**Q.1: A tower stands vertically on the ground. From a point on the ground which is 20m away from the foot of the tower, the angle of elevation of its top is found to be \(60^{\circ}\). Find the height of the tower. [Take \(\sqrt{3}=1.732\)] **

**Sol:**

Let AB be the tower standing on a level ground and 0 be the position of the observer. Then OA = 20 m and ∠OAB = 90° and ∠AOB = 60°

Let AB = h meters

From the right ⧍OAB, we have

\(\frac{AB}{OA}\) = Tan 60^{o} = √3

\(\frac{h}{20}\) = √3

h = (20 x √3)

h = 20 x 1.732

h = 34.64m

Hence the height of the tower is 20√3 m = 34.64m

**Q.2: A kite is flying at a height of 75m from the level ground, attached to a string inclined at \(60^{\circ}\) to the horizontal. Find the length of the string, assuming that there is no slack in it. [Take \(\sqrt{3}=1.732\)] **

**Sol:**

Let OB be the length of the string from the level of ground and 0 be the point of the observer, then,

AB = 75m and ∠OAB = 90° and ∠AOB = 60°, let OB = I meters.

From the right ⧍OAB:

\(\frac{OB}{AB}\) = cosec 60° = \(\frac{2}{√3}\)

\(\frac{I}{75}\) = \(\frac{2}{√3}\)

I = ( 75 x \(\frac{2}{√3}\) x \(\frac{√3}{√3}\) )

I = 25 x 2 x √3

I = 50√3

I = 86.6m

Hence, the length of the string 86.6m

**Q.3: The angle of depression from the top of a tower of a point A on the ground is \(30^{\circ}\). On moving a distance of 20 meters from the point A towards the foot of the tower to a point B, the angle of elevation of the top of the tower from the point B is \(60^{\circ}\). Find the height of the tower and its distance from the point A.**

**Sol:**

Let CD be the tower and BD be the ground

Then, ∠CBD = 30°, ∠CAD = 60°

∠BDC = 90°, AB = 20 m, CD = h metre and AD = x metre.

From ∆BCD:

\(\frac{CD}{BD}\) = tan 30°

= \(\frac{CD}{BD}\)

\(\frac{h}{20 + x}\) = \(\frac{1}{√3}\) = √3h = 20 + x

√3h = 20 + x

=> x = √3h – 20. …………(1)

From right ∆CAD, we have

\(\frac{CD}{AD}\) = tan 60°

\(\frac{h}{x}\) = √3

\(\frac{h}{√3}\) = x ……….(2)

From (1) and (2) we get,

√3h – 20 = \(\frac{h}{√3}\)

=> 3h – 20√3 = h

=> h = 10√3 = 10×1.732 => 17.32

H = 17.32m and BD = 30m

Hence, the height of the tower = 17.32m and the distance of the tower from the point A = 30m.

**Q.4: A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height 5 m. At a point on the plane, the angle of elevation of the bottom of the flagstaff is \(30^{\circ}\) and that of the top of the flagstaff is \(60^{\circ}\). Find the height of the tower. [Use \(\sqrt{3}= 1.73\)]**

**Sol:**

Let AB be the tower h meter high. CA is the flag staff 5 meter high.

Let PB = x meter

In ∆ PBC,

∠CPB = 60°, ∠PBC = 90°

\(\frac{BC}{PB}\) = tan 60°

\(\frac{5+x}{x}\) = √3

5 + h = √3x ……….(1)

∠APB = 30° and ∠ABP = 90°

\(\frac{AB}{PB}\) = tan 30°

\(\frac{h}{x}\) = \(\frac{1}{√3}\)

√3h = x

Putting value of x in (1), we get

5 + h = √3 x √3h => 3h

2h = 5

Or h = 5/2m

Thus, height of tower = 2.5m

**Q.5: A statue 1.46m tall stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is \(60^{\circ}\) and from the same point, the angle of elevation of the top of the pedestal is \(45^{\circ}\). Find the height of the pedestal. [Use \(\sqrt{3}= 1.73\)]**

**Sol:**

Let SP be the statue and PB be the pedestal. Angles of elevation of S and P are 60° and 45° respectively.

Further suppose AB = x m, PB = h m

In right ∆ ABS,

\(\frac{SB}{AB}\) = tan 60° = √3

\(\frac{h+1.46}{x}\) = √3 ……(1)

In right ∆ PAB,

\(\frac{PB}{AB}\) = tan 45° = 1

h = x ……….(2)

putting x = h in (1)

\(\frac{h + 1.46}{h}\) = √3

=> h + 1.46 = √3 h

h = 0.73 x 2.732 = 2m (nearly)

Thus, the height of the pedestal = 2m.

**Q.6: The angle of elevation of the top of a tower from a point on the same level as the foot of the tower is \(30^{\circ}\). On advancing 150m towards the foot of the tower, the angle of elevation becomes \(60^{\circ}\). Show that the height of the tower is 129.9 metres. [Given \(\sqrt{3}=1.732\)]**

**Sol:**

Let AB be the tower and let the angle of elevation of its top at C be 30°. Let D be a point at a distance 150m from C such that the angle of elevation of the top of the tower at D is 60°. Let h m be the height of the tower and AD = x m.

In ∆ CAB, we have

Tan 30° = \(\frac{AB}{AC}\)

=> \(\frac{1}{√3}\) = \(\frac{h}{x+150}\) ……….(1)

In ∆ DAB, we have

Tan 60° = \(\frac{AB}{AD}\) => √3 = \(\frac{h}{x}\)

X = \(\frac{h}{√3}\) ……(2)

Putting the x = \(\frac{h}{√3}\) in (1), we get

=> \(\frac{1}{√3}\) = \(\frac{√3h}{h+150√3}\)

=> h+150√3 = 3h => 3h – h = 150√3

2h = 150√3

h = 75√3

h = (75×1.732)m

h = 129.9

Hence the height of tower is 129.9 m

**Q.7: On a horizontal plane there is a vertical tower with a flagpole on the top of the tower. At a point, 9 meters away from the foot of the tower, the angle of elevation of the top and bottom of the flagpole are \(60^{\circ}\:and\:30^{\circ}\) respectively. Find the height of the tower and the flagpole mounted on it. [Take \(\sqrt{3}=1.73\)]**

**Sol:**

Let AB be the tower and BC be the flagpole, Let O be the point of observation. Then, OA = 9 m, ∠AOB = 30° and ∠AOC = 60°

From right angled ∆BOA:

\(\frac{AB}{OA}\) = tan 30°

\(\frac{AB}{9}\) = \(\frac{1}{√3}\) => AB = 3√3

From right angled ∆OAC

\(\frac{AC}{OA}\) = tan 60°

BC = (AC – AB) => 6√3m.

Thus, AB = 3√3m => 5.196m

BC = 6√3m => 10.392m

Hence, height of the tower= 5.196 m and,

height of the flagpole = 10.392 m

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