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## Download the PDF of RD Sharma For Class 10 Maths Chapter 12 Some Applications Of Trigonometry here

Some Applications of Trigonometry is the 12th chapter of the RD Sharma Class 10 textbook. It contains only one exercise of problems based on using the trigonometric results to find heights and distances. For knowing the appropriate stepwise procedure and approach to these problems, the RD Sharma Solutions for Class 10 is the right place to dig in. All the solutions are developed based on the latest CBSE marking schemes.

### RD Sharma Solutions for Class 10 Maths Chapter 11 Some Applications Of Trigonometry – exercise wise

### Access the RD Sharma Solutions For Class 10 Chapter 11 – Some Applications Of Trigonometry

Exercise 12.1 Page No: 12.29

**1. A tower stands vertically on the ground. From a point on the ground, 20 m away from the foot of the tower, the angle of elevation of the top the tower is 60Â°. What is the height of the tower?**

**Solution:**

Given:

Distance between the foot of the tower and point of observation = 20 m = BC

Angle of elevation of the top of the tower = 60Â° = Î¸

And, Height of tower (H) = AB

Now, from fig. ABC

Î”ABC is a right angle triangle,

So,

**2. The angle of elevation of a ladder against a wall is 60Â° and the foot of the ladder is 9.5 m away from the wall. Find the length of the ladder.**

**Solution:**

Given:

Distance between the wall and foot of the ladder = 9.5 m

Angle of elevation (Î¸) = 60Â°

Length of the ladder = L = AC

Now, from fig. ABC

Î”ABC is a right angle triangle,

So,

Thus, length of the ladder (L) = 19 m

**3. A ladder is placed along a wall of a house such that its upper end is touching the top of the wall. The foot of the ladder is 2 m away from the wall and the ladder is making an angle of 60Â° with the level of the ground. Determine the height of the wall.**

**Solution:**

Given,

Distance between the wall and the foot of the ladder = 2m = BC

Angle made by ladder with ground (Î¸) = 60Â°

Height of the wall (H) = AB

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

So,

**4. An electric pole is 10 m high. A steel wire tied to top of the pole is affixed at a point on the ground to keep the pole up right. If the wire makes an angle of 45Â° with the horizontal through the foot of the pole, find the length of the wire.**

**Solution:**

Given,

Height of the electric pole = 10 m = AB

The angle made by steel wire with ground (horizontal) Î¸ = 45Â°

Let length of wire = L = AC

So, form the fig. formed we have ABC as a right triangle.

**5. A kite is flying at a height of 75 meters from the ground level, attached to a string inclined at 60Â° to the horizontal. Find the length of the string to the nearest meter.**

**Solution:**

Given,

Height of kite flying from the ground level = 75 m = AB

Angle of inclination of the string with the ground (Î¸) = 60Â°

Let the length of the string be L = AC

So, form the fig. formed we have ABC as a right triangle.

Hence,

**6. A ladder 15 metres long reaches the top of a vertical wall. If the ladder makes an angle of 60 ^{o} with the wall, find the height of the wall. **

**Solution: **

Given,

The length of the ladder = 15m = AO

Angle made by the ladder with the wall = 60^{o}

Let the height of the wall be h metres.

And the horizontal ground taken as OX.

Then from the fig. we have,

In right Î”ABO, using trigonometric ratios

cos (60^{o}) = AB/AO

1/2 = h/ 15

h = 15/2

h = 7.5m

Hence, the height of the wall is 7.5m

**7. A vertical tower stands on a horizontal place and is surmounted by a vertical flag staff. At a point on the plane 70 meters away from the tower, an observer notices that the angles of elevation of the top and bottom of the flag-staff are respectively 60Â° and 45Â°. Find the height of the flag staff and that of the tower.**

**Solution:**

Given,

A vertical tower is surmounted by flag staff.

Distance between observer and the tower = 70 m = BC

Angle of elevation of bottom of the flag staff = 45Â°

Angle of elevation of top of the flag staff = 60Â°

Let the height of the flag staff = h = AD

Height of tower = H = AB

If we represent the above data in the figure then it forms right angle triangles Î”ABC and Î”BCD

When Î¸ is angle in right angle triangle we know that

tan Î¸ = opp. Side/ Adj. side

Now,

tan 45^{o} = AB/ BC

1 = H/ 70

âˆ´ H =70

Again,

h = 70 (1.732-1)

âˆ´ h = 51.24 m

Therefore, the height of tower = 70 m and the height of flag staff = 51.24 m

**8. A vertically straight tree, 15 m high, is broken by the wind in such a way that its top just touches the ground and makes an angle of 60Â° with the ground. At what height from the ground did the tree break?**

**Solution:**

Given,

The initial height of tree H = 15 m = AB + AC

Let us assume that it is broken at point A.

