Hello ivyrianne,
I would begin by drawing a diagram:
View attachment 999
$T$ is the height of the tower.
$x$ is the horizontal distance between the tower and the tree.
$a$ is the vertical distance between the bottom of the tower and the bottom of the tree.
$b$ is the vertical distance between the top of the tower and the top of the tree.
$h$ is the height of the tree.
$\theta$ is the angle of inclination of the hill.
$\alpha$ is the angle of depression from the top of the tower to the top of the tree.
$\beta$ is the angle of depression from the top of the tower to the bottom of the tree.
Now, from the diagram, we observe the following relationships:
(1) $$a+h+b=T$$
(2) $$\tan(\theta)=\frac{a}{x}$$
(3) $$\tan(\alpha)=\frac{b}{x}$$
(4) $$\tan(\beta)=\frac{h+b}{x}$$
Solving (1) and (4) for $h+b$ and equating, we find:
$$T-a=x\tan(\beta)$$
Solving (2) for $a$, and substituting into the above, we may write:
$$T-x\tan(\theta)=x\tan(\beta)$$
Solving this for $x$, there results:
(5) $$x=\frac{T}{\tan(\beta)+\tan(\theta)}$$
Now, solving (4) for $h$, we get:
$$h=x\tan(\beta)-b$$
Solving (3) for $b$ and substituting into the above, we get:
$$h=x\tan(\beta)-x\tan(\alpha)$$
$$h=x\left(\tan(\beta)-\tan(\alpha) \right)$$
Substituting for $x$ from (5), we find:
$$h=\left(\frac{T}{\tan(\beta)+\tan(\theta)} \right)\left(\tan(\beta)-\tan(\alpha) \right)$$
$$h=\frac{T\left(\tan(\beta)-\tan(\alpha) \right)}{\tan(\beta)+\tan(\theta)}$$
Plugging in the given data:
$$\theta=18^{\circ},\,\alpha=26^{\circ},\,\beta=38^{\circ},\,T=40\text{ m}$$
we have:
$$h=\frac{(40\text{ m})\left(\tan\left(38^{\circ} \right)-\tan\left(26^{\circ} \right) \right)}{\tan\left(38^{\circ} \right)+\tan\left(18^{\circ} \right)}\approx10.6147758253\text{ m}$$
