mathematics teachers enrichment program mtep … in right triangles in right 4abc, ... triangle add...

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Mathematics Teachers Enrichment ProgramMTEP 2012

Trigonometry and Bearings

Trigonometry in Right Triangles

In right 4ABC, AC is called the hypotenuse.

The vertices are labelled using capital letters. The sides oppo-site the vertices are labelled with the corresponding lower caseletter. In 4ABC, AB = c, BC = a, and AC = b.

A

B C

c b

a

Solving Right Triangles

When asked to solve a right triangle, you are expected to find all missing angles and allmissing side lengths. Sometimes you will only be asked to solve for specific missing infor-mation.

If you are given two sides in a right triangle, you do not require trigonometry to find thethird side. You can use Pythagoras’ Theorem. In the above triangle, a2 + c2 = b2.

If you are given two angles in a right triangle or any triangle for that matter, you do notrequire trigonometry to find the third angle. You can use the fact that the angles in atriangle add to 180◦. In the above triangle, ˆABC + ˆBAC + ˆACB = B + A+ C = 180◦.

We can use basic trigonometry in a right triangle to find other missing information. In theabove triangle:

sinA =side opposite A

hypotenuse=a

bsinC =

side opposite C

hypotenuse=c

b

cosA =side adjacent to A

hypotenuse=c

bcosC =

side adjacent to C

hypotenuse=a

b

tanA =side opposite A

side adjacent to A=a

ctanC =

side opposite C

side adjacent to C=c

a

We can use basic trigonometry in a right triangle to find a missing side provided that weknow one other side and an angle other than the right angle. We can find a missing anglein a right triangle provided we know two of the side lengths.

Two Definitions

The angle of elevation is the angle above the horizontal that an observer must up lookto see an object that is higher than the observer. The angle of depression is the anglebelow the horizontal that an observer must look down to see an object that is lower thanthe observer.

Horizontal

Look u

p to a

n obje

ct

Angle of Elevation

Horizontal

Angle of Depression

Look down to an object

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Example 1

The picture to the right shows Arch 22 in Banjul. An observerpositions himself in front of the arch, 45 m directly in front.The angle of elevation to the top of the arch from this pointis 38◦. Determine the height of the arch to the nearest metre.

Solution:

Let h represent the height of the arch. Represent the giveninformation on a diagram. Using basic trigonometry,

h

45= tan 38◦

h = 45 tan 38◦

h = 45(0.7813)

h = 35 m

The arch is approximately 35 m high.

h

45 m38

Example 2

A 5 m ladder rests against a building so that the foot of the ladder is 3.5 m out from thebottom of the wall. Determine, correct to the nearest degree, the angle that the foot ofthe ladder makes with the ground.

Solution:

Let θ represent the angle the foot of the ladder makes withthe ground. Represent the given information on a diagram.Using basic trigonometry,

cos θ =3.5

5

cos θ = 0.7

θ = 45.6◦

ladder5 mwall

ground3.5 m

The angle that the ladder makes with the ground is approximately 46◦.

2


Example 3

Two observers Sona and Yoro, 20 m apart observe a kite in the same vertical plane andfrom the same side of the kite. The angles of elevation from Sona and Yoro are 25◦ and55◦ respectively. Determine the height of the kite to the nearest metre.

Solution:

Let h represent the height of the kite and x represent thedistance from the closest observer to a point on the groundjust under the kite. Represent the given information on adiagram and label the vertices as shown.

Using basic trigonometry,

In 4KSB, tan 25◦ =h

x+ 20and in 4KY B, tan 55◦ =

h

x

h = (x+ 20) tan 25◦ and h = x tan 55◦

Since h = h, (x+ 20) tan 25◦ = x tan 55◦

x tan 25◦ + 20 tan 25◦ = x tan 55◦

20 tan 25◦ = x tan 55◦ − x tan 25◦

20 tan 25◦ = x(tan 55◦ − tan 25◦)

20 tan 25◦

(tan 55◦ − tan 25◦)= x

Then, h = x tan 55◦ =20 tan 25◦

(tan 55◦ − tan 25◦)× tan 55◦=13.8 m

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The height of the kite is approximately 14 m. We will look at this problem again a littlelater.

Solving Non-Right Triangles

An acute triangle is a triangle in which all angles are less than 90◦. An obtuse triangle isa triangle in which one angle is more than 90◦.

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When asked to solve a non-right triangle, you are expected to find all missing angles andall missing side lengths. Sometimes you will only be asked to solve for specific missinginformation. The Cosine Rule and Sine Rule are used to find unknowns in acute andobtuse triangles.

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The Cosine Rule

The Cosine Rule is used to find the length of a missing side when two sides and the anglecontained between them is given. It is also used to find missing angles when all three sidesare known.