And, the angle made by broken part with the ground (Î¸) = 60Â°

Height from ground to broken points = h = AB

So, we have

H = AC + h

âŸ¹ AC = (H – h) m

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

So,

Rationalizing denominator, we have

Therefore, the height of broken point from the ground is 15(2âˆš3 – 3)m

**9. A vertical tower stands on a horizontal plane and is surmounted by a vertical flag staff of height 5 meters. At a point on the plane, the angles of elevation of the bottom and the top of the flag staff are respectively 30Â° and 60Â°. Find the height of the tower.**

**Solution:**

Given,

Height of the flag staff = 5 m =AB

Angle of elevation of the top of flag staff = 60Â°

Angle of elevation of the bottom of the flagstaff = 30Â°

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 30^{o} = BC/DC

1/âˆš3 = h/x

x = hâˆš3 â€¦.. (i)

Now, in Î”ACD,

tan 60^{o} = AC/DC

âˆš3 = (5 + h)/ x

âˆš3x = 5 + h

âˆš3(hâˆš3) = 5 + h [using (i)]

3h = 5 + h

2h = 5

h = 5/2 = 2.5m

Therefore, the height of the tower = 2.5 m

**10**. **A person observed the angle of elevation of a tower as 30Â°. He walked 50 m towards the foot of the tower along level ground and found the angle of elevation of the top of the tower as 60Â°. Find the height of the tower.**

**Solution:**

Given,

The angle of elevation of the tower before he started walking = 30^{o}

Distance walked by the person towards the tower = 50m

The angle of elevation of the tower after he walked = 60^{o}

Let height of the tower (AB) = h m

Let the distance BC = x m

From the fig. in Î”ABC,

tan 60^{o} = AB/ BC

âˆš3 = h/x

x = h/âˆš3 â€¦.(i)

Now, in Î”ABD

tan 30^{o} = AB/ BD

1/âˆš3 = h/ (50 + x)

âˆš3h = 50 + x

âˆš3h = 50 + (h/âˆš3) [using (i)]

3h = 50âˆš3 + h

2h = 50âˆš3

h = 25âˆš3 = 25(1.73) = 43.25m

Therefore, the height of the tower = 43.25m

**11. The shadow of a tower, when the angle of elevation of the sun is 45Â°, is found to be 10 m longer than when it was 60Â°. Find the height of the tower.**

**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 Î”ABC

tan 60^{o} = AB/BC

âˆš3 = h/x

x = h/âˆš3â€¦. (i)

Next, in Î”ABD

tan 45^{o} = AB/BD

1 = h/(10 + x)

10 + x = h

10 + (h/âˆš3) = h [using (i)]

10âˆš3 + h = âˆš3h

h(âˆš3 -1) =10âˆš3

h = 10âˆš3/ (âˆš3 -1)

After rationalising the denominator, we have

h = [10âˆš3 x (âˆš3 + 1)]/ (3 – 1)

h = 5âˆš3(âˆš3 + 1)

h = 5(3 + âˆš3) = 23.66 [âˆš3 = 1.732]

Therefore, the height of the tower is 23.66 m.

**12. A parachute is descending vertically and makes angles of elevation of 45Â° and 60Â° at two observing points 100 m apart from each other on the left side of himself. Find the maximum height from which he falls and the distance of point where he falls on the ground from the just observation point.**

**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 = 45Â° = Î±

Angle of elevation from point C = 60Â° = Î²

Let B be the point just vertically down the parachute

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 Î¸ then

From (a) and (b)

x = 136.6 m in (b)

h = âˆš3 Ã— 136.6 = 236.6m

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

**13. On the same side of a tower, two objects are located. When observed from the top of the tower, their angles of depression are 45Â° and 60Â°. If the height of the tower is 150 m, find the distance between the objects.**

**Solution:**

Given,

The height of the tower (AB) = 150m

Angles of depressions of the two objects are 45^{o} and 60^{o}.

