Let AC be the broken part of the tree.
∴ Total height of the tree = AB+AC
In right ΔABC,
cos 30° = BC/AC
⇒ √3/2 = 8/AC
⇒ AC = 16/√3
Also,
tan 30° = AB/BC
⇒ 1/√3 = AB/8
⇒ AB = 8/√3
Total height of the tree = AB+AC = 16/√3 + 8/√3 = 24/√3
There are two slides of height 1.5 m and 3 m. (Given)
Let AB is 1.5 m and PQ be 3 m slides.
ABC is the slide inclined at 30° with length AC and PQR is the slide inclined at
60° with length PR.
In right ΔABC,
sin 30° = AB/AC
⇒ 1/2 = 1.5/AC
⇒ AC = 3m
also,
In right ΔPQR,
sin 60° = PQ/PR
⇒ √3/2 = 3/PR
⇒ PR = 2√3 m
Hence, length of the slides are 3 m and 2√3 m respectively.
Let AB be the height of the tower and C is the point elevation which is 30 m away from the foot of the tower.
In right ΔABC,
tan 30° = AB/BC
⇒ 1/√3 = AB/30
⇒ AB = 10√3
Thus, the height of the tower is 10√3 m.
Let BC be the height of the kite from the ground,
AC be the inclined length of the string from the ground and A is the point where string of the kite is tied.
In right ΔABC,
sin 60° = BC/AC
⇒ √3/2 = 60/AC
⇒ AC = 40√3 m
Thus, the length of the string from the ground is 40√3 m
Let the boy initially standing at point Y with inclination 30° and then he approaches the building to
the point X with inclination 60°.
∴ XY is the distance he walked towards the building.
also, XY = CD.
Height of the building = AZ = 30 m
AB = AZ - BZ = (30 - 1.5) = 28.5 m
In right ΔABD,
tan 30° = AB/BD
⇒ 1/√3 = 28.5/BD
⇒ BD = 28.5√3 m
also,
In right ΔABC,
tan 60° = AB/BC
⇒ √3 = 28.5/BC
⇒ BC = 28.5/√3 = 28.5√3/3 m
∴ XY = CD = BD - BC = (28.5√3 - 28.5√3/3) = 28.5√3(1-1/3) = 28.5√3 × 2/3 = 57/√3 m.
Thus, the distance boy walked towards the building is 57/√3 m.
Let BC be the 20 m high building.
D is the point on the ground from where the elevation is taken.
Height of transmission tower = AB = AC - BC
In right ΔBCD,
tan 45° = BC/CD
⇒ 1 = 20/CD
⇒ CD = 20 m
also,
In right ΔACD,
tan 60° = AC/CD
⇒ √3 = AC/20
⇒ AC = 20√3 m
Height of transmission tower = AB = AC - BC = (20√3 - 20) m = 20(√3 - 1) m.
Let AB be the height of statue.
D is the point on the ground from where the elevation is taken.
Height of pedestal = BC = AC - AB
In right ΔBCD,
tan 45° = BC/CD
⇒ 1 = BC/CD
⇒ BC = CD.
also,
In right ΔACD,
tan 60° = AC/CD
⇒ √3 = AB+BC/CD
⇒ √3CD = 1.6 m + BC
⇒ √3BC = 1.6 m + BC
⇒ √3BC - BC = 1.6 m
⇒ BC(√3-1) = 1.6 m
⇒ BC = 1.6/(√3-1) m
⇒ BC = 0.8(√3+1) m
Thus, the height of the pedestal is 0.8(√3+1) m.
Let CD be the height of the tower equal to 50 m (Given)
Let AB be the height of the building.
BC be the distance between the foots of the building and the tower.
Elevation is 30° and 60° from the tower and the building respectively.
In right ΔBCD,
tan 60° = CD/BC
⇒ √3 = 50/BC
⇒ BC = 50/√3
also,
In right ΔABC,
tan 30° = AB/BC
⇒ 1/√3 = AB/BC
⇒ AB = 50/3
Thus, the height of the building is 50/3.
Let AB and CD be the poles of equal height.
