Tag: Angle of elevation

Bearing and Distance

The bearing and distance is an example of applied knowledge of trigonometry. The navigation used on land, sea or air was done using bearing (at least before GPS was invented and widely used). In this topic, we will learn how to use the bearing to calculate the distance or position of one place to another.

We will use the convention for bearing as follows:
– three-figure or three-digit
– North or N is the reference for \({\small 0^{\large{\circ}}}\)
– measured in the clockwise direction from the North

I have put together some of the questions I received in the comment section below. You can try these questions also to further your understanding on this topic.

To check your answer, you can look through the solutions that I have posted either in Youtube videos or Instagram posts.

You can subscribe, like or follow my youtube channel and IG account. I will keep updating my IG daily post, preferably.

Furthermore, you can find some examples and more practices below! =).
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QUESTIONS FROM STUDENTS:

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EXAMPLE:

\({\small 1.\enspace}\) A, B and C are three towns located on a horizontal ground. It is given that AC \(=\) 22 km and BC \(=\) 18 km. C is at a bearing of \({\small 025^{\large{\circ}}}\) from A and \({\small 337^{\large{\circ}}}\) from B.
\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Show that \({\small \angle {ACB} = 48^{\large{\circ}}}\).
\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) Calculate
\({\small\hspace{2.8em}\left(i\right).\hspace{0.7em}}\) the distance AB,
\({\small\hspace{2.8em}\left(ii\right).\hspace{0.7em}}\) the bearing of A from B,
\({\small\hspace{2.8em}\left(iii\right).\hspace{0.5em}}\) the area of triangle ABC,
\({\small\hspace{2.8em}\left(iv\right).\hspace{0.7em}}\) the shortest distance from B to AC.
\({\small\hspace{1.2em}\left(c\right).\hspace{0.8em}}\) A helicopter, H, is hovering directly above a point D, nearest to B on AC. If the angle of elevation of H seen from B is \({\small 10^{\large{\circ}}}\), calculate the height of H above D.
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Show solution:

\({\small\hspace{1.2em}\left(a\right).\enspace}\) Proof. Draw a vertical line below point C and name a point on the vertical line, M. Also, name a point north of point A as Q and a point north of point B as V.
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\(\\[5pt] \hspace{1.2em}\) We can find that,
\(\\[5pt] \hspace{1.2em}\) \({\small \angle{CBV} = \ 360^{\large{\circ}} \ – 337^{\large{\circ}} \ }\)
\(\\[10pt] {\small\hspace{4.2em} = \ 23^{\large{\circ}}}\)
\(\\[5pt] \hspace{1.2em}\) By alternate angle theorem,
\(\\[5pt] \hspace{1.2em}\) \({\small \angle{ACM} = \ \angle{QAC}}\)
\(\\[5pt] {\small\hspace{4.4em} = \ 25^{\large{\circ}}}\).
\(\\[7pt] \hspace{1.2em}\) and
\(\\[5pt] \hspace{1.2em}\) \({\small \angle{BCM} = \ \angle{VBC}}\)
\(\\[10pt] {\small\hspace{4.4em} = \ 23^{\large{\circ}}}\).
\(\\[5pt] \hspace{1.2em}\) Hence,
\(\\[5pt] \hspace{1.2em}\) \({\small \angle{ACB} = \ \angle{ACM} \ + \ \angle{BCM}}\)
\(\\[5pt] {\small\hspace{4.1em} = \ 25^{\large{\circ}} \ + \ 23^{\large{\circ}} }\)
\(\\[18pt] {\small\hspace{4.1em} = \ 48^{\large{\circ}}}\) (Q.E.D.)
\({\small \hspace{1.2em}\left(b\right).\; (i). \enspace AC = 22 \ \textrm{km}}\)
\({\small \hspace{4.7em} BC = 18 \ \textrm{km}}\)
\({\small \hspace{3.5em} \angle {ACB} = 48^{\circ}}\)
\(\\[10pt]{\small \hspace{4.7em} AB =\; ?}\)
\(\\[12pt]\hspace{1.2em}\) By using Cosine Rule,
\(\\[7pt]\hspace{1.9em} {\small {AB}^2 = {AC}^2 + {BC}^2 – 2. \ {AC} . \ {BC} \cos \angle {ACB}}\)
\(\\[7pt]{\small\hspace{2.1em} AB\: = \, \sqrt{ 22^2 + 18^2 – 2 \times 22 \times 18 \times \cos 48^{\large{\circ}}}}\)
\(\\[15pt] \hspace{3.8em}{\small \approx \:16.67 \ \mathrm{km}}\)
\(\\[10pt] {\small \hspace{1.2em}\left(b\right).\; (ii). \enspace}\) The bearing of A from B \({\small =\: ?}\)
\(\\[12pt] \hspace{1.2em}\) Find \({\small \angle CAB}\) by using \({\small Sine \ Rule}\).
\(\\[10pt] \hspace{2.8em}{\large\frac{\sin \angle CAB}{BC} = \frac{\sin \angle ACB}{AB}}\)
\(\\[10pt] {\small\hspace{3.5em} \sin \angle CAB} = {\large\frac{BC \ \times \ \sin \angle ACB}{AB}}\)
\(\\[10pt] \hspace{7.5em} = {\large\frac{18 \ \mathrm{km} \ \times \ \sin 48^{\LARGE{\circ}}}{16.67 \ \mathrm{km}}}\)
\(\\[10pt] \hspace{2.6em} {\small \angle CAB\: \approx \: \sin^{-1}{(0.80)}}\)
\(\\[15pt] \hspace{5.5em} {\small\approx 53^{\large{\circ}}}\) or \({\small(180^{\large{\circ}} – 53^{\large{\circ}}) = 127^{\large{\circ}}}\)
\(\hspace{1.2em}\) Note that 127\(^{\circ}\) is not a solution for \(\angle\)CAB since it must be an acute angle.
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\(\\[5pt] \hspace{1.2em}\) Hence, the bearing of A from B
\(\\[5pt]\hspace{3.1em} {\small = \ \angle {QAC} \ + \ \angle {CAB}}\)
\(\\[5pt] {\small\hspace{3.1em} = \ 25^{\large{\circ}} \ + \ 53^{\large{\circ}} }\)
\(\\[18pt] {\small\hspace{3.1em} = \ 78^{\large{\circ}}}\)
\(\\[10pt] {\small \hspace{1.2em}\left(b\right).\; (iii). \enspace}\) The area of \(\triangle\)ABC
\(\\[10pt] \hspace{4.8em} = \frac{1}{2}ba \sin C\)
\(\\[10pt] \hspace{4.8em} = \frac{1}{2} \times 22 \times 18 \times \sin 48^{\circ}\)
\(\\[18pt]\hspace{4.8em} {\small \approx \:147.14 \,\mathrm{km^2}}\)
\({\small \hspace{1.2em}\left(b\right).\; (iv). \enspace}\) The shortest distance from B to AC \({\small =\; ?}\)
\( \hspace{1.2em}\) Let the foot of perpendicular distance from B to AC \({\small =\ D}\)
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\(\\[10pt]{\small \hspace{2em} BD =\; ?}\)
\(\\[15pt] \hspace{3em} {\small \sin \angle {ACB}} = \ {\large\frac{BD}{BC}}\)
\(\\[15pt] \hspace{4.2em} {\small \sin 48^{{\large \circ}}} = \ {\large\frac{BD}{18}}\)
\(\\[10pt]{\small\hspace{3.8em} BD \ = \ \sin 48^{{\large \circ}} \times 18 \ \mathrm{km}}\)
\(\\[18pt]{\small\hspace{5.5em} \approx \ 13.38\,\mathrm{km}} \)
\({\small\hspace{1.2em}\left(c\right).\enspace HD =\; ?}\)
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\(\\[15pt] \hspace{3em} {\small \tan \angle {HBD}} = \ {\large\frac{HD}{BD}}\)
\(\\[15pt] \hspace{4.3em} {\small \tan 10^{{\large \circ}}} = \ {\large\frac{HD}{13.38}}\)
\(\\[10pt]{\small\hspace{3.8em} HD \ = \ \tan 10^{{\large \circ}} \times 13.38 \ \mathrm{km}}\)
\({\small\hspace{5.6em} \approx \ 2.36\,\mathrm{km}} \)

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\({\small 2.\enspace}\) A ship leaves port at noon and has a bearing of \({\small 207^{\large{\circ}}}\). If the ship is traveling at 20 miles per hour, how many miles south and how many miles west is the ship from its departure point at 6 pm?