In Î”ABD

tan 45^{o} = AB/ BD

1 = 150/ BD

BD = 150m

Next, in Î”ABC

tan 60^{o} = AB/ BC

âˆš3 = 150/ BC

BC = 150âˆš3/ 3

BC = 50âˆš3 = 50(1.732) = 86.6 m

So,

DC = BD â€“ BC = 150 â€“ 86.6 = = 63.4

Therefore, the distance between two objects = 63.4 m

**14. The angle of elevation of a tower from a point on the same level as the foot of the tower is 30Â°. On advancing 150 meters towards the foot of the tower, the angle of elevation of the tower becomes 60Â°. Show that the height of the tower is 129.9 metres.**

**Solution:**

Given,

The angle of elevation of top tower from first point D, Î± = 30Â°

On advancing through D to C by 150 m, then CD = 150 m

Angle of elevation of top of the tower from second point C, Î² = 60Â°

Let height of tower AB = H m

Representing the above data in form of figure then it form a figure as shown with âˆ B = 90Â°

Substituting (b) in (a), we have

H = 129.9

Therefore, the height of the tower = 129.9 m

**15. The angle of elevation of the top of a tower as observed from a point in a horizontal plane through the foot of the tower is 32Â°. When the observer moves towards the tower a distance of 100 m, he finds the angle of elevation of the top to be 63Â°. Find the height of the tower and the distance of the first position from the tower. [Take tan 32Â° = 0.6248 and tan 63Â° = 1.9626]**

**Solution: **

Let the height of the tower = h m

And the distance BC = x m

Then, from the fig.

In Î”ABC

tan 63^{o} = AB/BC

1.9626 = h/x

x = h/ 1.9626

x = 0.5095 h â€¦. (i)

Next, in Î”ABD

tan 32^{o} = 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) â€¦.. [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

Distance of the first position from the tower = 100 + x = 146.69 m

**16. The angle of elevation of the top of a tower from a point A on the ground is 30Â°. On moving a distance of 20 meters towards the foot of the tower to a point B the angle of elevation increases to 60Â°. Find the height of the tower and the distance of the tower from the point A.**

**Solution:**

Given,

Angle of elevation of top of the tower from point A, Î± = 30Â°.

Angle of elevation of top of tower from point B, Î² = 60Â°.

And, the distance between A and B, AB = 20 m

Let height of tower CD = ‘h’ m and distance between second point B from foot of the tower be ‘x’ m

Representing the above data in form of figure with âˆ D = 90Â°,

Substituting (b) in (a), we have

x = 10 m

Therefore,

Height of the tower = 17.32 m

Distance of the tower from point A = (20 + 10) = 30 m

**17. From the top of a building 15 m high the angle of elevation of the top of tower is found to be 30Â°. From the bottom of the same building, the angle of elevation of the top of the tower is found to be 60Â°. Find the height of the tower and the distance between the tower and the building.**

**Solution:**

Given,

The height of the building = 15 m

The angle of elevation from the top of the building to top of the tower = 30^{o}

The angle of elevation from the bottom of the building to top of the tower = 60^{o}

Let the height of the tower = h m

And the distance between tower and building = x m

So, AD = x [since, AD || BE]

Similarly, AB = DE

Then, from fig.

In Î”BEC,

tan 60^{o} = CE/BE

âˆš3 = h/x

x = h/âˆš3 â€¦.(i)

Next, in Î”CDA,

tan 30^{o} = CD/DA

1/ âˆš3 = (h – 15)/x

âˆš3h – 15âˆš3 = x

âˆš3h – 15âˆš3 = h/âˆš3 [using (i)]

3h â€“ 45 = h

2h = 45

h = 45/2 = 22.5

Putting h in (i), we get

x = 22.5/(âˆš3)

x = 22.5/(1.73) = 12.975

Therefore, the height of the building = 22.5 m

And, distance between tower and building = 12.975 m

**18. On a horizontal plane there is a vertical tower with a flag pole on the top of the tower. At a point 9 m away from the foot of the tower the angle of elevation of the top and bottom of the flag pole are 60Â° and 30Â° respectively. Find the height of the tower and the flag pole mounted on it.**

