# Problems on Trigonometric Identities | Word Problems Involving Trigonometric Identities

Check problems on trigonometric identities along with the solutions. Find the step by step procedure to know the trigonometric identities problems. Refer to all the solutions present in the below sections to prepare for the exam. Scroll down the page to get the Trigonometric Identities Word Problems and study material. Know the various formulae involved in solving trigonometric identities below. Assess your knowledge level taking the help of the Practice Problems on Trigonometric Identities available.

Problem 1:

The angle of inclination from a point on the ground 30 feet away to the top of Lakeland’s Armington Clocktower is 60Â°. Find the height of the clock tower which is nearest to the foot?

Solution:

As given in the question,

The angle of inclination (Î¸) = 60Â°

The height from the ground(a) = 30 feet

To find the height of the Clocktower to the nearest foot, we use the formula

tan Î¸ = h/a

tan 60Â°Â = h / 30

h = 30 tan 60Â°

h = 9.60 â‰… 10 feet

Therefore, the height of the clocktower to the nearest foot is 10 feet.

Hence, the final solution is 10 feet.

Problem 2:

Mary wants to determine the California redwood tree height, there are two sightings available from the ground which one is 200 feet directly behind the other. If the angles of inclination (Î˜) are 45Â° and 30Â° respectively, how tall is the tree to the nearest foot?

Solution:

As given in the question,

Length at which trees are slighting = 200

The angle of inclinations = 45Â° and 30Â°

To find the inclination on the first tree, we apply the Pythagorean theorem,

tan 45Â° = h/x

h = x tan 45Â° is the (1) equation

tan 30Â° = h/(200+x)

h = (200 + x) tan 30 is the (2) equation

From both the equations,

x tan 45Â° = (200 + x) tan 30Â°

x tan 45Â° = 200 tan 30Â° + x tan 30Â°

x tan 45Â° – x tan 30Â° = 200 tan 30

Divide the equation with tan 45Â° – tan 30Â°

x (tan 45Â° – tan 30Â°) / tan 45Â° – tan 30Â° = 200 tan 30Â° / tan 45Â° – tan 30Â°

x = 273.21 feet

h = 273.21 tan 45

h = 273. 21 feet

h â‰… 273 feet

Therefore, the height of the tree to the nearest foot = 273 feet

Thus, the final solution = 273 feet

Problem 3:

A tree that is standing vertically on the level ground casts the 120 foot long shadow. The angle of elevation from the end of the shadow of the top of the tree is 21.4Â°. Find the height of the tree to the nearest foot?

Solution:

As given in the question,

Length of the foot-long shadow = 120

The inclination of the tree = 21.4Â°

To find the height of the tree to the nearest foot, we apply the Pythagorean theorem

tan Î¸ = 0/a

tan 214Â° = h/120

h = 120 tan 214Â°

h = 47.03

h â‰… 47 feet

The height of the tree to the nearest foot = 47 feet

Thus, the final solution is 47 feet

Problem 4:

The broadcast tower which is for the radio station WSAZ (“Carl” and “Jeff”‘s home of algebra) has 2 enormous flashing red lights on it. Of the 2 enormous flashing lights, one is at the very top and the other one is few feet below the top. From that point to the base of the tower it is 5000 feet away from level ground, the top light angle of elevation is 7.970Â° and the light angle of elevation of the second light is 7.125Â°. Find the distance between the nearest foot and the lights.

Solution:

As given in the question,

Height of the tower = 5000 feet

The angle of elevation of top light = 7.970

The angle of elevation of second light = 7.125Â°

To find the distance between the nearest foot and the lights, we have to use the Pythagorean theorem

tan Î¸ = h/5000

h = 5000 tan 7.97Â°

tan Î² = h-x/5000

h-x = 5000 tan (7.125)Â°

x = h – 5000 tan (7.125)Â°

x = 5000 tan 7.97Â° – 5000 tan 7.125Â°

x = 5000 (tan 7.97Â° – tan 7.125Â°)

x = 75.04 feet

x â‰… 75Â  feet

Therefore, 75 feet is the distance between the nearest foot and the lights

Thus, the final solution is 75 feet.

Problem 5:

Find the solution of tan (Î¸) = sin (Î¸) sec (Î¸)?