Show solution:

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\(\\[5pt] \hspace{1.2em}\) The traveling time \(=\) 6 pm – noon
\(\\[15pt] \hspace{8.1em}\) \(=\) 6 hours
\(\\[5pt] \hspace{1.2em}\) The distance traveled \(=\) speed \(\times\) time
\(\\[5pt] \hspace{9.4em}\) \(=\) 20 mph \(\times\) 6 h
\(\\[15pt]\hspace{9.4em}\) \(=\) 120 miles
\(\hspace{1.2em}\) The distance in the south direction is the vertical component of the right-angled triangle.
\(\\[5pt] \hspace{5em}\) \(=\) 120 \({\small \times \ \cos 27^{\large{\circ}}}\)
\(\\[15pt]\hspace{5em}\) \({\small \approx \ }\) 106.92 miles
\(\hspace{1.2em}\) The distance in the west direction is the horizontal component of the right-angled triangle.
\(\\[5pt] \hspace{5em}\) \(=\) 120 \({\small \times \ \sin 27^{\large{\circ}}}\)
\( \hspace{5em}\) \({\small \approx \ }\) 54.48 miles

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\({\small 3.\enspace}\) From the lookout tower A, a column of smoke is sighted due south. From a second lookout tower B, 5 miles west of A, the smoke is observed with a bearing of \({\small 117^{\large{\circ}}}\). How far is the fire from tower B? How far is it from tower A?

Show solution:

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\(\\[5pt] \hspace{1.2em}\) The distance of the smoke from tower B,
\(\\[15pt] \hspace{2.1em}\) \(= \ {\large \frac{5}{\cos 27^{\large{\circ}}}}\)
\(\\[15pt] \hspace{2.1em}\) \(\approx \ \) 5.61 miles
\(\\[5pt] \hspace{1.2em}\) The distance of the smoke from tower A,
\(\\[5pt] \hspace{2.1em}\) \(= \ \) 5 \({\small \times \ \tan 27^{\large{\circ}}}\)
\( \hspace{2.1em}\) \(\approx \ \) 2.55 miles

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\({\small 4.\enspace}\) Pittsburgh is 117 miles due south of Kansas City. A plane leaves Kansas City at noon and flies 200 mph with a bearing of \({\small 201^{\large{\circ}}}\). When, to the nearest minute, will the plane be due west of Pittsburgh?

Show solution:

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\( \hspace{1.2em}\) The distance of the airplane from Pittsburgh,
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\(\\[15pt] \hspace{2.1em}\) \(= \ {\large \frac{117}{\cos 21^{\large{\circ}}}}\)
\(\\[15pt] \hspace{2.1em}\) \(\approx \ \) 125.32 miles
\(\hspace{1.2em}\) The time of the airplane due west of Pittsburgh,
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\(\\[15pt] \hspace{2.1em}\) \(= \ {\large \frac{\textrm{distance}}{\textrm{speed}}}\)
\(\\[15pt] \hspace{2.1em}\) \(= \ {\large \frac{125.32 \ \textrm{miles}}{200 \ \textrm{miles per hour}}}\)
\( \hspace{2.1em}\) \(\approx \ \) 38 minutes

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\({\small 5.\enspace}\) Two ships A and B leave port at the same time with ship A sailing with a bearing of \({\small 023^{\large{\circ}}}\) at the speed of 11 mph and ship B sailing with a bearing of \({\small 113^{\large{\circ}}}\) at 15 mph. Approximate the bearing from ship B to ship A one hour later.

Show solution:

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\( \hspace{1.2em}\) In 1 hour, the distance traveled by ship A \(=\) 11 miles and by ship B \(=\) 15 miles.
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\( \hspace{1.2em}\) The angle between ship A, port and ship B,
\(\\[5pt]\hspace{3.1em} {\small = \ 113^{\large{\circ}} \ – \ 23^{\large{\circ}} }\)
\(\\[15pt] {\small\hspace{3.1em} = \ 90^{\large{\circ}}}\)
\( \hspace{1.2em}\) Since we have a right-angled triangle, we can use tangent function to quickly find \({\small \angle B}\). And then we can add up all of the angles to get the bearing.
\( \hspace{1.2em}\) The bearing of B from A,
\(\\[5pt]\hspace{3.1em} {\small = \ 113^{\large{\circ}} \ + \ 180^{\large{\circ}} \ + \ \tan^{-1}{({\large \frac{11}{15}})} }\)
\( {\small\hspace{3.1em} \approx \ 329^{\large{\circ}}} \)

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PRACTICE MORE WITH THESE QUESTIONS BELOW!

\({\small 1.\enspace}\) Two ships, A and B leave port at 13 00 hours. Ship A travels at a constant speed of 18 km per hour on a bearing of \({\small 070^{\large{\circ}}}\). Ship B travels at a constant speed of 25 km per hour on a bearing of \({\small 152^{\large{\circ}}}\). Calculate the distance between A and B at 15 00 hours.
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\({\small 2. \enspace}\) A plane is 120 miles north and 85 miles east of an airport. If the pilot wants to fly directly to the airport, what bearing should be taken?

