From the worked example in the text, the bank-to-first-bridge distance is \(x \approx 1.32\) mi (obtained from \(\dfrac{x}{\sin 50^{\circ}} = \dfrac{0.86}{\sin 30^{\circ}}\)) and the bank-to-cliff distance is \(y \approx 1.55\) mi. Continue applying the Law of Sines \(\dfrac{a}{\sin A} = \dfrac{b}{\sin B}\) in each successive triangle along a path from the bank (or another known landmark) toward the church, propagating computed sides into adjacent triangles, until the side joining the church and City Hall is the unknown side of a triangle whose other elements are now known.

Trace a path of triangles from the known \(0.86\) mi side between City Hall and the first bridge to a triangle that contains both the fire hall and the school. In each triangle along the path, apply \(\dfrac{a}{\sin A} = \dfrac{b}{\sin B}\) to obtain an unknown side from a side that has already been computed plus the measured angles. The last application of the Law of Sines (in the triangle that includes both the fire hall and the school) yields the desired fire-hall-to-school distance.

In triangle \(A C D\), the side \(C D = 20\) m is known, together with the angle at \(C\) (between \(C D\) and \(C A\)) and the angle at \(D\) (between \(D C\) and \(D A\)); the third angle at \(A\) is \(180^{\circ}\) minus those two. By the Law of Sines, \(\dfrac{A C}{\sin \angle A D C} = \dfrac{20}{\sin \angle C A D}\), which gives \(A C\). The same procedure in triangle \(B C D\) gives \(B C\). The angle at \(C\) in triangle \(A B C\) is the (signed) difference of the two angles measured at \(C\), and by the Law of Cosines \((A B)^2 = (A C)^2 + (B C)^2 - 2 \cdot A C \cdot B C \cdot \cos(\angle A C B)\). Taking the square root yields \(A B\).

Let stations \(A\) and \(B\) on the near bank be separated by the measured baseline, and let \(C\) be the top of the cliff. The angles of elevation from \(A\) and from \(B\) to \(C\) are given. In the triangle \(A B C\) (in the plane that contains the baseline and the cliff top), the interior angles can be determined from the elevation angles and the baseline geometry. By the Law of Sines, \(\dfrac{A C}{\sin \angle A B C} = \dfrac{A B}{\sin \angle A C B}\), giving the slant distance \(A C\). The height of the cliff is \(h = A C \cdot \sin(\theta_A)\), where \(\theta_A\) is the angle of elevation from station \(A\) to the cliff top.

(a) Let \(x\) be the horizontal distance from the closer observation point to the foot of the mountain, so the farther point is at horizontal distance \(x + d\). From the two right triangles formed, \(\tan \beta = \dfrac{h}{x}\) and \(\tan \alpha = \dfrac{h}{x + d}\), hence \(x = h \cot \beta\) and \(x + d = h \cot \alpha\). Subtracting gives \(d = h(\cot \alpha - \cot \beta)\), so \(h = \dfrac{d}{\cot \alpha - \cot \beta}\). (b) Combining the cotangents: \(\cot \alpha - \cot \beta = \dfrac{\cos \alpha}{\sin \alpha} - \dfrac{\cos \beta}{\sin \beta} = \dfrac{\cos \alpha \sin \beta - \sin \alpha \cos \beta}{\sin \alpha \sin \beta} = \dfrac{\sin(\beta - \alpha)}{\sin \alpha \sin \beta}\). Substituting into the result of part (a) gives \(h = d \dfrac{\sin \alpha \sin \beta}{\sin(\beta - \alpha)}\). (c) With \(\alpha = 25^{\circ}\), \(\beta = 29^{\circ}\), \(d = 800\) ft: \(\cot 25^{\circ} \approx 2.1445\) and \(\cot 29^{\circ} \approx 1.8040\), so \(\cot 25^{\circ} - \cot 29^{\circ} \approx 0.3405\) and \(h \approx \dfrac{800}{0.3405} \approx 2349\) ft. Using formula (b): \(\sin 25^{\circ} \approx 0.4226\), \(\sin 29^{\circ} \approx 0.4848\), \(\sin 4^{\circ} \approx 0.0698\), so \(h \approx 800 \cdot \dfrac{0.4226 \cdot 0.4848}{0.0698} \approx 2349\) ft. The two formulas are algebraically equivalent, so they yield the same height (about \(2349\) ft, to within rounding).

Let \(T\) denote the top of the mountain, and let \(A\) and \(B\) be the two landmarks at the base. The slant distance from \(T\) to \(A\) satisfies \(\sin 42^{\circ} = \dfrac{2430}{T A}\), so \(T A = \dfrac{2430}{\sin 42^{\circ}} \approx \dfrac{2430}{0.6691} \approx 3631.7\) ft. Similarly, \(T B = \dfrac{2430}{\sin 39^{\circ}} \approx \dfrac{2430}{0.6293} \approx 3861.6\) ft. The angle at \(T\) between the sight lines \(T A\) and \(T B\) is \(68^{\circ}\). Applying the Law of Cosines to triangle \(T A B\): \((A B)^2 = (T A)^2 + (T B)^2 - 2 \cdot T A \cdot T B \cdot \cos 68^{\circ} \approx 3631.7^2 + 3861.6^2 - 2 (3631.7)(3861.6)(0.3746) \approx 13189000 + 14912000 - 10510000 \approx 17591000\). Therefore \(A B \approx 4194\) ft.

Identify each triangle in the sketch. A triangle with one known side and two known angles is solved by the Law of Sines \(\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}\); a triangle with two known sides and the included angle is solved by the Law of Cosines \(c^2 = a^2 + b^2 - 2 a b \cos C\). Solve one triangle at a time, using each computed side as input to the next neighboring triangle through their shared edge. After all sides have been determined, scale them onto graph paper to produce an accurate map of the two adjacent lots.

The Great Trigonometric Survey was a multi-decade scientific project to map the Indian subcontinent with unprecedented precision. It began in \(1802\) under William Lambton and was continued from \(1823\) by George Everest. The team carefully measured a long initial baseline and then extended a chain of triangles across the country, using massive theodolites and applying corrections for atmospheric refraction and the curvature of the Earth. Among its most famous results is the determination of the height of Peak XV (later renamed Mount Everest), which the Survey computed to be approximately \(29000\) ft.