Question

A plane flies 545 kilometers with a bearing of 316° from Naples to Elgin (see figure). The plane then flies 680 kilometers from Elgin to Canton (Canton is due west of Naples). Find the bearing of the flight from Elgin to Canton. (Round to the nearest whole number.)

Answer 1

Answer:

• 235°

Step-by-step explanation:

• Use of low of sines
• Consider triangle CEN
• Bearing of angle counted from zero to the vector of flight

We have

• ∠CNE = complementary with 44° ⇒ m∠CNE = 90° - 44° = 46°
• NE = b = 545 km
• EC = a = 680 km
• ∠ECN = x

Find the  measure of angle x:

• sin 46° / a = sin x / b
• sin 46° / 680 = sin x / 545
• sin x = 545 sin 46 deg / 680
• x = arcsin (545 sin 46 deg / 680)
• x = 35° (rounded)

Bearing of the flight from Elgin to Canton is:

• 270° - 35° = 235° as the vector is in the third quadrant

Answer 2

Answer:

235°

Step-by-step explanation:

Bearing: The angle (in degrees) measured clockwise from north.

The given scenario can be modeled as a triangle.

Given sides of the triangle:

• a = 680 km
• b = 545 km

As due north is at a right angle to due west, the angle in the right corner of the triangle is:

⇒ 90° - 44° = 46°

Use the Sine Rule to find angle y marked on the attached diagram (the angle opposite side b).

Sine Rule

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

(where A, B and C are the angles and a, b and c are the sides opposite the angles)

$\implies \sf \dfrac{\sin 46^{\circ}}{680}=\dfrac{\sin y}{545}$

$\implies \sf \sin y= \dfrac{545\sin 46^{\circ}}{680}$

$\implies \sf y= \sin^{-1} \left(\dfrac{545\sin 46^{\circ}}{680}\right)$

$\implies \sf y=35.20682806...^{\circ}$

Interior angles of a triangle sum to 180°

⇒ x° + y° + 46° = 180°

⇒ x° + 35.2068...° + 46° = 180°

⇒ x° = 98.79317194...°

Consecutive Interior Angles Theorem

When a straight line intersects two parallel straight lines, the consecutive interior angles formed sum to 180°.  Therefore the angle between due north and side a of the triangle is:

⇒ 180° - 44° = 136°

To find the bearing of the flight from Elgin to Canton, find the angle labelled green on the attached diagram:

$\implies \sf Bearing =136^{\circ}+ 98.79317194...^{\circ}$

$\implies \sf Bearing = 234.7931719...^{\circ}$

$\implies \sf Bearing = 235^{\circ}$

As bearings are usually given as a three-figure bearings, the bearing of the flight from Elgin to Canton is 235°.

Learn more about bearings here:

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