Understand the Solar System,and the math behind it.

Answer

Full solution

1. Start from the one relationship that matters: in a long thin right triangle with short leg $b$ and long leg $d$, $\tan p=b/d$, and for a tiny angle $p\approx b/d$ (radians).

2. Write the first angle as $p=b/d$. For the second observation replace $b$ by $3b$ and $d$ by $1.5\,d$: $p'=\dfrac{3b}{1.5\,d}$.

3. Simplify the factors: $\dfrac{3}{1.5}=2$, so $p'=2\cdot\dfrac bd=2p$. The second angle is twice the first.

4. The claimed $4.5=3\times1.5$ treats both changes as if they made the angle bigger. A bigger baseline does, but a bigger distance makes the angle smaller, so that factor belongs under the fraction bar.

5. So the statement is false; the correct factor is $2$.

Answer

Full solution

1. Sketch the level ground, the mast and the sight line. A zenith angle is measured from straight up, so the $72.0^\circ$ sits between the vertical and the sight line, not between the ground and the sight line.

2. The angle between the sight line and the horizontal is therefore $90^\circ-72.0^\circ=18.0^\circ$. Now the mast height is the leg opposite this angle and the $85\ \mathrm{m}$ is the leg adjacent to it.

3. Use tangent in degree mode: $h=85\tan18.0^\circ=85(0.3249)\approx27.6\ \mathrm{m}$. A tall mast would need a much larger angle above the horizon, so this small height fits an angle of only $18^\circ$.

4. Each wrong choice is a specific slip: (A) treats $72.0^\circ$ as an angle above the horizon, $85\tan72.0^\circ$; (C) uses sine instead of tangent, $85\sin18.0^\circ$; (D) is $85/\sin72.0^\circ$, the length of the sight line rather than the height; (E) is $85/\sin18.0^\circ$, dividing by the sine of the angle above the horizon.