Finding the Length of a Missing Side Finding the Size of a Missing Angle

a2 = b2 + c2 − 2bc cosA cosA =b2 + c2 − a2

2bc

b2 = a2 + c2 − 2ac cosB cosB =a2 + c2 − b2

2ac

c2 = a2 + b2 − 2ab cosC cosC =a2 + b2 − c2

2ab

The Sine Rule

The Sine Rule is used to find the length of a missing side when two angles and one sidelength are given. It is also used to find the size of a missing angle when two sides are givenand the angle opposite one of the given sides is also given.

Finding the Size of a Missing Angle Finding the Length of a Missing Side

sinA

a=

sinB

b=

sinC

c

a

sinA=

b

sinB=

c

sinC

One must be careful in using the Sine Rule to find an unknown angle. If sin θ = 0.5 thenθ = 30◦ or θ = 150◦. The following example will illustrate the caution.

Example 4

In 4ABC, AB = 8, BC = 10 and ˆACB = 30◦. Determine the size of ˆCAB.

Solution: Using the Sine Rule:

sinA

10=

sin 30◦

8

sinA =10 sin 30◦

8= 0.625

A=39◦ or A=141◦

B=111◦ or B=9◦

There are two possible values for the size of ˆCAB, 39◦ or 141◦.There are two possible triangles. They are illustrated to theright.

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Example 5 (Example 3 Revisited)

Two observers Sona and Yoro, 20 m apart observe a kite in the same vertical plane andfrom the same side of the kite. The angles of elevation from Sona and Yoro are 25◦ and55◦ respectively. Determine the height of the kite to the nearest metre.

Solution:

Let h represent the height of the kite and a represent thedistance from the closest observer to the kite. Represent thegiven information on a diagram and label the vertices as shown.

Since SY B is a straight angle,

ˆKY S = 180◦ − ˆKY B = 180◦ − 55◦ = 125◦.

The angles in a triangle sum to 180◦. Then in 4KY S,

ˆSKY = 180◦ − ˆKSY − ˆKY S = 180◦ − 25◦ − 125◦ = 30◦.

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Using the Sine Rule in 4KY S, KY

sin ˆKSY=

SY

sin ˆSKYa

sin 25◦ =20

sin 30◦

a =20 sin 25◦

sin 30◦ =16.9

Using basic trigonometry in 4KY B,h

a= sin 55◦ or h = a sin 55◦.

Substituting for a, h =

(20 sin 25◦

sin 30◦

)× sin 55◦=13.8 m.

The height of the kite is approximately 14 m. There are many ways to approach thisproblem.

Bearings

A bearing is used to represent the direction of one point relative to another point. Abearing is measured in a clockwise direction from North.

Example 6

Y is 25 km from X on a bearing of 075◦. Determine thebearing to X from Y .

Solution: Draw a representative sketch.

XA ‖ Y B since both lines are drawn to north. Therefore,ˆAXY + ˆBYX add to 180◦. It follows that ˆBYX = 105◦.

The required bearing is the reflex angle ˆBYX = 360◦−105◦ =255◦

Drawing the axes in at the points is very useful in helping tofind information.

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Example 7

On a bearing of 046◦, it is 8.1 km from Senegambia to Bakau. On a bearing of 122◦, it is10.6 km from Senegambia to Abuko.

a) Determine the direct distance from Bakau to Abuko.

b) Determine the bearing from Bakau to Abuko.

Solution:

Represent the information on a reasonable diagram.

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ˆNSB = 46◦ is the bearing from Senegambia to Bakau.ˆNSA = 122◦ is the bearing from Senegambia to Abuko.

Then ˆBSA = 122◦ − 46◦ = 76◦.

Since NS ‖MT , using the properties of parallel lines, ˆNSB = ˆTBS = 46◦.

By the Cosine Rule, |BA|2 = 8.12+10.62−2(8.1)(10.6)cos 76◦ = 136.43 and |BA| = 11.7 km.

By the Sine Rule,sin ˆSBA

10.6=

sin 76◦

11.7and sin ˆSBA = 10.6 ×

(sin 76◦

11.7

)= 0.8791. Then

ˆSBA = 62◦.

Let ˆTBA = x◦. Then x = ˆSBA− ˆTBS = 62◦ − 46◦ = 16◦.

It follows that the bearing from Bakau to Abuko is (180− x)◦ = 180◦ − 16◦ = 164◦.

Therefore the distance from Bakau to Abuko is 11.7 km on a bearing of 164◦.

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Example 8

Recently on a morning walk, John headed out on a bearing of 045◦ and walked 5 km. Atthat point he changed his direction to a bearing of 110◦ and walked a further 3 km.

a) Determine John’s distance from his original starting point.

b) Determine the bearing that John must take to walk directly to his original startingpoint.

