From the top of a hill the angle of depression of a point A on the ground is 30 degrees. After walking three quarters of the way down the straight slope, the angle of depression of A is 15 degrees. If the slope makes an angle θ with the horizontal, find tan θ.

Step-by-step solution

1. 1

   Set up coordinates and normalise the slope length. Because the answer is an angle, only ratios matter, so take the whole slope length to be $L=1$. The summit $B$ then sits at height

   $h=\sin\theta$

   above the ground, and descending a distance $s$ along the slope lowers you by $s\sin\theta$ while moving you $s\cos\theta$ horizontally toward $A$.

2. 2

   Use the first sighting to fix the horizontal distance. Let $d$ be the horizontal distance from $B$ to $A$. The angle of depression at $B$ is $30^{\circ}$, so

   $\tan 30^{\circ}=\frac{h}{d}=\frac{\sin\theta}{d}\quad\Longrightarrow\quad d=\sqrt3\,\sin\theta$

3. 3

   Write the second sighting. After descending $\tfrac34$ of the slope, the observer is at height $\sin\theta-\tfrac34\sin\theta=\tfrac14\sin\theta$ and at horizontal distance $d-\tfrac34\cos\theta$ from $A$. Hence

   $\tan 15^{\circ}=\frac{\tfrac14\sin\theta}{\sqrt3\sin\theta-\tfrac34\cos\theta}$

   The unknown $d$ has been eliminated, leaving one equation in $\theta$ alone.

4. 4

   Solve for tan θ. Cross-multiplying and collecting the $\sin\theta$ terms:

   $\tan 15^{\circ}\left(\sqrt3\sin\theta-\tfrac34\cos\theta\right)=\tfrac14\sin\theta$

   $\sin\theta\left(\sqrt3\tan 15^{\circ}-\tfrac14\right)=\tfrac34\tan 15^{\circ}\cos\theta$

   $\tan\theta=\frac{\tfrac34\tan 15^{\circ}}{\sqrt3\tan 15^{\circ}-\tfrac14}$

5. 5

   Substitute the exact value of tan 15°. Since $\tan 15^{\circ}=2-\sqrt3$,

   $\tan\theta=\frac{3(2-\sqrt3)}{4\sqrt3(2-\sqrt3)-1}=\frac{3(2-\sqrt3)}{8\sqrt3-12-1}=\frac{6-3\sqrt3}{8\sqrt3-13}$

6. 6

   Rationalise the denominator. Multiply top and bottom by $8\sqrt3+13$; the denominator becomes $(8\sqrt3)^{2}-13^{2}=192-169=23$, and the numerator is

   $(6-3\sqrt3)(8\sqrt3+13)=48\sqrt3+78-72-39\sqrt3=9\sqrt3+6$

   $\boxed{\tan\theta=\frac{6+9\sqrt3}{23}\approx 0.9386}$

7. 7

   Check the answer for consistency. Numerically $\theta=\arctan(0.9386)=43.19^{\circ}$. The setup requires $A$ to lie beyond the foot of the slope, i.e. $d=\sqrt3\sin\theta>\cos\theta$, which means $\tan\theta>\tfrac{1}{\sqrt3}$, i.e. $\theta>30^{\circ}$ — satisfied ✓. Substituting back: with $\theta=43.19^{\circ}$, $d=\sqrt3(0.6845)=1.1854$ and the second sighting gives $\arctan\!\left(\tfrac{0.1711}{1.1854-0.5468}\right)=\arctan(0.2679)=15.00^{\circ}$ ✓.