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Geometry and Trigonometry - Applications in 3D Geometry – Navigation

Grade 11IB_AA

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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True bearings are measured clockwise from North (000∘000^\circ) and are always written as three digits (e.g., 045∘045^\circ or 210∘210^\circ). In 3D navigation, bearings typically represent horizontal direction on the xyxy-plane, while altitude or elevation introduces the zz-coordinate.

Diagram showing a true bearing of 060 degrees measured clockwise from North.
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The angle of elevation is the angle between the horizontal ground plane and the line of sight looking up at an object. This creates a right-angled triangle where the vertical height hh and horizontal distance dd are related by h=dtan⁡θh = d \tan \theta.

Right-angled triangle showing angle of elevation theta, horizontal distance d, and height h.
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To solve 3D problems, decompose the scenario into 2D triangles. Use the horizontal plane (bearings) to find ground distances and then use vertical planes (elevation/depression) to find heights or direct distances (hypotenuses).

3D representation of a point above a horizontal ground plane with an observer at O and an object at P.
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The Sine Rule and Cosine Rule are essential when the horizontal movement involves non-right-angled triangles. For a triangle with sides a,b,ca, b, c and opposite angles A,B,CA, B, C, use these to link positions before applying 3D trigonometric ratios.

📐Formulae

d=(x2−x1)2+(y2−y1)2+(z2−z1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}

asin⁡A=bsin⁡B=csin⁡C\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}

a2=b2+c2−2bccos⁡Aa^2 = b^2 + c^2 - 2bc \cos A

cos⁡A=b2+c2−a22bc\cos A = \frac{b^2 + c^2 - a^2}{2bc}

Height=Horizontal Distance×tan⁡(θ) (where θ is the angle of elevation)\text{Height} = \text{Horizontal Distance} \times \tan(\theta) \text{ (where } \theta \text{ is the angle of elevation)}

💡Examples

Problem 1:

An observer at point AA sees a drone at point DD. The drone is at an altitude of 200200 m. From point AA, the bearing of the drone is 040∘040^\circ and the angle of elevation is 30∘30^\circ. Calculate the horizontal distance from the observer to the point GG directly below the drone, and the direct distance ADAD.

Solution:

  1. Let GG be the point on the ground directly below the drone DD. We have a right-angled triangle △ADG\triangle ADG where ∠AGD=90∘\angle AGD = 90^\circ.
  2. The angle of elevation is ∠DAG=30∘\angle DAG = 30^\circ. The altitude (opposite side) is DG=200DG = 200 m.
  3. Using tan⁡(30∘)=DGAG\tan(30^\circ) = \frac{DG}{AG}: AG=200tan⁡(30∘)=2001/3=2003≈346.4 mAG = \frac{200}{\tan(30^\circ)} = \frac{200}{1/\sqrt{3}} = 200\sqrt{3} \approx 346.4 \text{ m}
  4. Using sin⁡(30∘)=DGAD\sin(30^\circ) = \frac{DG}{AD}: AD=200sin⁡(30∘)=2000.5=400 mAD = \frac{200}{\sin(30^\circ)} = \frac{200}{0.5} = 400 \text{ m}

Explanation:

To solve 3D navigation problems, first identify the vertical triangle (Drone-Ground-Observer). Use the angle of elevation and the vertical height to find the horizontal distance (adjacent side) and the line-of-sight distance (hypotenuse).

Problem 2:

A ship travels from port PP on a bearing of 060∘060^\circ for 5050 km to point QQ. It then changes course and travels on a bearing of 150∘150^\circ for 3030 km to point RR. Find the distance PRPR.

Solution:

  1. Draw the bearings. The angle between the first path PQPQ and the North line is 60∘60^\circ.
  2. At QQ, draw a new North line. The interior angle between the path QPQP and the North line at QQ is 180∘−60∘=120∘180^\circ - 60^\circ = 120^\circ (consecutive interior angles).
  3. The bearing of RR from QQ is 150∘150^\circ.
  4. The angle ∠PQR=360∘−(120∘+150∘)=90∘\angle PQR = 360^\circ - (120^\circ + 150^\circ) = 90^\circ.
  5. Since △PQR\triangle PQR is right-angled at QQ, use Pythagoras: PR=502+302=2500+900=3400≈58.3 kmPR = \sqrt{50^2 + 30^2} = \sqrt{2500 + 900} = \sqrt{3400} \approx 58.3 \text{ km}

Explanation:

In 2D navigation segments of 3D problems, use the properties of parallel North lines to find the interior angles of the triangle formed by the journey. In this case, the interior angle at QQ turned out to be 90∘90^\circ, simplifying the calculation.

