3D application of basic trigonometry https://imgur.com/a/D3ANJ (can't be uploaded because of size)
I first thought that ON//BC, because B is due east of O and C is due north of B.
However, that results in me getting an incorrect value of OT(correct value = 39.3)

Let $OB=a$.
Thus, $$OT=a\tan40^{\circ}$$ and
$$OC\tan25^{\circ}=a\tan40^{\circ},$$
which gives
$$OC=\frac{a\tan40^{\circ}}{\tan25^{\circ}}$$ and by Pythagoras theorem we obtain:
$$70^2+a^2=\left(\frac{a\tan40^{\circ}}{\tan25^{\circ}}\right)^2$$ and from here we can find a value of $a$:
$$a=\frac{70}{\sqrt{\frac{\tan^240^{\circ}}{\tan^225^{\circ}}-1}}$$ and $OT$.
I got $OT=39.262...$