**Solution:**

Let BC be the tower and AB be the flagstaff on the tower

Distance of the point of observation from foot of the tower DC = 9 m

Angle of elevation of top of flagstaff is 60Â°

Angle of elevation of bottom of flagstaff is 30Â°

Let height of the tower = h m = BC

Height of the pole = x m = AB

From fig, we have

In Î”BCD,

tan 30^{o} = BC/DC

1/âˆš3 = h/9

h = 9/âˆš3 = 3âˆš3

Next, in Î”ACD

tan 60^{o} = AC/DC

âˆš3 = (x + h)/9

x + h = 9âˆš3

x + 3âˆš3 = 9âˆš3

x = 6âˆš3 m

Therefore,

Height of the tower = 3âˆš3 m

And, height of the pole = 6âˆš3 m

**19. A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle of 30Â° with the ground. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.**

**Solution:**

Let the initial height of the tree be AC.

And, due to storm the tree is broken at B.

Let the bent portion of the tree be AB = x m and the remaining portion BC = h m

So, the height of the tree AC = (x + h) m

And, given DC = 8m

Now, in Î”BCD

tan 30^{o} = BC/DC

1/âˆš3 = h/8

h = 8/âˆš3

Next, in Î”BCD

cos 30^{o} = DC/BD

âˆš3/2 = 8/x

x = 16/âˆš3 m

So, x + h = 16/âˆš3 + 8/âˆš3

= 24/âˆš3 = 8âˆš3

Therefore, the height of the tree is 8âˆš3 m.

**20. From a point P on the ground the angle of elevation of a 10 m tall building isÂ 30Â°. A flag is hoisted at the top of the building and the angle of elevation of the top of the flag staff from P isÂ 45Â°. Find the length of the flag staff and the distance of the building from the point P.**

**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 = 30^{o}

Angle of elevation of top of the flag staff = 45^{o}

From the fig.

In Î”BQP,

tan 30^{o} = BQ/PQ

1/âˆš3 = 10/x

x = 10âˆš3 m

Next,

In Î”AQP,

tan 45^{o} = AQ/PQ

1 = (h + 10)/x

h + 10 = x = 10âˆš3

h = (10âˆš3 â€“ 10) = 10(1.732) â€“ 10 = 17.32 â€“ 10

= 7.32 m

Therefore, the distance of point P from building = x = 10âˆš3 = 10(1.732) = 17.32m

**21. A 1.6 m tall girl stands at a distance of 3.2 m from a lamp post and casts a shadow of 4.8 m on the ground. Find the height of the lamp post by using (i) trigonometric ratio (ii) properties of similar triangles.**

**Solution:**

Let AC be the lamp post of height â€˜hâ€™

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

InÂ Î”BDE,

tan Î¸ = 1.6/4.8

tan Î¸ = 1/3

Next, InÂ Î”ABC

tan Î¸ = 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

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

**22. A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30Â° to 60Â° as he walks towards the building. Find the distance he walked towards the building.**

**Solution: **

Given,

The height of the tall boy (AS) = 1.5 m

The length of the building (PQ) = 30 m

Let the initial position of the boy be S. And, then he walks towards the building and reached at the point T.

From the fig. we have

AS = BT = RQ = 1.5 m

PR = PQ â€“ RQ = (30 â€“ 1.5)m = 28.5 m

In Î”PAR,

tan 30^{o} = PR/AR

1/ âˆš3 = 28.5/ AR

AR = 28.5âˆš3

In Î”PRB,

tan 60^{o} = PR/BR

âˆš3 = 28.5/ BR

BR = 28.5/âˆš3 = 9.5âˆš3

So, ST = AB = AR â€“ BR = 28.5âˆš3 â€“ 9.5âˆš3 = 19âˆš3

Therefore, the distance which the boy walked towards the building is 19âˆš3 m.

**23.** **The shadow of a tower standing on level ground is found to be 40 m longer when Sunâ€™s altitude is 30Â° than when it was 60Â°. Find the height of the tower.**

**Solution: **

When the sunâ€™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, DB is 40 m longer than BC.

So, BD = (40 + x) m

And two right triangles ABC and ABD are formed.

In Î”ABC,

tan 60^{o} = AB/ BC

âˆš3 = h/x

x = h/âˆš3 â€¦ (i)

In Î”ABD,

tan 30^{o} = AB/ BD

1/ âˆš3 = h/ (x + 40)

x + 40 = âˆš3h

h/âˆš3 + 40 = âˆš3h [using (i)]

h + 40âˆš3 = 3h

2h = 40âˆš3

h = 20âˆš3

Therefore, the height of the tower is 20âˆš3 m.