\({\small 3. \enspace}\) Ohio is 190 miles due north of Virginia. Stamford is due east of Ohio and is 460 miles from Virginia. What is the bearing angle heading from Virginia to Stamford? What is the bearing angle from Stamford to Virginia?

\({\small 4. \enspace}\) D is 30 km due east of A. A ship sails from A on a certain bearing for 15 km to B. It then sails to C, which is 40 km from A and due north of D. Finally it sails due south to D. AB is the bisector of \({\small \angle {EAC}}\). Calculate the total distance for the whole journey.
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\({\small 5. \enspace}\) An observer at a radar station plots an aircraft due north of him and at a range of 50 km, measured horizontally. The aircraft keeps on traveling due west at a constant height. The bearing of the plane from the observer 10 minutes later is \({\small 300^{\large{\circ}}}\). Calculate the speed of the aircraft in km/h.

\({\small 6. \enspace}\) N is north of S. NS is the tangent to the circle, radius 1 km, at T. A man walks 2 km along the circle from T to P. What is the distance and bearing of P from T?
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\({\small 7. \enspace}\) In the diagram, A, B, C and D are four points on a horizontal field. A is north of B. The bearing of D from A is \({\small 115^{\large{\circ}}}\) and the bearing of D from B is \({\small 25^{\large{\circ}}}\). \({\small \angle {BDC} = 130^{\large{\circ}}}\), BD \(=\) 150 m and CD \(=\) 110 m.
\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Find the value of \({\small \angle {ADB}}\).
\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) Find the bearing of B from D.
\({\small\hspace{1.2em}\left(c\right).\hspace{0.8em}}\) Find the bearing of C from D.
\({\small\hspace{1.2em}\left(d\right).\hspace{0.8em}}\) Calculate the shortest distance of point A to line DB.
\({\small\hspace{1.2em}\left(e\right).\hspace{0.8em}}\) Calculate the length of BC.
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\({\small 8. \enspace}\) In the diagram (not drawn to scale), E, G and C represents three points at the Esplanade, Gardens by the Bay and the City Hall respectively. Given that EG \(=\) 1125 m and CE \(=\) 957 m, the bearing of G from E is \({\small 135^{\large{\circ}}}\) and \({\small \angle {CEG} = 60^{\large{\circ}}}\). Find
\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) the bearing of E from C,
\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) the distance between C and G,
\({\small\hspace{1.2em}\left(c\right).\hspace{0.8em}}\) \({\small \angle {EGC}}\),
\({\small\hspace{1.2em}\left(d\right).\hspace{0.8em}}\) the bearing of C from G.
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\({\small 9. \enspace}\) In the diagram, Q is due east of P. Given that PQ \(=\) 55 m and RQ \(=\) 32 m and \({\small \angle {PRQ} = 105^{\large{\circ}}}\). Find
\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) The bearing of R from P,
\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) The area of \({\small \triangle {PRQ}}\).
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\({\small 10.\enspace}\) In the diagram, A, B, C and D are four corners of a field. AD \(=\) 53 m, DC \(=\) 87 m, BC \(=\) 66 m, \({\small \angle {ABC} = 90^{\large{\circ}}}\) and \({\small \angle {ACB} = 58^{\large{\circ}}}\).
\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Calculate AC.
\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) The bearing of B from C is \({\small 340^{\large{\circ}}}\). Calculate
\({\small\hspace{2.8em}\left(i\right).\hspace{0.7em}}\) the bearing of A from C,
\({\small\hspace{2.8em}\left(ii\right).\hspace{0.7em}}\) the bearing of A from B.
\({\small\hspace{1.2em}\left(c\right).\hspace{0.8em}}\) Calculate
\({\small\hspace{2.8em}\left(i\right).\hspace{0.7em}}\) the angle of \({\small \angle {ADC}}\),
\({\small\hspace{2.8em}\left(ii\right).\hspace{0.7em}}\) the area of \({\small \triangle {ADC}}\),
\({\small\hspace{2.8em}\left(iii\right).\hspace{0.5em}}\) the shortest distance from A to DC.
\({\small\hspace{1.2em}\left(d\right).\hspace{0.8em}}\) An eagle is hovering vertically above A. The angle of elevation of the eagle from D is \({\small 42^{\large{\circ}}}\). Calculate the greatest angle of elevation of the eagle from a point on DC.
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As always, if you have any particular questions to discuss, leave it in the comment section below. Cheers =) .