O is the point between them from where the height of elevation taken.
BD is the distance between the poles.
AB = CD,
OB + OD = 80 m
Now,
In right ΔCDO,
tan 30° = CD/OD
⇒ 1/√3 = CD/OD
⇒ CD = OD/√3 ... (i)
also,
In right ΔABO,
tan 60° = AB/OB
⇒ √3 = AB/(80-OD)
⇒ AB = √3(80-OD)
AB = CD (Given)
⇒ √3(80-OD) = OD/√3
⇒ 3(80-OD) = OD
⇒ 240 - 3 OD = OD
⇒ 4 OD = 240
⇒ OD = 60
Putting the value of OD in equation (i)
CD = OD/√3 ⇒ CD = 60/√3 ⇒ CD = 20√3 m
also,
OB + OD = 80 m ⇒ OB = (80-60) m = 20 m
Thus, the height of the poles are 20√3 m and distance from the point of elevation are 20 m and 60 m respectively.
Let AB be the building of height 7 m and EC be the height of tower.
A is the point from where elevation of tower is 60° and the angle of depression of its foot is 45°
EC = DE + CD
also, CD = AB = 7 m.
and BC = AD
In right ΔABC,
tan 45° = AB/BC
⇒ 1= 7/BC
⇒ BC = 7 m = AD
also,
In right ΔADE,
tan 60° = DE/AD
⇒ √3 = DE/7
⇒ DE = 7√3 m
Height of the tower = EC = DE + CD
= (7√3 + 7) m = 7(√3+1) m.
Let AB be the lighthouse of height 75 m.
Let C and D be the positions of the ships.
30° and 45° are the angles of depression from the lighthouse.
In right ΔABC,
tan 45° = AB/BC
⇒ 1= 75/BC
⇒ BC = 75 m
also,
In right ΔABD,
tan 30° = AB/BD
⇒ 1/√3 = 75/BD
⇒ BD = 75√3 m
The distance between the two ships = CD = BD - BC = (75√3 - 75) m = 75(√3-1) m.
Let the initial position of the balloon be A and final position be B.
Height of balloon above the girl height = 88.2 m - 1.2 m = 87 m
Distance travelled by the balloon =
DE = CE - CD
In right ΔBEC,
tan 30° = BE/CE
⇒ 1/√3= 87/CE
⇒ CE = 87√3 m
also,
In right ΔADC,
tan 60° = AD/CD
⇒ √3= 87/CD
⇒ CD = 87/√3 m = 29√3 m
Distance travelled by the balloon = DE = CE - CD = (87√3 - 29√3) m = 29√3(3 - 1) m = 58√3 m.
Let AB be the tower.
D is the initial and C is the final position of the car respectively.
Angles of depression are measured from A.
BC is the distance from the foot of the tower to the car.
In right ΔABC,
tan 60° = AB/BC
⇒ √3 = AB/BC
⇒ BC = AB/√3 m
also,
In right ΔABD,
tan 30° = AB/BD
⇒ 1/√3 = AB/(BC + CD)
⇒ AB√3 = BC + CD
⇒ AB√3 = AB/√3 + CD
⇒ CD = AB√3 - AB/√3
⇒ CD = AB(√3 - 1/√3)
⇒ CD = 2AB/√3
Here, distance of BC is half of CD. Thus, the time taken is also half.
Time taken by car to travel distance CD = 6 sec.
Time taken by car to travel BC = 6/2 = 3 sec.
Let AB be the tower.
C and D be the two points with distance 4 m and 9 m from the base respectively.
In right ΔABC,
tan x = AB/BC
⇒ tan x = AB/4
⇒ AB = 4 tan x ... (i)
also,
In right ΔABD,
tan (90°-x) = AB/BD
⇒ cot x = AB/9
⇒ AB = 9 cot x ... (ii)
Multiplying eqn (i) and (ii)
AB2 = 9 cot x × 4 tan x
⇒ AB2 = 36
⇒ AB = ± 6
Height cannot be negative. Therefore, the height of the tower is 6 m. Hence, Proved.
