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📐 Key Concepts: Oblique Triangles

Law of Sines:

$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$

Law of Cosines:

$$c^2 = a^2 + b^2 - 2ab\cos C$$

Use Law of Sines when you know AAS or ASA. Use Law of Cosines when you know SAS or SSS.

Problem 1: Height of Tower (Right and Oblique Triangles)

Towers A and B stand on a horizontal plane, tower B being 150 m high. The angle of elevation of the top of tower A as seen from point C (in the same vertical plane) is 50°. The angle of depression of C viewed from the top of tower B is 27.5°. The angle subtended at the top of tower B by the top of tower A and C is 50°. Find the height of tower A.

Tower Problem – Diagram Tower Problem – Diagram Tower Problem – Diagram
Solution Solution Solution Solution

Step 1 — Find BC (horizontal distance from tower B to point C) using the depression angle from top of B:

$$BC = \frac{BT_B}{\tan27.5°} = \frac{150}{0.5206} \approx 288.1 \text{ m}$$

Length $CT_B = \sqrt{BC^2 + 150^2} = \sqrt{288.1^2 + 150^2} \approx 325.0$ m.

Step 2 — Set up triangle $C$–$T_B$–$T_A$. From $C$: elevation angle to $T_B = 27.5°$; elevation angle to $T_A = 50°$. Since the towers are on the same side of $C$, angle at $C$ in the triangle $= 50° - 27.5° = 22.5°$. The angle at $T_B$ between lines to $C$ and $T_A = 50°$ (given). Therefore angle at $T_A = 180° - 50° - 22.5° = 107.5°$.

Step 3 — Law of Sines:

$$\frac{CT_A}{\sin50°} = \frac{CT_B}{\sin107.5°} \Rightarrow CT_A = \frac{325.0 \times \sin50°}{\sin107.5°} = \frac{325.0 \times 0.7660}{0.9537} \approx 261.0 \text{ m}$$

Step 4 — Height of tower A:

$$h_A = CT_A \times \sin50° = 261.0 \times 0.7660 \approx \boxed{200 \text{ m}}$$

Problem 2: Elevation of Mountain Peak B

Mountain peak A is 1,040 m above sea level. From peak A, the angle of elevation of peak B is 10°. A pilot flying directly over peak A at 2,820 m above sea level measures the angle of depression to peak B as 46°12'. Determine the elevation of peak B.

Mountain Peak – Diagram Mountain Peak – Diagram Mountain Peak – Diagram
Solution Solution Solution Solution

Let $d$ = horizontal distance between peak A and peak B, and $h_B$ = elevation of peak B.

From peak A (at 1040 m), angle of elevation to B = 10°:

$$h_B - 1040 = d\tan10° \quad \cdots (1)$$

From the pilot (at 2820 m directly above A), angle of depression to B = 46°12' = 46.2°:

$$2820 - h_B = d\tan46.2° \quad \cdots (2)$$

Add (1) and (2):

$$1780 = d(\tan10° + \tan46.2°) = d(0.1763 + 1.0538) = 1.2301\,d$$
$$d = \frac{1780}{1.2301} \approx 1447 \text{ m}$$

Elevation of peak B:

$$h_B = 1040 + 1447\tan10° = 1040 + 1447(0.1763) \approx 1040 + 255 = \boxed{1295 \text{ m above sea level}}$$

Problem 3: Distance Across a Building (Law of Cosines)

Two survey points A and B are on opposite sides of a building and cannot be seen from each other. A third point C is chosen such that CA = 100 m, CB = 120 m, and the angle ACB = 55°. Find the distance AB.

Apply the Law of Cosines with $a = CA = 100$ m, $b = CB = 120$ m, included angle $C = 55°$, solving for side $c = AB$:

$$c^2 = a^2 + b^2 - 2ab\cos C = 100^2 + 120^2 - 2(100)(120)\cos 55°$$
$$c^2 = 10000 + 14400 - 24000(0.5736) = 24400 - 13766 = 10634$$
$$c = \sqrt{10634} \approx \boxed{103.1 \text{ m}}$$

Problem 4: Finding a Missing Angle (Law of Sines)

In triangle ABC, side $a = 15$ m, side $b = 20$ m, and angle $A = 32°$. Find angle B, angle C, and side $c$.

Step 1 — Find angle B using Law of Sines:

$$\frac{\sin B}{b} = \frac{\sin A}{a} \Rightarrow \sin B = \frac{20\sin 32°}{15} = \frac{20(0.5299)}{15} = 0.7066$$
$$B = \arcsin(0.7066) \approx 44.9°$$

Step 2 — Find angle C:

$$C = 180° - A - B = 180° - 32° - 44.9° = 103.1°$$

Step 3 — Find side c:

$$c = \frac{a\sin C}{\sin A} = \frac{15\sin 103.1°}{sin 32°} = \frac{15(0.9744)}{0.5299} \approx \boxed{27.6 \text{ m}}$$