Rowena has a very long, level backyard. From a particular point, she determines that the angle of elevation to a large tree in her backyard is $15^{\circ}$. She moves $40 \mathrm{~m}$ closer and determines that the new angle of elevation is $30^{\circ}$. How tall, in metres, is the tree?

![](https://cdn.mathpix.com/cropped/2024_04_30_790b6557ee0304df033eg-2.jpg?height=292&width=702&top_left_y=1025&top_left_x=771)
We label the diagram as shown and call the height of the tree $h \mathrm{~m}$.

![](https://cdn.mathpix.com/cropped/2024_04_30_264fdb419065898dfdf2g-03.jpg?height=320&width=731&top_left_y=1905&top_left_x=754)

Since $\triangle B C T$ is right-angled at $C$ and $\angle T B C=30^{\circ}$, then $B C=\sqrt{3} T C=\sqrt{3} h \mathrm{~m}$.

Since $\triangle A C T$ is right-angled at $C$, then $\tan \left(15^{\circ}\right)=\frac{h}{40+\sqrt{3} h}$.

This gives $40 \tan \left(15^{\circ}\right)+\sqrt{3} \tan \left(15^{\circ}\right) h=h$ from which we obtain $h=\frac{40 \tan \left(15^{\circ}\right)}{1-\sqrt{3} \tan \left(15^{\circ}\right)}=20$.

Therefore, the height of the tree is $20 \mathrm{~m}$.