\\ & \therefore MQ=\frac{300}{\sqrt{3}}\\ & \text{in}\Delta PQN,\tan 60^{\circ }=\sqrt{3}=\frac{PQ}{NQ}\\ & \therefore PQ=300\sqrt{3}\\ & \text{Tree}\text{grew}=PQ-MQ\\ & \therefore \text{Tree}\text{grew}\\ & =300\sqrt{3}-\frac{300}{\sqrt{3}}\\ & =300\frac{2}{\sqrt{3}}\\ & =3\times 100\times \frac{2}{\sqrt{3}}\\ & =200\sqrt{3}\text{m}\end{array}$$