Solution:

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The diagram shows the given information and some information which is referred to below.

Let John start at point A, go to point B and then on to point C. We are required tofind the distance from C to A and the corresponding bearing. Let d represent the distancefrom C to A.

Let ˆNAB = 45◦ represent the bearing from A to B.Let ˆPBC = 110◦ represent the bearing from B to C.

Since PBS is a straight angle, ˆSBC = 180◦ − ˆPBC = 180◦ − 110◦ = 70◦.

Using properties of parallel lines and the fact that NA ‖ BS, ˆNAB = ÂBS = 45◦. Simi-larly, since MC ‖ BS, ˆSBC = ˆMCB = 70◦.

We know that ÂBC = ÂBS + ˆSBC = 45◦ + 70◦ = 115◦.

Using the Cosine Rule, d2 = 52 + 32 − 2(5)(3) cos 115◦ = 46.68 and d = 6.83 km.

Using the Sine Rule,sin ˆBCA

5=

sin 115◦

dand sin ˆBCA = 5×

(sin 115◦

d

)=0.6633. Then

ˆBCA = 41.5◦. The bearing from C to A is calculated 360◦ − 70◦ − 41.5◦ = 248.5◦.

The distance from C to A is 6.8 km and the bearing is 249◦.

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Great and Small Circles

A great circle is a circle on the surface of a sphere that has the same diameter as thesphere, dividing the sphere into two equal hemispheres. A small circle of a sphere is thecircle constructed by a plane crossing the sphere, parallel to a great circle, but not throughthe centre of the sphere. Small circles always have a smaller radius than the sphere. Whenan orange is sliced in half, two hemispheres are created and the flat bottom of one of thehemispheres is an example of a great circle. If you make slices parallel to the bottom ofthe hemisphere, the radius of each slice would get smaller than that of the previous slice.These slices would illustrate small circles.

The Earth is generally considered to be spherical in shape with a radius of approximately6 400 km. The Earth rotates about a polar axis, a straight line through the Earth’s centre,which goes through the north and south poles.

When great circles are drawn through the north and south poles, the lines formed onthe surface of the Earth are called meridian lines or lines of longitude. The GreenwichMeridian is used as the standard from which other meridians are measured up to 180◦Wand 180◦E. The Greenwich Meridian is also known as the Prime Meridian.

When circles are drawn perpendicular to the polar axis, the circles formed on the surfaceof the Earth are called lines of latitude. The line of latitude whose centre is also the centreof the Earth is a great circle called the Equator. All other lines of latitude are small circles.These circles are called parallels of latitude and are measured in degrees north or south ofthe equator. Lines of latitude vary from 90◦N to 90◦S.

The angle formed by the radius drawn from the centre of the Earth to the Equator andthe centre of the Earth to the circumference of the small circle determines the degrees ofthe parallel of latitude. If it is north of the Equator, it is measured in ◦N and if it is southof the Equator, it is measured in ◦S.

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In the above diagram, OQ is the radius of the Earth drawn to the Equator, OA is theradius of the Earth drawn to the parallel of latitude at 42◦ S that passes through A.

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Example 9

Determine the length of the parallel of latitude 42◦S. Also determine how fast a personstanding at point A is travelling as a result of the Earth’s rotation. Assume the radius ofthe Earth is approximately 6 400 km and the circumference of the Earth is 40 000 km.

Solution:

4OPA is right angled. Let R be the radius of the Earth. Let r represent the radius ofthe small circle which makes up latitude 42◦S.

Since OQ ‖ AP , ˆQOA = ˆOAP = 42◦.

Then|AP ||OA|

= cos 42◦ and |AP | = |OA| cos 42◦. But |AP | = r and |OA| = R so

r = R cos 42◦.

In general r = R cos θwhere r is the radius of the small circle on latitude θ (N or S)and R is the radius of the Earth.

Back to our example, r = R cos 42◦ and C = 2πr where C is the circumference of thesmall circle. Therefore the the length of latitude 42◦S is C = 2πr = 2πR cos 42◦ =(2πR) cos 42◦ = 40 000 cos 42◦ = 29 700 km (three significant digits). Note that 2πR is thecircumference of the Earth.

To calculate the speed at which a person at point A is moving, we note that one completerevolution takes 24 hours. Using speed = distance÷ time = 29 700÷ 24 = 1 240 km/h.

∴ the length of the parallel of latitude 42◦S is 29 700 km and a person standing at pointA is moving at 1 240 km/h, due to the Earth’s rotation.

A Key Idea Before Proceeding

AB is an arc on the circumference of the circle with centre C, shownto the right. The angle at the centre of the circle is called a sectorangle.

The length of the arc isθ

360◦ × 2πr where r is the radius of the

circle and θ is the sector angle.

