Question 1 :
The angle of elevation of the top of the building at a distance of 50 m from its foot on a horizontal plane is found to be 60 degree. Find the height of the building.
Question 2 :
A ladder placed against a wall such that it reaches the top of the wall of height 6 m and the ladder is inclined at an angle of 60 degree. Find how far the ladder is from the foot of the wall.
Question 3 :
A string of a kite is 100 meters long and it makes an angle of 60° with horizontal. Find the height of the kite,assuming that there is no slack in the string.
Question 4 :
From the top of the tower 30 m height a man is observing the base of a tree at an angle of depression measuring 30 degree. Find the distance between the tree and the tower.
Question 5 :
A man wants to determine the height of a light house. He measured the angle at A and found that tan A = 3/4. What is the height of the light house if A is 40 m from the base ?
Question 6 :
He A ladder is leaning against a vertical wall makes an angle of 20° with the ground. The foot of the ladder is 3 m from the wall.Find the length of ladder.
Question 7 :
A kite flying at a height of 65 m is attached to a string inclined at 31° to the horizontal. What is the length of string ?
Question 8 :
The length of a string between a kite and a point on the ground is 90 m. If the string is making an angle θ with the level ground such that tan θ = 15/8, how high will the kite be ?
Question 9 :
An airplane is observed to be approaching the air point. It is at a distance of 12 km from the point of observation and makes an angle of elevation of 50 degree. Find the height above the ground.
Question 10 :
A balloon is connected to a meteorological station by a cable of length 200 m inclined at 60 degree angle . Find the height of the balloon from the ground. (Imagine that there is no slack in the cable)
Question 1 :
The angle of elevation of the top of the building at a distance of 50 m from its foot on a horizontal plane is found to be 60°. Find the height of the building.
Solution :
Now we need to find the length of the side AB.
tanθ = Opposite side/Adjacent side
tan 60° = AB/BC
√3 = AB/50
√3 x 50 = AB
AB = 50√3
Approximate value of √3 is 1.732
AB = 50 (1.732)
AB = 86.6 m
So, the height of the building is 86.6 m.
Question 2 :
A ladder placed against a wall such that it reaches the top of the wall of height 6 m and the ladder is inclined at an angle of 60°. Find how far the ladder is from the foot of the wall.
Solution :
Here AB represents height of the wall, BC represents the distance between the wall and the foot of the ladder and AC represents the length of the ladder.
In the right triangle ABC, the side which is opposite to angle 60° is known as opposite side (AB), the side which is opposite to 90° is called hypotenuse side (AC) and remaining side is called adjacent side (BC).
Now, we need to find the distance between foot of the ladder and the wall. That is, we have to find the length of BC.
tan θ = Opposite side/Adjacent side
tan60° = AB/BC
√3 = 6/BC
BC = 6/√3
BC = (6/√3) x (√3/√3)
BC = (6√3)/3
BC = 2√3
Approximate value of √3 is 1.732
BC = 2 (1.732)
BC = 3.464 m
So, the distance between foot of the ladder and the wall is 3.464 m.
Question 3 :
A string of a kite is 100 meters long and it makes an angle of 60° with horizontal. Find the height of the kite, assuming that there is no slack in the string.
Solution :
Now we need to find the height of the side AB.
Sin θ = Opposite side/Hypotenuse side
sinθ = AB/AC
sin 60° = AB/100
√3/2 = AB/100
(√3/2) x 100 = AB
AB = 50 √3 m
So, the height of kite from the ground 50 √3 m.
Question 4 :
From the top of the tower 30 m height a man is observing the base of a tree at an angle of depression measuring 30 degree. Find the distance between the tree and the tower.
Solution :
Here AB represents height of the tower, BC represents the distance between foot of the tower and the foot of the tree.
Now we need to find the distance between foot of the tower and the foot of the tree (BC).
tan θ = Opposite side/Adjacent side
tan 30° = AB/BC
1/√3 = 30/BC
BC = 30√3
Approximate value of √3 is 1.732
BC = 30 (1.732)
BC = 81.96 m
So, the distance between the tree and the tower is 51.96 m.
Question 5 :
A man wants to determine the height of a light house. He measured the angle at A and found that tan A = 3/4. What is the height of the light house if A is 40 m from the base ?
Solution :
Now we need to find the height of the light house (BC).
tanA = Opposite side/Adjacent side
tanA = BC/AB
Given : tanA = 3/4
3/4 = BC/40
3 x 40 = BC x 4
BC = (3 x 40)/4
BC = (3 x 10)
BC = 30 m
So, the height of the light house is 30 m.
