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Right Triangle in Real-Life

An Application to Right Triangle Trigonometry

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Introduction• A triangle has three sides and three

angles. If three parts of a triangle, at least one of which is side, are given then other three parts can be uniquely determined.

• Finding other unknown parts, when three parts are known is called ‘solution of triangle’.

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Example:

Fasten your seatbelts A small plane takes off from an airport and rises uniformly at an angle of 4°30’ with the horizontal ground. After it has traveled over a horizontal distance of 600m, what is the altitude of the plane to the nearest meter?

x

600m

4°30’

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Solution:

Let x = the altitude of the plane as it travels 600m horizontallySince we have the values of an acute angle and its adjacent side, we will use x

600m

4°30’

4°30’

600m

x

_ _ _tan

_ _ _

opposite side of

adjacent side of

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Let us solve

tan 4 30 '600

x

m

600 (tan 4 30 ')

600 (0.0787)

47.22 47

x m

x m

x m m

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Let us solve

Answer: The altitude of the plane after it has traveled over a horizontal distance of 600m is 47m.

tan 4 30 '600

x

m

600 (tan 4 30 ')

600 (0.0787)

47.22 47

x m

x m

x m m

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Let us solve some problems

Sail away

A ship sailed from a port with a bearing of S22°E. How far south has the ship traveled after covering a distance of 327km?

x327km

22°

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__x__ 327 km

x=327km (cos 22°)

x=327km (0.927)

x=303.18 km

cos 22° =

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Good Morning

From the tip of a shadow by the vertical object such as a tree, the angle of elevation of the top of the object is the same as the angle of elevation of the sun. What is the angle of elevation of the sun if a 7m tall tree casts a shadow of 18m?

Θ7m

18m

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tanA= oppadj

tanA= 718

The measure of the angle of elevation of the sun is 21° 15’

7 18

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Happy Landing

A plane is flying at an altitude of 1.5km. The pilot wants to descend into an airport so that the path of the plane makes an angle of 5° with the ground. How far from the airport (horizontal distance) should the descent begin?

1.5km5°

x

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Solve for the other leg; call it "x"

EQUATION: -----------

tan5°= __x__ 1.5 km

x= 1.5km (tan 5°)

x= 1.5km (0.00878)

x= 17.1450 km

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A ladder is placed 5 feet from the foot of the wall. The top of the ladder reaches a point 12 feet above the ground. Find the length of the ladder

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Solution: The ladder, the wall and the ground from a right triangle. The length of the ladder is c, the length of the hypotenuse of the right triangle. The distance from the foot of the ladder to the wall is a=5, and the distance from the ground to the top of the ladder is b= 12. Use the Theorem of Pythagoras

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Solve for the triangle given: γ=90° α=57°20’ b=100

α57°20’

y=90° ᵝ=?

b= 100

C=?

a=?

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To get ᵝ

57°20’ + 90° + ᵝ = 180

ᵝ = 32° 40’

α + y + ᵝ

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To get side a:

Tan 32°40’ = 100 a

a tan 32°40’ = 100tan 32°40 tan 32°40

a=

a= 155.97

100tan 32°40

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To get c:sin c = a c

Sin 57°20’ = 155.97 c

c= 155.97sin 57°20’

c= 185.27

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α57°20’

y=90° ᵝ=32° 40’

b= 100

C=185.27

a= 155.97

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Edgardo S. Mata&

Ronaldo O. Abinal Jr.

Eng.Raymond Tandaan

Subj.Teacher