Triangle Solving – Problem 30: The Zhenguo Temple Pagoda is a seven-story brick tower

The Zhenguo Temple Pagoda is a seven-story brick tower. A student wants to estimate its height $MN$. A building $AB$ of height $7.5\,\text{m}$ lies due north of the pagoda. From a point $C$ on level ground (points $B,C,N$ are collinear), the angles of elevation to the top of the building $A$ and the top of the pagoda $M$ are $15^{\circ}$ and $60^{\circ}$, respectively. From point $A$, the angle of elevation to $M$ is $30^{\circ}$. Approximate the height $MN$. (Use $\sqrt{3}\approx1.73$.)

A. $31.42\,\text{m}$ B. $33.26\,\text{m}$ C. $35.48\,\text{m}$ D. $37.52\,\text{m}$

Step 1. Let the height of the pagoda be $MN=h$. Since the angle of elevation to $A$ from $C$ is $15^{\circ}$, $$\tan 15^{\circ}=\frac{AB}{BC}\Rightarrow BC=\frac{7.5}{\tan 15^{\circ}}.$$

Step 2. Since the angle of elevation to $M$ from $C$ is $60^{\circ}$, $$\tan 60^{\circ}=\frac{MN}{CN}=\frac{h}{CN}\Rightarrow CN=\frac{h}{\sqrt{3}}.$$

Step 3. From $A$, the angle of elevation to $M$ is $30^{\circ}$, and $A$ is $7.5\,\text{m}$ above the ground, so $$\tan 30^{\circ}=\frac{h-7.5}{BN}\Rightarrow BN=\sqrt{3}(h-7.5).$$

Step 4. Since $B,C,N$ are collinear with $B$ between $C$ and $N$, $CN=BN-BC$, hence $$\frac{h}{\sqrt{3}}=\sqrt{3}(h-7.5)-\frac{7.5}{\tan 15^{\circ}}.$$

Step 5. Using $\tan 15^{\circ}=2-\sqrt{3}\Rightarrow \frac{1}{\tan 15^{\circ}}=2+\sqrt{3}$, we have $BC=7.5(2+\sqrt{3})$.

Step 6. Substitute and simplify to obtain $$h=\frac{15\sqrt{3}}{2}(1+\sqrt{3})=\frac{15}{2}(3+\sqrt{3})\approx 35.48\,\text{m}.$$

Step 7. Therefore the correct choice is C.

C