Solution:

sinÎ¸/ cosÎ¸ = sinÎ¸ secÎ¸

sinÎ¸. (1/cosÎ¸) = sinÎ¸ secÎ¸

sinÎ¸ secÎ¸ = sinÎ¸ secÎ¸

tanÎ¸ = sinÎ¸ secÎ¸

= sinÎ¸(1/cosÎ¸)

= sinÎ¸/cosÎ¸

= tanÎ¸

âˆ´ Hence it is proved

Problem 6:

Prove that (sec(Î¸)-tan(Î¸))(sec(Î¸)+tan(Î¸))=1

Solution:

secÂ²(Î¸)-tanÂ²(Î¸)=1

1/cosÂ²(Î¸)-sinÂ²(Î¸)/cos(Î¸)=1

(1-sinÂ²(Î¸))/cosÂ²(Î¸)=1

cosÂ²(Î¸)/cosÂ²(Î¸)=1

(1-sinÂ²(Î¸))/cosÂ²(Î¸)

1/cosÂ²(Î¸)-sinÂ²(Î¸)/cosÂ²(Î¸)=secÂ²(Î¸)-tanÂ²(Î¸)

(sec(Î¸)-tan(Î¸))(sec(Î¸)+tan(Î¸))

Therefore, (sec(Î¸)-tan(Î¸))(sec(Î¸)+tan(Î¸)) = 1

âˆ´Hence, it is proved

Problem 7:

Prove that sec(Î¸)/(1-tan(Î¸))=1/(cos(Î¸)-sin(Î¸))

Solution:

1/((cos(Î¸)-sin(Î¸)).1/cos(Î¸).1/cos(Î¸)

(1/cos(Î¸))/((cos(Î¸)-sin(Î¸))-(cos(Î¸))=sec(Î¸)/(1-tan(Î¸))

Therefore, sec(Î¸)/(1-tan(Î¸))=1/(cos(Î¸)-sin(Î¸))

âˆ´Hence, it is proved

Problem 8 :

An aeroplane over the Pacific sights an atoll at a 20Â° angle of depression. If the plane is 435 ma above water, how many kilometres is it from a point 435m directly above the centre of a troll?

Solution:

As given in the question,

The angle of depression = 20Â°

Height of the plane above water = 435ma

Height above the centre of a troll = 435m

To find the kilometers, we use the pythegorean theorem

tan = Î¸/A

tan 20Â° = 435/x

x = 435/tan20Â°

x = 1.195 km

Therefore, 1.195 kilometres is it from a point 435m directly above the centre of a troll

Thus, the final solution is 1.195 kilometres

Problem 9:

The force F (in pounds)on the back of a person when he or she bends over an acute angle Î¸ (in degrees) is given by F = 0.2W sin(Î¸ + 70)/sin12Â° where w is the weight in pounds of the person

a) Simplify the formula or F.

b) Find the force on the back of a person where an angle of 30Â° weight is 50 pounds if he bends an angle of 30Â°

c) How many pounds should a person weigh for his book to endure a force of 400 lbs if he bends 40Â°?

Solution:Â

a. F = 0.2W sin (Î¸ + 90)/sin 12Â°

= 0.2W [sinÎ¸ cos 90 + cosÎ¸ sin90]/sin 12Â°

= 0.2W [Î¸(sinÎ¸) + (cosÎ¸) (1)]/sin 12Â°

= 0.2W [0 + cosÎ¸]/sin 12Â°

F = 0.2W(cosÎ¸)/sin 12Â°

The value of F is 0.2W(cosÎ¸)/sin 12Â°

b. W = 50, Î¸ = 30Â°, F=?

F = 0.2W cosÎ¸/sin 12Â°

F = 0.2 (50) (cos30Â°)/sin 12Â°

F = 41.45

The force on the back of a person wherein the angle of 30e weight is 50 pounds if he bends an angle of 30Â° is 41.45

c. F = 400, Î¸ = 40Â°, W = ?

400 = 0.2(w)(cos 40Â°)/sin 12Â°

400(sin 12Â°)/0.2 cos 40 = 0.2 (w) (cos 40Â°)/0.2 cos40Â°

542.82 = w

Therefore, the weigh for his book to endure a force of 400 lbs if he bends 40Â° is 542.82 pounds

Problem 10:

An observer standing on the top of vertical cliff pots a house in the adjacent valley at an angle of depression of 12Â°. The cliff is 60m tall. How far is the house from the base of the cliff?

