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Trigonometry - Bearings and 3D Trigonometry

Grade 12IGCSE

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

🔑Concepts

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Bearings are measured from North, in a clockwise direction, and always expressed as three digits (e.g., 045°).

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To solve bearing problems, use interior angles of parallel lines (North lines are parallel) to find missing angles.

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The Sine Rule is applied when you have a known side-angle pair and one other piece of information.

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The Cosine Rule is used when you have two sides and the included angle (SAS) or all three sides (SSS).

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In 3D Trigonometry, identify right-angled triangles within the 3D shape by projecting lines onto planes.

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The angle between a line and a plane is the angle between the line and its orthogonal projection on the plane.

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3D Pythagoras theorem extends to find the diagonal of a cuboid: d2=l2+w2+h2d^2 = l^2 + w^2 + h^2.

📐Formulae

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

a2=b2+c2−2bccos⁡Aa^2 = b^2 + c^2 - 2bc \cos A (Cosine Rule for sides)

cos⁡A=b2+c2−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)

💡Examples

Problem 1:

A ship sails 10 km from port P on a bearing of 060° to point Q. It then sails 15 km from Q on a bearing of 150° 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=240∘−150∘=90∘PQR = 240^\circ - 150^\circ = 90^\circ.
  2. Since it is a right-angled triangle, use Pythagoras: PR=102+152=100+225=325≈18.03PR = \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. θ=tan⁡−1(0.5)≈26.6∘\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.