Let AC be the broken part of the tree.
∴ Total height of the tree = AB+AC
In right ΔABC,
cos 30° = BC/AC
⇒ √3/2 = 8/AC
⇒ AC = 16/√3
Also,
tan 30° = AB/BC
⇒ 1/√3 = AB/8
⇒ AB = 8/√3
Total height of the tree = AB+AC = 16/√3 + 8/√3 = 24/√3
There are two slides of height 1.5 m and 3 m. (Given)
Let AB is 1.5 m and PQ be 3 m slides.
ABC is the slide inclined at 30° with length AC and PQR is the slide inclined at
60° with length PR.
In right ΔABC,
sin 30° = AB/AC
⇒ 1/2 = 1.5/AC
⇒ AC = 3m
also,
In right ΔPQR,
sin 60° = PQ/PR
⇒ √3/2 = 3/PR
⇒ PR = 2√3 m
Hence, length of the slides are 3 m and 2√3 m respectively.
Let AB be the height of the tower and C is the point elevation which is 30 m away from the foot of the tower.
In right ΔABC,
tan 30° = AB/BC
⇒ 1/√3 = AB/30
⇒ AB = 10√3
Thus, the height of the tower is 10√3 m.
Let BC be the height of the kite from the ground,
AC be the inclined length of the string from the ground and A is the point where string of the kite is tied.
In right ΔABC,
sin 60° = BC/AC
⇒ √3/2 = 60/AC
⇒ AC = 40√3 m
Thus, the length of the string from the ground is 40√3 m
Let the boy initially standing at point Y with inclination 30° and then he approaches the building to
the point X with inclination 60°.
∴ XY is the distance he walked towards the building.
also, XY = CD.
Height of the building = AZ = 30 m
AB = AZ - BZ = (30 - 1.5) = 28.5 m
In right ΔABD,
tan 30° = AB/BD
⇒ 1/√3 = 28.5/BD
⇒ BD = 28.5√3 m
also,
In right ΔABC,
tan 60° = AB/BC
⇒ √3 = 28.5/BC
⇒ BC = 28.5/√3 = 28.5√3/3 m
∴ XY = CD = BD - BC = (28.5√3 - 28.5√3/3) = 28.5√3(1-1/3) = 28.5√3 × 2/3 = 57/√3 m.
Thus, the distance boy walked towards the building is 57/√3 m.
Let BC be the 20 m high building.
D is the point on the ground from where the elevation is taken.
Height of transmission tower = AB = AC - BC
In right ΔBCD,
tan 45° = BC/CD
⇒ 1 = 20/CD
⇒ CD = 20 m
also,
In right ΔACD,
tan 60° = AC/CD
⇒ √3 = AC/20
⇒ AC = 20√3 m
Height of transmission tower = AB = AC - BC = (20√3 - 20) m = 20(√3 - 1) m.
Let AB be the height of statue.
D is the point on the ground from where the elevation is taken.
Height of pedestal = BC = AC - AB
In right ΔBCD,
tan 45° = BC/CD
⇒ 1 = BC/CD
⇒ BC = CD.
also,
In right ΔACD,
tan 60° = AC/CD
⇒ √3 = AB+BC/CD
⇒ √3CD = 1.6 m + BC
⇒ √3BC = 1.6 m + BC
⇒ √3BC - BC = 1.6 m
⇒ BC(√3-1) = 1.6 m
⇒ BC = 1.6/(√3-1) m
⇒ BC = 0.8(√3+1) m
Thus, the height of the pedestal is 0.8(√3+1) m.
Let CD be the height of the tower equal to 50 m (Given)
Let AB be the height of the building.
BC be the distance between the foots of the building and the tower.
Elevation is 30° and 60° from the tower and the building respectively.
In right ΔBCD,
tan 60° = CD/BC
⇒ √3 = 50/BC
⇒ BC = 50/√3
also,
In right ΔABC,
tan 30° = AB/BC
⇒ 1/√3 = AB/BC
⇒ AB = 50/3
Thus, the height of the building is 50/3.
Let AB and CD be the poles of equal height.
