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.
Also, Read:
Problem 1:
The angle of inclination from a point on the ground 30 feet away to the top of Lakeland’s Armington Clocktower is 60°. Find the height of the clock tower which is nearest to the foot?
Solution:
As given in the question,
The angle of inclination (θ) = 60°
The height from the ground(a) = 30 feet
To find the height of the Clocktower to the nearest foot, we use the formula
tan θ = h/a
tan 60° = h / 30
h = 30 tan 60°
h = 9.60 ≅ 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 (Θ) are 45° and 30° 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 = 45° and 30°
To find the inclination on the first tree, we apply the Pythagorean theorem,
tan 45° = h/x
h = x tan 45° is the (1) equation
tan 30° = h/(200+x)
h = (200 + x) tan 30 is the (2) equation
From both the equations,
x tan 45° = (200 + x) tan 30°
x tan 45° = 200 tan 30° + x tan 30°
x tan 45° – x tan 30° = 200 tan 30
Divide the equation with tan 45° – tan 30°
x (tan 45° – tan 30°) / tan 45° – tan 30° = 200 tan 30° / tan 45° – tan 30°
x = 273.21 feet
h = 273.21 tan 45
h = 273. 21 feet
h ≅ 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 21.4°. 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 = 21.4°
To find the height of the tree to the nearest foot, we apply the Pythagorean theorem
tan θ = 0/a
tan 214° = h/120
h = 120 tan 214°
h = 47.03
h ≅ 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 7.970° and the light angle of elevation of the second light is 7.125°. 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 = 7.125°
To find the distance between the nearest foot and the lights, we have to use the Pythagorean theorem
tan θ = h/5000
h = 5000 tan 7.97°
tan β = h-x/5000
h-x = 5000 tan (7.125)°
x = h – 5000 tan (7.125)°
x = 5000 tan 7.97° – 5000 tan 7.125°
x = 5000 (tan 7.97° – tan 7.125°)
x = 75.04 feet
x ≅ 75 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 (θ) = sin (θ) sec (θ)?
Solution:
sinθ/ cosθ = sinθ secθ
sinθ. (1/cosθ) = sinθ secθ
sinθ secθ = sinθ secθ
tanθ = sinθ secθ
= sinθ(1/cosθ)
= sinθ/cosθ
= tanθ
∴ Hence it is proved
Problem 6:
Prove that (sec(θ)-tan(θ))(sec(θ)+tan(θ))=1
Solution:
sec²(θ)-tan²(θ)=1
1/cos²(θ)-sin²(θ)/cos(θ)=1
(1-sin²(θ))/cos²(θ)=1
cos²(θ)/cos²(θ)=1
(1-sin²(θ))/cos²(θ)
1/cos²(θ)-sin²(θ)/cos²(θ)=sec²(θ)-tan²(θ)
(sec(θ)-tan(θ))(sec(θ)+tan(θ))
Therefore, (sec(θ)-tan(θ))(sec(θ)+tan(θ)) = 1
∴Hence, it is proved
Problem 7:
Prove that sec(θ)/(1-tan(θ))=1/(cos(θ)-sin(θ))
Solution:
1/((cos(θ)-sin(θ)).1/cos(θ).1/cos(θ)
(1/cos(θ))/((cos(θ)-sin(θ))-(cos(θ))=sec(θ)/(1-tan(θ))
Therefore, sec(θ)/(1-tan(θ))=1/(cos(θ)-sin(θ))
∴Hence, it is proved
Problem 8 :
An aeroplane over the Pacific sights an atoll at a 20° 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 = 20°
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 = θ/A
tan 20° = 435/x
x = 435/tan20°
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 θ (in degrees) is given by F = 0.2W sin(θ + 70)/sin12° 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 30° weight is 50 pounds if he bends an angle of 30°
c) How many pounds should a person weigh for his book to endure a force of 400 lbs if he bends 40°?
Solution:
a. F = 0.2W sin (θ + 90)/sin 12°
= 0.2W [sinθ cos 90 + cosθ sin90]/sin 12°
= 0.2W [θ(sinθ) + (cosθ) (1)]/sin 12°
= 0.2W [0 + cosθ]/sin 12°
F = 0.2W(cosθ)/sin 12°
The value of F is 0.2W(cosθ)/sin 12°
b. W = 50, θ = 30°, F=?
F = 0.2W cosθ/sin 12°
F = 0.2 (50) (cos30°)/sin 12°
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 30° is 41.45
c. F = 400, θ = 40°, W = ?
400 = 0.2(w)(cos 40°)/sin 12°
400(sin 12°)/0.2 cos 40 = 0.2 (w) (cos 40°)/0.2 cos40°
542.82 = w
Therefore, the weigh for his book to endure a force of 400 lbs if he bends 40° 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 12°. 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 = 12°
Height of the cliff = 60m
To find, the distance of the house from the base of the cliff, we apply the Pythagorean theorem
tan 12° = 60/x
x = 60/tan 12°
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 24°, the angle of depression of the base of building B is 34° How tall is each building?
Solution:
As given in the question,
The angle of elevation of the top of the building = 24°
The angle of depression of the base of the building = 34°
The distance of both buildings = 35m
To find the height of each building, we apply the Pythagoras theorem,
tan 24° = c/35
c = 15.6
tan 56° = 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 3°C, and the daily temperature is around 18°. 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 18°. 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(2Π/24 t) + 10.5
T (t) = 7.5sin (Π/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 θ 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 θ = 5/6
θ = cos¯¹ (5/6)
θ = 33.56°
Problem 14:
The acceleration of the piston is given by a = 5.0(sinωt + cos2ωt). At what positive values of ωt less than 2Π does a = 0?
Solution:
Let ωt be the crank angle(product of time and angular velocity) in piston acceleration, a be the accelaration of a piston.
a = 5.0(sinωt + cos2ωt)
0 = 5.0(sinωt + cos2ωt)
0 = sinωt + cos2ωt
0 = sinωt + 1 – 2sin²ωt
2sin²ωt – sinωt – 1 = 0
(2sinωt + 1) (sinωt – 1) = 0
2sinωt + 1 = 0
2sinωt = -1
sinωt = -1/2
ωt = sin¯¹(-1/2)
ωt = (7π/6, 11π/6)
sinωt – 1 = 0
sinωt = 1
ωt = sin¯¹(1)
ωt = π/2
{π/2, 7π/6, 11π/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 θ 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
cosθ = 34/42.5
θ = cos¯¹(34/42.5)
θ = 36.87°