**24. From a point on the ground the angle of elevation of the bottom and top of a transmission tower fixed at the top of 20 m high building areÂ 45Â°Â andÂ 60Â°Â respectively. Find the height of the transmission tower.**

**Solution:**

Given,

Height of the building = 20 m = AB

Let height of tower above building = h = BC

Height of tower + building = (h + 20) m [from ground] = CA

Angle of elevation of bottom of tower,Â Î± = 45Â°

Angle of elevation of top of tower,Â Î² = 60Â°

Let distance between tower and observation point = x m

Then, from the fig. we have

x = 20 m

Next,

h =Â 20(1.73 âˆ’ 1) = 20 x .732 = 14.64

Therefore, the height of the tower is 14.64m

**25. The angle of depression of the top and bottom of 8 m tall building from the top of a multistoried building areÂ 30Â°Â andÂ 45Â°. Find the height of the multistoried building and the distance between the two buildings.**

**Solution:**

Let height of multistoried building ‘h’ m = AD

Height of the tall building = 8 m = BE

Angle of depression of top of the tall building from the multistoried building = 30Â°

Angle of depression of bottom of the tall building from the multistoried building = 30Â°

And, let the distance between the two buildings = ‘x’ m = ED

So, BC = x and CD = 8 m [As BCDE forms a rectangle]

AD = AC + CD

So, AC = (h â€“ 8) m

From the fig. we have

InÂ Î”BCA

tan 30^{o} = AC/ BC

1/âˆš3 = (h – 8)/x

x = âˆš3(h – 8)â€¦â€¦ (i)

InÂ Î”ADE

tan 45^{o} = AD/ ED

1 = h/ x

h = x

h = âˆš3(h – 8) [using (i)]

h = âˆš3h – 8âˆš3

(âˆš3 – 1)h = 8âˆš3

h = 8âˆš3/ (âˆš3 -1)

Rationalising the denominator by (âˆš3 + 1), we have

h = 8âˆš3(âˆš3 + 1)/ (3 – 1)

h = 4âˆš3(âˆš3 + 1)

h = 12 + 4âˆš3

= 4(3 + âˆš3)

Thus, x = h = 4(3 + âˆš3)

Therefore, the height of the building is 4(3 + âˆš3)m and the distance between the building is also 4(3 + âˆš3)m

**26. A statue 1.6 m tall stands on the top of a pedestal. From a point on the ground, angle of elevation of the top of the statue isÂ 60Â°Â and from the same point the angle of elevation of the top of the pedestal isÂ 45Â°. Find the height of the pedestal.**

**Solution:**

Letâ€™s assume AB as the statue, BC be the pedestal and D be the point on ground from where eleveation angles are measured.

Given,

Angle of elevation of the top of statue is 60^{o}

Angle of elevation of the top of the pedestal is 45^{o}

So, from the fig. we have

InÂ Î”BCD,

tan 45^{o} = BC/ CD

1 = BC/ CD

CD = BC

Next, InÂ Î”ADE

tan 60^{o} = (AB + BC)/ CD

âˆš3 = (AB + BC)/ BC [As CD = BC, found above]

BCâˆš3 = AB + BC

AB = (âˆš3 – 1)BC

BC = AB/ (âˆš3 – 1)

BC = 1.6/ (âˆš3 – 1)

Rationalising the denominator, we have

BC = 1.6(âˆš3 + 1)/ 2

BC = 0.8(âˆš3 + 1) m

Therefore, the height of pedestal is 0.8(âˆš3 + 1) m

**27. A T.V. tower stands vertically on a bank of a river of a river. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower isÂ 60Â°. From a point 20 m away this point on the same bank, the angle of elevation of the top of the tower isÂ 30Â°. Find the height of the tower and the width of the river.**

**Solution:**

Let AB be the T.V tower of height ‘h’ m on the bank of river and ‘D’ be the point on the opposite side of the river. An angle of elevation at the top of the tower isÂ 30Â°

Let AB = h and BC = x

And, given CD = 20 m

From fig. we have

InÂ Î”ACB

Next, inÂ Î”DBA,

h = 10âˆš3 m

And,

x = h/âˆš3

x = 10âˆš3/ âˆš3

x = 10

Therefore, the height of the tower isÂ 10âˆš3 mÂ and width of the river is 10 m.

