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Trigonometry - Bearings and 3D Trigonometry

Grade 12A Level

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

πŸ”‘Concepts

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Bearings are measured from North (000∘000^\circ), in a clockwise direction, and are always written as three digits. For example, East is 090∘090^\circ and South is 180∘180^\circ. The bearing of point BB from point AA is the angle measured clockwise from the North line at AA to the line ABAB.

Diagram showing a bearing of 060 degrees from point A to point B.
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The angle between two planes is found by identifying two lines, one in each plane, that meet at a point on the intersection line and are both perpendicular to that intersection line. The angle between these two lines is the angle between the planes.

Geometric representation of the angle between two perpendicular intersecting lines on a plane.
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Angle of elevation and depression: The angle of elevation is the angle measured upwards from the horizontal line to the line of sight of an object. The angle of depression is measured downwards from the horizontal line.

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3D Pythagoras' Theorem: In a cuboid with dimensions xx, yy, and zz, the length of the space diagonal dd is given by d=x2+y2+z2d = \sqrt{x^2 + y^2 + z^2}.

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When solving 3D problems, look for right-angled triangles within the 3D shape. Often, you must first calculate a length on one plane (like the base) to find a side for a triangle in a different plane (like a vertical cross-section).

πŸ“Formulae

asin⁑A=bsin⁑B=csin⁑C\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} (Sine Rule)

a2=b2+c2βˆ’2bccos⁑Aa^2 = b^2 + c^2 - 2bc \cos A (Cosine Rule for sides)

cos⁑A=b2+c2βˆ’a22bc\cos A = \frac{b^2 + c^2 - a^2}{2bc} (Cosine Rule for angles)

Area=12absin⁑C\text{Area} = \frac{1}{2}ab \sin C (Area of a non-right triangle)

d=x2+y2+z2d = \sqrt{x^2 + y^2 + z^2} (3D Distance/Pythagoras)

πŸ’‘Examples

Problem 1:

A ship sails 10 km from port P on a bearing of 060Β° to point Q. It then sails 15 km from Q on a bearing of 150Β° to point R. Calculate the distance PR.

Solution:

  1. Find the internal angle PQR. The bearing from Q back to P is 60+180=240∘60 + 180 = 240^\circ. The bearing from Q to R is 150∘150^\circ. The angle PQR=240βˆ˜βˆ’150∘=90∘PQR = 240^\circ - 150^\circ = 90^\circ.
  2. Since it is a right-angled triangle, use Pythagoras: PR=102+152=100+225=325β‰ˆ18.03PR = \sqrt{10^2 + 15^2} = \sqrt{100 + 225} = \sqrt{325} \approx 18.03 km.

Explanation:

By finding the relationship between the two bearings at point Q, we determine that the path forms a right-angled triangle, allowing for the use of the Pythagorean theorem.

Problem 2:

In a cuboid with length 8cm, width 6cm, and height 5cm, find the angle that the space diagonal makes with the base.

Solution:

  1. Find the length of the diagonal of the base (d_base): dbase=82+62=64+36=10d_{base} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = 10 cm.
  2. The space diagonal, the base diagonal, and the height form a right-angled triangle.
  3. Let the angle be θ\theta. tan⁑θ=heightbase diagonal=510=0.5\tan \theta = \frac{\text{height}}{\text{base diagonal}} = \frac{5}{10} = 0.5.
  4. ΞΈ=tanβ‘βˆ’1(0.5)β‰ˆ26.6∘\theta = \tan^{-1}(0.5) \approx 26.6^\circ.

Explanation:

To find the angle between a line (space diagonal) and a plane (the base), you first find the projection of that line onto the plane (the base diagonal) and then use SOH CAH TOA.

Problem 3:

A vertical flagpole TPTP of height hh stands at the corner PP of a horizontal rectangular field PQRSPQRS. The length PQ=40PQ = 40 m and QR=30QR = 30 m. The angle of elevation of the top of the pole TT from point RR is 25∘25^\circ. Calculate the height hh of the flagpole.

A 3D diagram showing a rectangular base PQRS and a vertical pole TP at corner P.

Solution:

  1. First, find the distance PRPR on the horizontal ground using Pythagoras' Theorem: PR=PQ2+QR2=402+302=1600+900=2500=50Β mPR = \sqrt{PQ^2 + QR^2} = \sqrt{40^2 + 30^2} = \sqrt{1600 + 900} = \sqrt{2500} = 50\text{ m}
  2. In the vertical right-angled triangle β–³TPR\triangle TPR, we know the base PR=50PR = 50 m and the angle ∠TRP=25∘\angle TRP = 25^\circ.
  3. Use the tangent ratio: tan⁑(25∘)=TPPR=h50\tan(25^\circ) = \frac{TP}{PR} = \frac{h}{50}
  4. Solve for hh: h=50Γ—tan⁑(25∘)β‰ˆ50Γ—0.4663=23.32Β mh = 50 \times \tan(25^\circ) \approx 50 \times 0.4663 = 23.32\text{ m}

Explanation:

This problem requires identifying a right-angled triangle on the ground to find a length that connects to a vertical triangle containing the unknown height.

Problem 4:

A hiker walks 55 km from point AA on a bearing of 040∘040^\circ to point BB. They then walk 88 km on a bearing of 110∘110^\circ to point CC. Calculate the distance ACAC.

Bearings diagram showing points A, B, and C with North lines at A and B.

Solution:

  1. Determine the interior angle ∠ABC\angle ABC. Let the North line at BB be NBN_B.
  2. The angle from the line BABA to the North line NBN_B is 180βˆ˜βˆ’40∘=140∘180^\circ - 40^\circ = 140^\circ (interior angles between parallel North lines).
  3. The angle from NBN_B to BCBC is 110∘110^\circ.
  4. The interior angle ∠ABC\angle ABC is 360βˆ˜βˆ’(140∘+110∘)=110∘360^\circ - (140^\circ + 110^\circ) = 110^\circ.
  5. Apply the Cosine Rule to β–³ABC\triangle ABC: AC2=AB2+BC2βˆ’2(AB)(BC)cos⁑(∠ABC)AC^2 = AB^2 + BC^2 - 2(AB)(BC) \cos(\angle ABC) AC2=52+82βˆ’2(5)(8)cos⁑(110∘)AC^2 = 5^2 + 8^2 - 2(5)(8) \cos(110^\circ) AC2=25+64βˆ’80(βˆ’0.3420)=89+27.36=116.36AC^2 = 25 + 64 - 80(-0.3420) = 89 + 27.36 = 116.36
  6. AC=116.36β‰ˆ10.79Β kmAC = \sqrt{116.36} \approx 10.79\text{ km}

Explanation:

Bearings are used to find the internal angle of a triangle. Parallel North lines allow us to find angles using the properties of transversals.