# Some Applications of Trigonometry Chapter 9 Extra Questions and Solutions For Class 10 CBSE Mathematics

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## Solution.

Given,

Distance between the wall and the foot of the ladder = 2m = BC. Angle made by ladder with ground (θ) = 60°. Height of the wall (H) = AB.

Now, the fig. of ABC forms a right angle triangle.

So,

tan θ = (Opposite Side / Adjacent Side) = tan 60 = AB/BC = √3 = AB/2 = AB = 2 √3 m.

So, the height ( AB = H) of the wall is 2 √3 m.

## Solution.

Distance between the wall and foot of the ladder = 9.5 m. Angle of elevation (θ) = 60°. Length of the ladder = L = AC.

Now, from fig. ABC ,  ΔABC is a right angle triangle ,  So,

### YOU ARE READING: Some Applications of Trigonometry Chapter 9 Extra Questions and Solutions For Class 10 CBSE Mathematics

Therefore, the length of the ladder AC is 19 m.

## Solution.

Let the height of the balloon from the ground = h m.

It is given that , the length of the cable = 215 m and the inclination of the cable is 60o.

In right-angled triangle ΔABC,

sin 60o = AB/ AC

√3 / 2 = h / 215

h = 215 (√3/2) = 185.9

Therefore, the height of the balloon from the ground is 186m.

## Solution.

It is given that ,

The length of the ladder = 15m = AO. Angle made by the ladder with the wall = 60o. Let the height of the wall be h metres and the horizontal ground taken as OX.

Then from the fig. we have,

In right ΔABO, using trigonometric ratios,

cos (60o) = AB/AO

1/2 = h/ 15

h = 15/2

h = 7.5m

### YOU ARE READING: Some Applications of Trigonometry Chapter 9 Extra Questions and Solutions For Class 10 CBSE Mathematics

Hence, the height of the wall is 7.5m.

## Solution.

Given,

The initial height of tree H = 15 m = AB + AC. We assume that it is broken at point A. And, the angle made by broken part with the ground (θ) = 60°. Height from ground to broken points = h = AB.

So, we have

H = AC + h

⟹ AC = (H – h) m = ( 15 – h ) m .

We get a right triangle formed by the above given data, So,

sin∅=oppositeside/hypotenuse

Now,

Sin60 = AB/ BC

√3/2=h/(H-h)

Therefore, the height of broken point from the ground is  15(2√3 – 3)m.

## Solution.

Given,

Height of the flag staff = 5 m = AB, Angle of elevation of the top of flag staff = 60°

Angle of elevation of the bottom of the flagstaff = 30°. Let height of tower be ‘h’ m = BC . And, let the distance of the point from the base of the tower = x m.

In right angle triangle BCD, we have

tan 30o = BC/DC

1/√3 = h/x

x = h√3 ….. (i)

Now, in ΔACD,

tan 60o = AC/DC

√3 = (5 + h)/ x

√3x = 5 + h

√3(h√3) = 5 + h [using (i)]

3h = 5 + h

2h = 5

h = 5/2 = 2.5m

### YOU ARE READING: Some Applications of Trigonometry Chapter 9 Extra Questions and Solutions For Class 10 CBSE Mathematics

Therefore, the height of the tower = 2.5 m.

## Solution.

Let the parachute at highest point A and let C and D be points which are 100 m apart on ground where from then CD = 100 m. Angle of elevation from point D = 45° = α . Angle of elevation from point C = 60° = β .

Let B be the point just vertically down the parachute.Now let us draw figure according to above data then it forms the figure as shown in which ABC and ABD are two triangles,

Maximum height of the parachute from the ground AB = H m . Distance of point where parachute falls to just nearest observation point = x m . If in right angle triangle one of the included angles is θ then …

Therefore, The maximum height of the parachute from the ground, H = 236.6m. Distance between the two points where parachute falls on the ground and just the observation is x = 136.6 m.

## Solution.

Given,

The angle of elevation of top tower from first point D, α = 30°. On moving through D to C by 150 m, then CD = 150 m and angle of elevation of top of the tower from second point C, β = 60°. Let height of tower AB = H m.

### YOU ARE READING: Some Applications of Trigonometry Chapter 9 Extra Questions and Solutions For Class 10 CBSE Mathematics

Therefore, it has been proved that the height of the tower is 129.9 m.

