This problem involves using trigonometry to find distances based on angles of elevation and given heights.

Given:

• The height of the house $AB = 70$ meters.
• The angle of elevation from point $A$ to point $C$ (the mountain peak) is $30^\circ$.
• The angle of elevation from point $B$ to point $C$ is $15^\circ$.
• The horizontal distance between $A$ and $B$ is denoted as $AB$, but the value of $AB$ is not explicitly provided.

Let's denote:

• $H$ as the horizontal distance from point $A$ to the mountain's base, $C$.
• $AC$ and $BC$ as the distances from points $A$ and $B$ to the mountain's peak, respectively.

Step-by-step solution:

1. Apply tangent trigonometry for $\triangle AHC$:

In this right-angled triangle: $\tan(30^\circ) = \frac{h_C}{H}$ Where:

• $h_C$ is the height of the mountain from point $A$ (the base of the house). Thus, $H = \frac{h_C}{\tan(30^\circ)}$ We know that $\tan(30^\circ) = \frac{1}{\sqrt{3}}$, so: $H = h_C \cdot \sqrt{3}$

2. Apply tangent trigonometry for $\triangle BHC$:

Now, for the observer at point $B$: $\tan(15^\circ) = \frac{h_C}{H - 70}$ Since $H - 70$ represents the horizontal distance from $B$ to the mountain's base.

Solving for $h_C$:

We can now solve this system of equations by eliminating $H$ and finding $h_C$.

Would you like me to proceed with the numerical calculations and provide more detailed steps?