SIN	COS	TAN	Word Problems	Elevation vs Depression
5	5	5	5	5
10	10	10	10	10
15	15	15	15	15
20	20	20	20	20
25	25	25	25	25
Final Question

Sin * = ____/_____

Sin * = ____/_____

Sin * = Opposit / Hypotenus

Go to Question (1) -->

Solve for angle E

Go to Question (1) -->

Solve for angle E

Sin E = Opp / Hyp

Sin E = 34 / 50

Sin E = 0.68

Sin^-1 0.68 = 42.8^o

Go to question 2 and

solve for angle F.

Go to question 2 and

solve for angle F.

Sin F = Opp / Hyp

Sin F = 20 / 26

Sin F= 0.769

Sin ^-1 0.769 = 50.3⁰

Go to Question 7

and Solve for x.

Go to Question 7

and Solve for x.

Sin 65^o = opp / hyp

Sin 65^o = 9cm / x

x = 9cm / (sin65^o)

x = 9 cm / .906 = 9.9 cm

Go to question 9

and solve for w .

Go to question 9

and solve for w .

Sin w = opp / hyp

Sin 15^o = 3.5 m / w

w = 3.5m / sin15^o

w = 3.5 m / 0.259 = 13.5 m

Complete the following Equation :

Cos x = _____ / ______

Complete the following Equation :

Cos x = _____ / ______

Cos x = Adjacent / Hypotenus

Go to Question 3

and Solve for angle C

Go to Question 3

and Solve for angle C

Cos c = Adj / Hyp

Cos c = 14/24

Cos c = 0.583

Cos^-1 0.583 = 54.3^o

Go to question 4

and Solve for angle D

Go to question 4

and Solve for angle D

Cos d = adj / hyp

Cos d = 18 / 40

Cos d = 0.45

Cos ^-1 0.45 = 63.3^o

Go to question 8

and solve for z .

Go to question 8

and solve for z .

cos Z = ADJ / HYP

cos 20^O = 22 m / Z

z = 22m / cos 20^o

z = 22 m / 0.940 = 23.4 m

Go to question 11

and solve for t .

Go to question 11

and solve for t .

cos t = adj / hyp

cos 47^o = 23.1m / t

t = 23.1m / cos47^o

t = 23.1m / 0.682 = 33.9m

Complete the following equation :

Tan x = _______ / _______

Complete the following equation :

Tan x = _______ / _______

Tan x = Opposit / Adjacent

Go to Question 5

and Solve for angle A :

Go to Question 5

and Solve for angle A :

Tan A = Opp / Adj

Tan A = 8 /14 (Divide )

Tan A = 0.571

A = Tan ^-10.571 A = 29.7^O

Go to Question 6 and

Solve for Angle B

Go to Question 6 and

Solve for Angle B

Tan B = Opp / Adj

Tan B = 30/28

Tan B = 1.071

Tan^-1 1.071 = 47.0^o

Go to question 9

and solve for y.

Go to question 9

and solve for y.

Tan y = opp / adj

Tan 55^o = y / 3.7m

y = tan 55^o x 3.7m

y = 1.428 x 3.7m = 5.3m

Go to question 12

and solve for s.

Go to question 12

and solve for s.

tan s = opp / adj

tan 49^o = 17.2cm / s

s = 17.2cm / tan49^o

s = 17.2cm / 1.150 = 15.0 cm

Identify the following Angle :

Angle = 49 degrees

Julien is standing 27 feet from the library and looking up at the library at a 50^o. How high is the building?

tan50^o = opp / adj

tan50^o = opp / 32m

opp = tan50^o x 32m

38.1 m

Kate is flying a kite. The string of her kite is 56 m long and the kite is 41 feet from ground level. Name the length of the hypotenus.

The hypotenus is 56m long.

Julien is standing 32 metres from the library and looking up at the library at a 50^o. How high is the building?

tan50^o = opp / adj

tan50^o = opp / 32m

opp = tan50^o x 32m

38.1 m

The cruise ship left land at a 21^o angle. The ship is now 81 km from the dock and the sea captain is wondering how far south they have traveled. Draw and Solve.

sin21^o = opp / hyp

sin21^o = opp / 81km

opp = sin21^o x 81 km

opp = 29.0 km

Identify the following triangle as an angle of elevation or depression.

angle 54^o = angle of elevation

The line that runs parallel to the ground and is intersected perpendicully is called___________ .

the horizon

Identify if the following angle is an angle of elevation or of depression .

A kite is out on 300 m of line. The angle of the string to the ground (horizon) is 60^o. Assuming flat terrain, how high off the ground is the kite?

Identify if the following angle is an angle of elevation or of depression .

A kite is out on 300 m of line. The angle of the string to the ground (horizon) is 60^o. Assuming flat terrain, how high off the ground is the kite?

The angle 60^o is an angle of elevation.

Solve the following question.

A kite is out on 300 ft of line. The angle of the string to the ground (horizon) is 60^o. Assuming flat terrain, how high off the ground is the kite?

Solve the following question.

A kite is out on 300 ft of line. The angle of the string to the ground (horizon) is 60^o. Assuming flat terrain, how high off the ground is the kite?

sin x = opp / adj

sin60^o = opp / 300ft

opp = sin60^o x 300ft

opp = 259.8ft

Is the following question an angle of depression or elevation?

A pilot is at position B. Her altitude reads 4,920 m above see level. She knows the house is 72 m above sea level. If she measures the angle of ______________ of 15^o, how far from her house (dotted line) is she?

Is the following question an angle of depression or elevation?

The angle 15^o is an angle of depression.

Solve the following word problem.

A pilot is at position B. Her altitude reads 5,058 m above see level. She knows the house is 36 m above sea level. If she measures the angle of depression of 22^o, how far from her house (dotted line) is she?

Solve the following word problem.

sin = opp / hyp

sin22^o = (5,058m - 36m) / hyp

hyp = 5,022m / sin22^o

hyp = 5,022m / .375

hyp = 13,044.2m

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