O is the point between them from where the height of elevation taken.
BD is the distance between the poles.
AB = CD,
OB + OD = 80 m
Now,
In right ΔCDO,
tan 30° = CD/OD
⇒ 1/√3 = CD/OD
⇒ CD = OD/√3 ... (i)
also,
In right ΔABO,
tan 60° = AB/OB
⇒ √3 = AB/(80-OD)
⇒ AB = √3(80-OD)
AB = CD (Given)
⇒ √3(80-OD) = OD/√3
⇒ 3(80-OD) = OD
⇒ 240 - 3 OD = OD
⇒ 4 OD = 240
⇒ OD = 60
Putting the value of OD in equation (i)
CD = OD/√3 ⇒ CD = 60/√3 ⇒ CD = 20√3 m
also,
OB + OD = 80 m ⇒ OB = (80-60) m = 20 m
Thus, the height of the poles are 20√3 m and distance from the point of elevation are 20 m and 60 m respectively.
Let AB be the building of height 7 m and EC be the height of tower.
A is the point from where elevation of tower is 60° and the angle of depression of its foot is 45°
EC = DE + CD
also, CD = AB = 7 m.
and BC = AD
In right ΔABC,
tan 45° = AB/BC
⇒ 1= 7/BC
⇒ BC = 7 m = AD
also,
In right ΔADE,
tan 60° = DE/AD
⇒ √3 = DE/7
⇒ DE = 7√3 m
Height of the tower = EC = DE + CD
= (7√3 + 7) m = 7(√3+1) m.
Let AB be the lighthouse of height 75 m.
Let C and D be the positions of the ships.
30° and 45° are the angles of depression from the lighthouse.
In right ΔABC,
tan 45° = AB/BC
⇒ 1= 75/BC
⇒ BC = 75 m
also,
In right ΔABD,
tan 30° = AB/BD
⇒ 1/√3 = 75/BD
⇒ BD = 75√3 m
The distance between the two ships = CD = BD - BC = (75√3 - 75) m = 75(√3-1) m.
Let the initial position of the balloon be A and final position be B.
Height of balloon above the girl height = 88.2 m - 1.2 m = 87 m
Distance travelled by the balloon =
DE = CE - CD
In right ΔBEC,
tan 30° = BE/CE
⇒ 1/√3= 87/CE
⇒ CE = 87√3 m
also,
In right ΔADC,
tan 60° = AD/CD
⇒ √3= 87/CD
⇒ CD = 87/√3 m = 29√3 m
Distance travelled by the balloon = DE = CE - CD = (87√3 - 29√3) m = 29√3(3 - 1) m = 58√3 m.
Let AB be the tower.
D is the initial and C is the final position of the car respectively.
Angles of depression are measured from A.
BC is the distance from the foot of the tower to the car.
In right ΔABC,
tan 60° = AB/BC
⇒ √3 = AB/BC
⇒ BC = AB/√3 m
also,
In right ΔABD,
tan 30° = AB/BD
⇒ 1/√3 = AB/(BC + CD)
⇒ AB√3 = BC + CD
⇒ AB√3 = AB/√3 + CD
⇒ CD = AB√3 - AB/√3
⇒ CD = AB(√3 - 1/√3)
⇒ CD = 2AB/√3
Here, distance of BC is half of CD. Thus, the time taken is also half.
Time taken by car to travel distance CD = 6 sec.
Time taken by car to travel BC = 6/2 = 3 sec.
Let AB be the tower.
C and D be the two points with distance 4 m and 9 m from the base respectively.
In right ΔABC,
tan x = AB/BC
⇒ tan x = AB/4
⇒ AB = 4 tan x ... (i)
also,
In right ΔABD,
tan (90°-x) = AB/BD
⇒ cot x = AB/9
⇒ AB = 9 cot x ... (ii)
Multiplying eqn (i) and (ii)
AB2 = 9 cot x × 4 tan x
⇒ AB2 = 36
⇒ AB = ± 6
Height cannot be negative. Therefore, the height of the tower is 6 m. Hence, Proved.