Question 6 :
A man wants to determine the height of a wall. A ladder is leaning against a vertical wall makes an angle of 20° with the ground. The foot of the ladder is 3 m from the wall. Find the length of ladder.
Solution :
Now we need to find the length of the ladder (AC).
Cos θ = Adjacent side/Hypotenuse side
Cos θ = BC/AC
Cos 20° = 3/AC
0.9396 = 3/AC
AC = 3/0.9396
AC = 3.192
So, the length of the ladder is 3.192 m.
Question 7 :
A kite flying at a height of 65 m is attached to a string inclined at 31° to the horizontal. What is the length of string ?
Solution :
Now we need to find the length of the string AC.
Sin θ = Opposite side/Hypotenuse side
Sin θ = AB/AC
Sin 31° = AB/AC
0.5150 = 65/AC
AC = 65/0.5150
AC = 126.2 m
Hence, the length of the string is 126.2 m.
Question 8 :
The length of a string between a kite and a point on the ground is 90 m. If the string is making an angle θ with the ground level such that tan θ = 15/8, how high will the kite be ?
Solution :
Now we need to find the length of the side AB.
Tan θ = 15/8 --------> cot θ = 8/15
csc θ = √(1+ cot²θ)
csc θ = √(1 + 64/225)
csc θ = √(225 + 64)/225
csc θ = √289/225
csc θ = 17/15 -------> sin θ = 15/17
But, sin θ = Opposite side/Hypotenuse side = AB/AC
AB/AC = 15/17
AB/90 = 15/17
AB = (15 x 90)/17
AB = 79.41
So, the height of the tower is 79.41 m.
Question 9 :
An airplane is observed to be approaching the air point. It is at a distance of 12 km from the point of observation and makes an angle of elevation of 50 degree. Find the height above the ground.
Solution :
Now we need to find the length of the side AB.
From the figure given above, AB stands for the height of the airplane above the ground.
sin θ = Opposite side/Hypotenuse side
sin 50° = AB/AC
0.7660 = h/12
0.7660 x 12 = h
h = 9.192 km
So, the height of the airplane above the ground is 9.192 km.
Question 10 :
A balloon is connected to a meteorological station by a cable of length 200 m inclined at 60 degree angle . Find the height of the balloon from the ground. (Imagine that there is no slack in the cable)
Solution :
Now we need to find the length of the side AB.
From the figure given above, AB stands for the height of the balloon above the ground.
sin θ = Opposite side/Hypotenuse side
sin θ = AB/AC
sin 60° = AB/200
√3/2 = AB/200
AB = (√3/2) x 200
AB = 100√3
Approximate value of √3 is 1.732
AB = 100 (1.732)
AB = 173.2 m
So, the height of the balloon from the ground is 173.2 m.
Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here.
If you have any feedback about our math content, please mail us :
v4formath@gmail.com
We always appreciate your feedback.
You can also visit the following web pages on different stuff in math.
WORD PROBLEMS
Word problems on simple equations
Word problems on linear equations
Word problems on quadratic equations
Area and perimeter word problems
Word problems on direct variation and inverse variation
Word problems on comparing rates
Converting customary units word problems
Converting metric units word problems
Word problems on simple interest
Word problems on compound interest
Word problems on types of angles
Complementary and supplementary angles word problems
Trigonometry word problems
Markup and markdown word problems
Word problems on mixed fractrions
One step equation word problems
Linear inequalities word problems
Ratio and proportion word problems
Word problems on sets and venn diagrams
Pythagorean theorem word problems
Percent of a number word problems
Word problems on constant speed
Word problems on average speed
Word problems on sum of the angles of a triangle is 180 degree
OTHER TOPICS
Time, speed and distance shortcuts
Ratio and proportion shortcuts
Domain and range of rational functions
Domain and range of rational functions with holes
Graphing rational functions with holes
Converting repeating decimals in to fractions
Decimal representation of rational numbers
Finding square root using long division
L.C.M method to solve time and work problems
Translating the word problems in to algebraic expressions
Remainder when 2 power 256 is divided by 17
Remainder when 17 power 23 is divided by 16
Sum of all three digit numbers divisible by 6
Sum of all three digit numbers divisible by 7
Sum of all three digit numbers divisible by 8
Sum of all three digit numbers formed using 1, 3, 4
Sum of all three four digit numbers formed with non zero digits