Problem 3:

A hiker at point AA observes a mountain peak MM on a bearing of 035∘035^\circ with an angle of elevation of 15∘15^\circ. After walking 55 km due East to point BB, the hiker finds the new bearing of the peak is 330∘330^\circ. Calculate the height of the mountain peak MM above the level of AA and BB.

3D diagram showing ground points A, B, H and mountain peak M.

Solution:

  1. Let HH be the point on the ground directly below peak MM. Let hh be the height MHMH.
  2. In △AMH\triangle AMH, AH=htan⁡15∘AH = \frac{h}{\tan 15^\circ}.
  3. In the horizontal △ABH\triangle ABH: ∠HAB=90∘−35∘=55∘\angle HAB = 90^\circ - 35^\circ = 55^\circ (since East is 090∘090^\circ). ∠HBA=330∘−270∘=60∘\angle HBA = 330^\circ - 270^\circ = 60^\circ (since West is 270∘270^\circ). ∠AHB=180∘−(55∘+60∘)=65∘\angle AHB = 180^\circ - (55^\circ + 60^\circ) = 65^\circ.
  4. Apply Sine Rule in △ABH\triangle ABH: AHsin⁡60∘=ABsin⁡65∘\frac{AH}{\sin 60^\circ} = \frac{AB}{\sin 65^\circ} AH=5×sin⁡60∘sin⁡65∘≈4.778 kmAH = \frac{5 \times \sin 60^\circ}{\sin 65^\circ} \approx 4.778 \text{ km}
  5. Substitute AHAH back into the height equation: h=AH×tan⁡15∘=4.778×tan⁡15∘≈1.28 kmh = AH \times \tan 15^\circ = 4.778 \times \tan 15^\circ \approx 1.28 \text{ km}

Explanation:

We first solve the horizontal triangle on the ground using the Sine Rule to find the distance from the observer's first position to the base of the mountain. Then, we use the right-angled triangle in the vertical plane to find the height.

Problem 4:

An airplane PP is flying at an altitude of 10,00010,000 m. At a specific moment, it is at a bearing of 020∘020^\circ from a control tower TT and at an angle of elevation of 25∘25^\circ. An observer at point SS, located 15,00015,000 m due South of the tower TT, also views the plane. Calculate the distance from the observer at SS to the plane PP.

Geometry diagram showing points T (Tower), S (South), G (Ground projection) and P (Plane).

Solution:

  1. Let GG be the ground position of the plane directly below PP. TGTG is the horizontal distance from the tower.
  2. In △PTG\triangle PTG, TG=10000tan⁡25∘≈21445 mTG = \frac{10000}{\tan 25^\circ} \approx 21445 \text{ m}.
  3. In the horizontal plane △STG\triangle STG: ST=15000ST = 15000, TG=21445TG = 21445, and ∠STG=180∘−20∘=160∘\angle STG = 180^\circ - 20^\circ = 160^\circ (since SS is South and GG is at 020∘020^\circ from TT).
  4. Use Cosine Rule to find SGSG: SG2=150002+214452−2(15000)(21445)cos⁡160∘SG^2 = 15000^2 + 21445^2 - 2(15000)(21445) \cos 160^\circ SG2≈225,000,000+459,887,025−(643,350,000×−0.9397)SG^2 \approx 225,000,000 + 459,887,025 - (643,350,000 \times -0.9397) SG≈1,289,444,000≈35909 mSG \approx \sqrt{1,289,444,000} \approx 35909 \text{ m}
  5. Use Pythagoras in vertical △PSG\triangle PSG to find SPSP: SP=SG2+PG2=359092+100002≈37275 mSP = \sqrt{SG^2 + PG^2} = \sqrt{35909^2 + 10000^2} \approx 37275 \text{ m}

Explanation:

First, find the ground projection distance using the tangent ratio. Next, use the Cosine Rule on the horizontal plane to find the distance between the observer and the point directly below the plane. Finally, apply the Pythagorean theorem in 3D (or a vertical triangle) to find the direct distance.