Solution:

As given in the question,

The angle of depression = 12Â°

Height of the cliff = 60m

To find, the distance of the house from the base of the cliff, we apply the Pythagorean theorem

tan 12Â° = 60/x

x = 60/tan 12Â°

x = 282m

282m is the distance of the house from the base of the cliff

Hence, the final solution is 282m

Problem 11:

Building A and B are across the street from each other which is 35m apart. From the point on the roof of building A, the angle of elevation of the top of building B is 24Â°, the angle of depression of the base of building B is 34Â° How tall is each building?

Solution:

As given in the question,

The angle of elevation of the top of the building = 24Â°

The angle of depression of the base of the building = 34Â°

The distance of both buildings = 35m

To find the height of each building, we apply the Pythagoras theorem,

tan 24Â° = c/35

c = 15.6

tan 56Â° = 35/a

a = 23.6m

b = a+c

b = 39.2m

A is 23.6m tall

B is 39.2m tall

Therefore, the height of building A is 23.6m tall

The height of building B is 39.2m tall

Thus the final solution is 23.6m, 39.2m

Problem 12:

In Johannesburg in June, the daily low temperature is usually around 3Â°C, and the daily temperature is around 18Â°. The temperature is typically halfway between the daily high and daily low at 10 am and 10 pm. and the highest temperatures are in the afternoon. Find out the trigonometric function which models the temperature T in Johannesburg t hours after midnight?

Solution:

As per the question,

To determine the trigonometric model, the temperature ‘T’ which is Celsius degree and the temperature in axis and then right over here is time in hours. To think about the range of temperatures, the daily temperature is around 3-degree celsius and the highest is 18Â°. The midpoint between 18 and 3 is 10.5 (21 divided by 2)

Let F(t) is the temperature t hours after 10 am

F(t) = 7.5sin(2Î /24 t) + 10.5

T (t) = 7.5sin (Î /12(t-10)) + 10.5

T(10) = F(0) where T(10) is temperature at 10 pm

F(0) is the temperature at 10 am

Therefore, T(10) = F(0) is the trigonometric function that models the temperature T in Johannesburg t hours after midnight

Problem 13:

A ladder is 6 meters long and reaches the wall at a point of 5m from the ground. What is the angle which the ladder will make with the wall?

Solution:

Let Î¸ be the inclination of the ladder which it makes with the wall

As given in the question,

Length of the ladder = 6m

The distance at which the ladder touches the wall = 5m

The angle at which the ladder will make with the wall

cos Î¸ = 5/6

Î¸ = cosÂ¯Â¹ (5/6)

Î¸ = 33.56Â°

Problem 14:

The acceleration of the piston is given by a = 5.0(sinÏ‰t + cos2Ï‰t). At what positive values of Ï‰t less than 2Î  does a = 0?

Solution:

Let Ï‰t be the crank angle(product of time and angular velocity) in piston acceleration, a be the accelaration of a piston.

a = 5.0(sinÏ‰t + cos2Ï‰t)

0 = 5.0(sinÏ‰t + cos2Ï‰t)

0 = sinÏ‰t + cos2Ï‰t

0 = sinÏ‰t + 1 – 2sinÂ²Ï‰t

2sinÂ²Ï‰t – sinÏ‰t – 1 = 0

(2sinÏ‰t + 1) (sinÏ‰t – 1) = 0

2sinÏ‰t + 1 = 0

2sinÏ‰t = -1

sinÏ‰t = -1/2

Ï‰t = sinÂ¯Â¹(-1/2)

Ï‰t = (7Ï€/6, 11Ï€/6)

sinÏ‰t – 1 = 0

sinÏ‰t = 1

Ï‰t = sinÂ¯Â¹(1)

Ï‰t = Ï€/2

{Ï€/2,Â 7Ï€/6, 11Ï€/6}

Problem 15:

A lighthouse at sea level is 34 mi from a boat. It is known that the top of the lighthouse is 42.5mi from the boat. Find the angle of depression from the top of the lighthouse.

Solution:

Let Î¸ be the angle of inclination from the top of the lighthouse to the boat

Let x be the horizontal distance from the base of the lighthouse to the boat, and

r be the distance from the top of the lighthouse to the boat.

As given in the question,

The length of the lighthouse at sea level = 34 mi

The distance of the top lighthouse from the boat = 42.5 mi

To find the angle of depression from the top of the lighthouse, we apply the Pythagorean theorem

cosÎ¸ = 34/42.5

Î¸ = cosÂ¯Â¹(34/42.5)

Î¸ = 36.87Â°

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