**28. From the top of a 7 m high building, the angle of elevation of the top of a cable isÂ 60Â°Â and the angel of depression of its foot isÂ 45Â°. Determine the height of the tower.**

**Solution:**

Given

Height of the building = 7 m = AB

Height of the cable tower = CD

Angle of elevation of the top of the cable tower from the top of the building = 60Â°

Angle of depression of the bottom of the buildingÂ from the top of the building= 45Â°

Then, from the fig. we see that

ED = AB = 7 m

And,

CD = CE + ED

So, InÂ Î”ABD, we have

AB/ BD = tan 45^{o}

AB = BD = 7

BD = 7

InÂ Î”ACE,

AE = BD = 7

And, tan 60^{o} = CE/AE

âˆš3 = CE/ 7

CE = 7âˆš3 m

So, CD = CE + ED = (7âˆš3 + 7)= 7(âˆš3 + 1) m

Therefore, the height of the cable tower is 7(âˆš3 + 1)m

**29. As observed from the top of a 75 m tall lighthouse, the angles of depression of two ships areÂ 30Â°Â andÂ 45Â°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.**

**Solution:**

Given;

Height of the lighthouse = 75m = ‘h’ m = AB

Angle of depression of ship 1,Â Î± = 30Â°

Angle of depression of bottom of the tall building,Â Î² = 45Â°

The above data is represented in form of figure as shown

Let distance between ships be ‘x’ m = CD

If in right angle triangle one of the included angle isÂ Î¸

BC = 75 … (2)

Substituting (2) in (1)

Therefore, the distance from the ships is 75(âˆš3 + 1) m

**30. The angle of elevation of the top of the building from the foot of the tower isÂ 30Â°Â and the angle of the top of the tower from the foot of the building isÂ 60Â°. If the tower is 50 m high, find the height of the building.**

**Solution:**

Let AB be the building and CD be the tower.

Given,

The angle of elevation of the top of the building from the foot of the tower is 30^{o}.

And, the angle of elevation of the top of the tower from the foot of the building is 60^{o}.

Height of the tower = CD = 50 m

From the fig. we have

InÂ Î”CDB,

CD/ BD = tan 60^{o}

50/ BD = âˆš3

BD = 50/âˆš3 â€¦. (i)

Next inÂ Î”ABD,

AB/ BD = tan 30^{o}

AB/ BD = 1/âˆš3

AB = BD/ âˆš3

AB = 50/âˆš3/ (âˆš3) [From (i)]

AB = 50/3

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

**31. From a point on a bridge across a river the angle of depression of the banks on opposite side of the river areÂ 30Â°Â andÂ 45Â°Â respectively. If the bridge is at the height of 30 m from the banks, find the width of the river.**

**Solution:**

Given,

The bridge is at a height of 30 m from the banks.

Let, A and B represent the points on the bank on opposite sides of the river. And, AB is the width of the river. P is a point on the bridge which is at the height of 30 m from the banks.

Now, from the fig, we have

AB = AD + DB

In right Î”APD,

âˆ A = 30^{o}

So, tan 30^{o} = PD/ AD

1/âˆš3 = PD/ AD

AD = âˆš3(30)

AD = 30âˆš3 m

Next, in right Î”PBD

âˆ B = 45^{o}

So, tan 45^{o} = PD/ BD

1 = PD/ BD

BD = PD

BD = 30 m

We know that, AB = AD + DB = 30âˆš3 + 30 = 30(âˆš3 + 1)

Hence, the width of the river = 30(1 + âˆš3)m

**32.** **Two poles of equal heights are standing opposite to each other on either side of the road which is 80 m wide. From a point between them on the road the angles of elevation of the top of the poles are 60Â° and 30Â° respectively. Find the height of the poles and the distances of the point from the poles.**

**Solution: **

Given,

Distance between the poles = 80 m = BD

Let the point of observation of the angles be O.

The angles of elevation to the top of the points is 60^{o} and 30^{o}

Let AB and CD be the poles and O is the point on the road.

From the fig. we have

In Î”APD,

AB/ BO = tan 60^{o}

AB/ BO = âˆš3

BO = AB/ âˆš3 â€¦.(i)

Next in Î”CDO,

CD/ DO = tan 30^{o}

CD/ (80 – BO) = 1/ âˆš3

âˆš3CD = 80 – BO

âˆš3AB = 80 â€“ (AB/âˆš3) [As AB = CD and using (i)]

3AB = 80âˆš3 â€“ AB

4AB = 80âˆš3

AB = 20âˆš3

So, BO = 20âˆš3/(âˆš3) = 20 m

And, DO = BD â€“ BO = 80 â€“ 20 = 60

Hence, the height of the poles is 20âˆš3 m and the point between the poles is 20m and 60 m far from these poles.

