What is the error solving this problem about instantaneous rate of change? I have the following problem:

A hot air balloon rising straight up from a level field is tracked by
a range finder $150$ meters from the liftoff point. At the moment that
the range finder’s elevation angle is $\frac{\pi}{4}$, the angle is
increasing at the rate of $0.14$ rad/min. How fast is the balloon
rising at that moment?

My development was:
Let $h$ the altitude of the hot air balloon, $\theta$ the angle.
Using trigonometry, i got: $\sin(\theta) \cdot 150\sqrt{2} = h$, where $150\sqrt{2}$ is the hypotenuse.
Using implicit derivation respect to the time or moment $t$, to get:
$\frac{d}{dt}\sin(\theta) \cdot 150\sqrt{2}=\frac{d}{dt}h$
Since $\sin(\theta)$ is a composition of the functions $\sin(x)$ and $\theta(t)$ i need to use chain rule, so i have: $\cos(\theta) \cdot \frac{d}{dt}\theta \cdot 150\sqrt{2}=\frac{d}{dt}h \implies \frac{d}{dt}h=21$.
But the correct answer is $42$ that is exactly the double of my answer, what is wrong with my development? Thanks in advance.

When you write $\sin(\theta) \cdot 150\sqrt{2} = h$ you are implicitly assuming that the hypotenuse isn't changing, which is false.  If you were going to set it up like this you would need to say $\sin(\theta)\cdot c = h$ where here $c$ is the hypotenuse.  You can then use the fact that $c^2 = 150^2 + h^2$ to eliminate $c$ and then proceed with differentiating with respect to $t$.
I should also note that another route you could take is to say $\tan(\theta) = \frac{h}{150}$ and go from there.  In that case you wouldn't need to worry about the hypotenuse.
