Answer:
Approximately [tex]35\; \rm m[/tex].
Step-by-step explanation:
In a right triangle, the tangent of an angle is the ratio between the length of the side opposite to this angle and the length of the side adjacent to this angle:
[tex]\displaystyle \tan{\theta} = \frac{\text{opposite}}{\text{adjacent}}[/tex].
Let [tex]x[/tex] denote the height of this kite in meters.
Refer to the diagram attached. Let [tex]\sf O[/tex] denote the point right below the kite in the same vertical plane as [tex]\sf P[/tex] and [tex]\sf Q[/tex]. Denote the kite as [tex]\sf K[/tex].
The kite [tex]\sf K[/tex], observer [tex]\sf Q[/tex], and point [tex]\sf O[/tex] are the three vertices of right triangle [tex]\triangle {\sf KQO}[/tex], where [tex]\angle{\sf KQO} = 45^\circ[/tex].
In this right triangle, [tex]\sf OK[/tex] is the side opposite to [tex]\angle{\sf KQO}[/tex], whereas [tex]\sf OQ[/tex] is the side adjacent to [tex]\angle{\sf KQO}\![/tex].
Therefore:
[tex]\displaystyle \tan(\angle{\sf KQO}) = \frac{\text{length of ${\sf OK}$}}{\text{length of ${\sf OQ}$}}[/tex].
[tex]\displaystyle \tan(45^\circ) = \frac{x}{\text{length of ${\sf OQ}$}}[/tex].
[tex]\text{length of ${\sf OQ}$} = x\, \tan(45^\circ)[/tex].
Similarly, in right triangle [tex]\triangle {\sf KPO}[/tex], [tex]\angle{\sf KPO} = 35^\circ[/tex]. In this right triangle, [tex]\sf OK[/tex] is the side opposite to [tex]\angle{\sf KPO}[/tex], whereas [tex]\sf OP[/tex] is the side adjacent to [tex]\angle{\sf KPO}\![/tex].
[tex]\displaystyle \tan(\angle{\sf KPO}) = \frac{\text{length of ${\sf OK}$}}{\text{length of ${\sf OP}$}}[/tex].
[tex]\displaystyle \tan(35^\circ) = \frac{x}{\text{length of ${\sf OP}$}}[/tex].
[tex]\text{length of ${\sf OP}$} = x\, \tan(35^\circ)[/tex]
The question states that observer [tex]\sf P[/tex] and observer [tex]\sf Q[/tex] are on the same side of the kite. Hence, the horizontal distance between [tex]\sf P\![/tex] and [tex]\sf Q\![/tex] would be the same as the difference between:
In other words:
[tex]\begin{aligned}& \text{horizontal distance between ${\sf P}$ and ${\sf Q}$} \\ &= (\text{horizontal distance between ${\sf P}$ and ${\sf K}$} \\ &\quad\quad -\text{horizontal distance between ${\sf Q}$ and ${\sf K}$})\end{aligned}[/tex].
[tex]\begin{aligned}& \text{length of ${\sf PQ}$} \\ &= \text{length of ${\sf OP}$} - \text{length of ${\sf OQ}$}\end{aligned}[/tex].
[tex]\displaystyle 15 = \frac{x}{\tan(35^\circ)} - \frac{x}{\tan(45^\circ)}[/tex].
Solve for [tex]x[/tex]:
[tex]\begin{aligned}x &= 15 \left/\left(\frac{1}{\tan({35}^\circ)} - \frac{1}{\tan({45}^\circ)}\right)\right. \approx 35\end{aligned}[/tex].