**33. A man sitting at a height of 20 m on a tall tree on a small island in the middle of a river observes two poles directly opposite to each other on the two banks of the river and in line with foot of tree. If the angles of depression of the feet of the poles from a point at which the man is sitting on the tree on either side of the river areÂ 60Â°Â andÂ 30Â°Â respectively. Find the width of the river.**

**Solution:**

From the given data, the fig. is made

Let width of river = PQ = (x + y) m

Height of tree (AB) = 20 m

So, in Î”ABP

tan 60^{o} = AB/ BP

âˆš3 = 20/ x

x = 20/ âˆš3 m

In Î”ABQ,

tan 30^{o} = AB/ BQ

1/ âˆš3 = 20/ y

y = 20âˆš3

So, (x + y) = 20/ âˆš3 + 20âˆš3

= [20 + 20(3)]/ âˆš3

= 80/âˆš3

Therefore, the width of the river is 80/âˆš3 m.

**34. A vertical tower stands on a horizontal plane and is surmounted by a flag staff of height 7m. From a point on the plane, the angle of elevation of the bottom of flag staff isÂ 30Â°Â and that of the top of the flag staff isÂ 45Â°. Find the height of the tower.**

**Solution:**

Given,

The length of the flag staff = 7 m

Angles of elevation of the top and bottom of the flag staff from point D is 45^{o }and 30^{o} respectively.

Let the height of tower (BC) = h m

And, let DC = x m

So, in Î”ABP

tan 30^{o} = BC/DC

1/ âˆš3 = h/ x

x = hâˆš3â€¦. (i)

And, in Î”ACD

tan 45^{o} = AC/ DC

1 = (7 + h)/ x

x = 7 + h

hâˆš3 = 7 + h [from (i)]

h(âˆš3 – 1) = 7

h = 7/(âˆš3 – 1)

Now, rationalising the denominator we get

h = 7(âˆš3 + 1)/ 2 = 7(1.732 + 1)/2 = 9.562

Therefore, the height of the tower is 9.56 m

**35. The length of the shadow of a tower standing on level plane is found to be 2x meters longer when the sun’s attitude isÂ 30Â°Â than when it wasÂ 30Â°. Prove that the height of tower isÂ x(âˆš3+1) meters.**

**Solution:**

From the question, the following fig. is made

Let the height of tower (AB) = h m

Let the distance BC = y m

Then, in Î”ABC

tan 45^{o} = AB/BC

1 = h/y

y = h

Next, in Î”ABD

tan 30^{o} = AB/BD

1/âˆš3 = h/ (2x + y)

2x + y = âˆš3h

2x + h = âˆš3h

2x = (âˆš3 – 1)h

h = 2x/ (âˆš3 – 1)

Therefore, the height of the tower is 2x (âˆš3 -1) m

- Hence Proved

**36. A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle ofÂ 30Â°Â with the ground. The distance from the foot of the tree to the point where the top touches the ground is 10 meters. Find the height of the tree.**

**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 30^{o} with the ground.

From the fig.

In Î”BCD, we have

tan 30^{o} = BC/ DC

1/âˆš3 = h/ 10

h = 10/ âˆš3

Next, in Î”BCD

cos 30^{o} = DC/BD

âˆš3/2 = 10/x

x = 20/âˆš3 m

So,

x + h = 20/âˆš3 + 10/âˆš3

= 30/âˆš3

= 10âˆš3 = 10(1.732) = 17.32

Therefore, the height of the tree is 17.32 m

**37. A balloon is connected to a meteorological ground station by a cable of length 215 m inclined atÂ 60Â°Â to the horizontal. Determine the height of the balloon from the ground. Assume that there is no slack in the cable.**

**Solution:**

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

Given, the length of the cable = 215 m and the inclination of the cable is 60^{o}.

In Î”ABC

sin 60^{o} = AB/ AC

âˆš3/2 = h/215

h = 215âˆš3/2 = 185.9

Hence, the height of the balloon from the ground is 186m (approx).