## Solution.

Let BC be the tower and AB be the flag pole on the tower. Distance of the point of observation from foot of the tower DC = 9 m. Angle of elevation of top of flag pole is 60° and angle of elevation of bottom of flag pole is 30°.

Let height of the tower = h m = BC . and the height of the pole = x m = AB .

From fig, we have

In ΔBCD,

tan 30o = BC/DC

1/√3 = h/9

h = 9/√3 = 3√3

Next, in ΔACD

tan 60o = AC/DC

√3 = (x + h)/9

x + h = 9√3

x + 3√3 = 9√3

x = 6√3 m

Therefore, Height of the tower = 3√3 m and height of the pole = 6√3 m.

## Solution.

Let the height of flag-staff(AB) = h m.

And, the distance PQ = x m.

Given,

Angle of elevation of top of the building = 30o.

Angle of elevation of top of the flag staff = 45o

From the fig.

In ΔBQP,

tan 30o = BQ/PQ

1/√3 = 10/x

x = 10√3 m

Next,

In ΔAQP,

tan 45o = AQ/PQ

1 = (h + 10)/x

h + 10 = x = 10√3

h = (10√3 – 10) = 10(1.732) – 10 = 17.32 – 10

= 7.32 m

### YOU ARE READING: Some Applications of Trigonometry Chapter 9 Extra Questions and Solutions For Class 10 CBSE Mathematics

Therefore, the distance of point P from building = x = 10√3 = 10(1.732) = 17.32m

## Solution.

When the sun’s altitude is the angle of elevation of the top of the tower from the tip of the shadow. . Let AB be h m and BC be x m. From the question, DC is 40 m longer than BC.

So, BD = (40 + x) m.

And two right triangles ABC and ABD are formed.

.

In ΔABC,

tan 60o = AB/ BC

√3 = h/x

x = h/√3 … (i)

In ΔABD,

tan 30o = AB/ BD

1/ √3 = h/ (x + 40)

x + 40 = √3h

h/√3 + 40 = √3h [using (i)]

h + 40√3 = 3h

2h = 40√3

h = 20√3

Therefore, the height of the tower is 20√3 m.

## Solution.

Let AC be the height of the tree which is (x + h) m. Given, the broken portion of the tree is making an angle of 30o with the ground.

From the fig.

In ΔBCD, we have

tan 30o = BC/ DC

1/√3 = h/ 10

h = 10/ √3

Next, in ΔBCD

cos 30o = DC/BD

√3/2 = 10/x

x = 20/√3 m

So,

x + h = 20/√3 + 10/√3

= 30/√3

= 10√3 = 10(1.732) = 17.32

### YOU ARE READING: Some Applications of Trigonometry Chapter 9 Extra Questions and Solutions For Class 10 CBSE Mathematics

Therefore, the height of the tree is 17.32 m.

## Solution.

Let the height of the tower(AB) = h m. Let the length of the shorter shadow be x m. Then, the longer shadow is (10 + x)m.

So, from fig. In ΔABC

tan 60o = AB/BC

√3 = h/x

x = h/√3…. (i)

Next, in ΔABD

tan 45o = AB/BD

1 = h/(10 + x)

10 + x = h

10 + (h/√3) = h [using (i)]

10√3 + h = √3h

h(√3 -1) =10√3

h = 10√3/ (√3 -1)

After rationalising the denominator, we have

h = [10√3 x (√3 + 1)]/ (3 – 1)

h = 5√3(√3 + 1)

h = 5(3 + √3) = 23.66      [√3 = 1.732]

Therefore, the height of the tower is 23.66 m.

## Solution.

Given,

The height of the tower (AB) = 150m. Angles of depressions of the two objects are 45o and 60o.

In ΔABD

tan 45o = AB/ BD

1 = 150/ BD

BD = 150m

Next, in ΔABC

tan 60o = AB/ BC

√3 = 150/ BC

BC = 150/√3

BC = 50√3 = 50(1.732) = 86.6 m

So,

DC = BD – BC = 150 – 86.6 = = 63.4

### YOU ARE READING: Some Applications of Trigonometry Chapter 9 Extra Questions and Solutions For Class 10 CBSE Mathematics

Therefore, the distance between two objects = 63.4 m.

## Solution.

Let the height of the tower = h m  and the distance BC = x m.