Notice that the arc length could also be thought of as the lengthsubtended by the reflex angle, 360◦ − θ. We are concerned with theshortest arc length.

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Example 10

Both Quito, Ecuador and Mbandaka, DR of Congo are located on the Equator. Quito isat 78.5◦W longitude and Mbandaka is at 18.5◦E longitude. Determine the distance fromQuito to Mbandaka along the Equator. Use 6 400 km as the radius of the Earth.

Solution:

On the diagram, G is the point where the prime meridianhits the equator, Q represents the location of Quito, Ecuadorand M represents the location of Mbandaka, DR of Congo.We are looking for the distance from Quito to Mbandakaalong the Equator.

Since Q is west of G and M is east of G, we add thetwo angles to create the sector angle. The sector angle,θ = 78.5◦ + 18.5◦ = 97.0◦. This angle is under 180◦ so wecan use it as the sector angle.

The distance from Quito to Mbandaka is

θ

360×2πr =

97

360×2π(6400)=10 800 km, to three significant figures.

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Example 11

Both Cape Town, South Africa and Mbandaka, DR of Congo lie on longitude 18.5◦E.Their latitudes are 34◦S and 0◦N respectively. Calculate the shortest distance betweenCape Town and Mbandaka. Use 40 000 km as the circumference of the Earth.

Solution:

On the diagram, C represents the location of Cape Town,South Africa and M represents the location of Mbandaka,DR of Congo. We are looking for the distance from CapeTown to Mbandaka along the great circle through the polaraxis that contains longitude 18.5◦E.

The second diagram is a cross-section which illustrates thegreat circle containing the North Pole (N), South Pole (S),Mbandaka, DR of Congo (M) and Cape Town, South Africa(C).

The sector angle is found easily since M is at the Equator.θ = 34.0◦. This angle is under 180◦ so we can use it as thesector angle.

The distance from Cape Town to Mbandaka is

θ

360×2πr =

34

360×40 000 = 3 780 km, to three significant figures.

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Example 12

Banjul, The Gambia and Mek’ele, Ethiopia lie along latitude 13.3◦N. Banjul is locatedat (13.3◦N, 16.6◦W) and Mek’ele is located at (13.3◦N, 39.6◦E). If an airplane flies alonglatitude 13.3◦N from Banjul to Mek’ele averaging 525 km/h, how long would the flighttake? Use 40 000 km as the circumference of the Earth.

Solution:

On the top diagram, the small circle containinglatitude 13.3◦N is shown. This will be helpful indetermining r, the radius of the small circle. Weknow r = R cos θ = R cos 13.3◦.

The second diagram shows the small circle withBanjul, located at B, the prime meridian passingthrough G and Mek’ele, located at M . The sectorangle is 16.6◦ + 39.6◦ = 56.2◦.

The distance along the parallel of latitude from Banjulto Mek’ele is

θ

360× 2πr =

56.2

360× 2πr

=56.2

360× 2πR cos 13.3◦

=56.2

360× (40 000) cos 13.3◦

= 6 080

Using time = distance ÷ speed, time =6080÷ 525 = 11.58 h.

The flight will take approximately 11.58 h or 11 hours and 35 minutes.

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Example 13

Barsakelmes Nature Reserve, near the Aral Sea, in Kazakhstan is located at 46.3◦N latitudeand 59.6◦E longitude. Gilford Pinchot National Forest in Washington State, USA is locatedat 46.3◦N latitude and 120.4◦W longitude.

a) Determine the distance between the nature reserve and the forest along the parallelof latitude.

b) Determine the distance between the nature reserve and the forest along a great circle.

Solution:

a) On the top diagram, the small circle con-taining latitude 46.3◦N is shown. Thetwo locations are 59.6◦ + 120.4◦ = 180◦

apart on the small circle containing latitude46.3◦N. We know r = R cos θ = R cos 46.3◦.The distance between the reserve and forest is

180

360× 2πr =

1

2× 2πR cos 46.3◦

=1

2× (40 000) cos 46.3◦

= 13 800 km. (3 significant figures)

The distance from the reserve to the forest is13 800 km along the parallel of latitude.

b) The lines of longitude are 180◦ apart. There-fore the forest and reserve are contained ona great circle passing through the North andSouth poles. The second diagram shows thelocation of the forest (F ) and the reserve(R) on the great circle. The sector angle is180◦ − 2 × 46.3◦ = 87.4◦. The distance fromF to R along the great circle is

87.4

360× 2πR =

87.4

360× 40 000=9 710 km.

The distance from the reserve to the forest is9 710 km along the great circle.

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The great circle distance between any two points on the Earth’s surface is the shortestpossible distance between the two points along the surface of the Earth.

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mathematics teachers enrichment program mtep … in right triangles in right 4abc, ... triangle add...

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