Then, from the fig.

In ΔABC

tan 63o = AB/BC

1.9626 = h/x

x = h/ 1.9626

x = 0.5095 h …. (i)

Next, in ΔABD

tan 32o = AB/ BD

0.6248 = h/ (100 + x)

h = 0.6248(100 + x)

h = 62.48 + 0.6248x

h = 62.48 + 0.6248(0.5095 h) ….. [using (i)]

h = 62.48 + 0.3183h

0.6817h = 62.48

h = 62.48/0.6817 = 91.65

Using h in (i), we have

x = 0.5095(91.65) = 46.69

Therefore,

The height of the tower is 91.65 m and distance of the first position from the tower = 100 + x = 146.69 m.

## Solution.

Let AC be the lamp post of height ‘h’ and DE is the tall girl and her shadow is BE. So, we have ED = 1.6 m, BE = 4.8 m and EC = 3.2

(i) By using trigonometric ratio

In ΔBDE,

tan θ = 1.6/4.8

tan θ = 1/3

Next, In ΔABC

tan θ = h/ (4.8 + 3.2)

1/3 = h/8

h = 8/3 m

ii) By using similar triangles

Since triangle BDE and triangle ABC are similar (by AA criteria), we have

### YOU ARE READING: Some Applications of Trigonometry Chapter 9 Extra Questions and Solutions For Class 10 CBSE Mathematics

Therefore, the height of the lamp post is h = 8/3 m.

## Solution.

Let’s assume AB as the statue, BC be the pedestal and D be the point on ground from where elevation angles are measured. It is given that the Angle of elevation of the top of statue is 60o and angle of elevation of the top of the pedestal is 45o.

So, from the fig. we have

In ΔBCD,

tan 45o = BC/ CD

1 = BC/ CD

CD = BC

tan 60o = (AB + BC)/ CD

√3 = (AB + BC)/ BC [As CD = BC, found above]

BC√3 = AB + BC

AB = (√3 – 1)BC

BC = AB/ (√3 – 1)

BC = 1.6/ (√3 – 1)

Rationalising the denominator, we have

BC = 1.6(√3 + 1)/ 2

BC = 0.8(√3 + 1) m

Therefore, the height of pedestal is 0.8(√3 + 1) m

## Solution.

Given

Height of the building = 7 m = AB and  height of the cable tower = CD.

Angle of elevation of the top of the cable tower from the top of the building = 60° and angle of depression of the bottom of the building from the top of the building= 45°.

Then, from the fig. we see that

ED = AB = 7 m

And,

CD = CE + ED

So, In ΔABD, we have

AB/ BD = tan 45o

AB = BD = 7

BD = 7

In ΔACE,

AE = BD = 7

And, tan 60o = CE/AE

√3 = CE/ 7

CE = 7√3 m

So, CD = CE + ED = (7√3 + 7)= 7(√3 + 1) m

### YOU ARE READING: Some Applications of Trigonometry Chapter 9 Extra Questions and Solutions For Class 10 CBSE Mathematics

Therefore, the height of the cable tower is 7(√3 + 1)m.

## Solution.

Let AB be the building and CD be the tower. It is given that The angle of elevation of the top of the building from the foot of the tower is 30o and the angle of elevation of the top of the building from the foot of the tower is 30o.

Height of the tower = CD = 50 m

From the fig. we have

In ΔCDB,

CD/ BD = tan 60o

50/ BD = √3

BD = 50/√3 …. (i)

Next in ΔABD,

AB/ BD = tan 30o

AB/ BD = 1/√3

AB = BD/ √3

AB = 50/√3/ (√3) [From (i)]

AB = 50/3

Therefore, the height of the building is 50/3 m.

## Solution.

From the question, the following fig. is made , Let the height of tower (AB) = h m and distance BC = y m.

Then, in ΔABC

tan 45o = AB/BC

1 = h/y

y = h

Next, in ΔABD

tan 30o = AB/BD

1/√3 = h/ (2x + y)

2x + y = √3h

2x + h = √3h

2x = (√3 – 1)h

h = 2x/ (√3 – 1) x (√3 + 1)/ (√3 + 1)

h = 2x (√3 + 1)/(3-1) = 2x (√3 + 1)/2 = x (√3 + 1)

Therefore, the height of the tower is x (√3 + 1) m

Hence Proved the height.

.