### RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance

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## RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance

**Exercise 14**

**Question 1:**

Let AB be the tower standing on a level ground and O 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

Hence the height of the tower is [latex]20sqrt { 3 } m=34.64m [/latex]

**Question 2:**

Let OB be the length of the string from the level of ground and O be the point of the observer, then, AB = 75m and ∠OAB = 90° and ∠AOB = 60°, let OB = l meters.

From the right ∆OAB, we have

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**Question 3:**

Let AB be the man,

AB= 1.6m, CD is the tower

AE CD, DE = AB

Let CE = h

In ∆ACE,

Height of tower = DE + DC = (1.6 + 25.98)m = 27.58 m

**Question 4:**

Let AB be the tree bent at the point C so that part CB takes the position CD, then CD = CB

Let AC = x meters

Then, CD = CB = (10 – x) m

and ∠ADC = 60°

Therefore, tree bent at the height of 4.64m from the bottom.

**Question 5:**

Let AB be the lamp post and CD be the boy, let CE be the shadow of CD

Let, ∠AEB = θ

From right ∆ECD, we get

From right ∆EAB, we get

Hence, the height of the lamp post = 2.5 m

**Question 6:**

Let CD be the height of the building

Then, ∠CAB = 30°, ∠CBD = 45°,

∠ADC = 90° and AB = 30m

CD = h meters and BD = x meters

From right ∆CAD, we have

From right ∆BCD, we have

From (1) and (2), we get

Putting h = 40.98m in (2), we get x = 40.98 m

Hence, height of building = 40.98m and

Distance of its base from the point A

= AB = (30+x) m

= (30+40.98) m = 70.98 m

**Question 7:**

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

From right ∆CAD, we have

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

**Question 8:**

Let AB and CD be the building and the tower respectively.

AB = 15 m, AE ⊥ CD

ED = AB = 15 m

Let EC = h m

And BD = AE = x m

In CAE,

∠CAE = 30°and ∠AEC = 90°

In CBD, ∠CBD = 60° and ∠CDB = 90°

Eliminating x from (1) and (2), we get

Height of tower = CE + ED = (h + 15) m

= (7.5 + 15) m = 22.5m

Hence, Height of the tower = 22.5 m and the distance between the tower and the building = 12.99 m

**Question 9:**

AB and CD are the two houses.

Window is at A.

In ∆ ABD, ∠B = 90°, AB = 15m

AE is drawn perpendicular to CD

Therefore, AE = BD = 15 m

Let CE = h m

In ∆ ACE,

∠CAE = 30°, ∠CEA = 90°

Height of opposite house = CE + ED

= (h + 15) m = (8.66 + 15) m = 23.66 m

Hence proved.

**Question 10:**

Let AB be the tower with height = h m

AC = flag staff = x m

PB = 30 m

In ∆ PBC,

∠CPB = 60° and ∠CBP = 90°

Putting value of h in (1), we get

Thus, height of tower = 30m and height of flag staff = 21.96 m

**Question 11:**

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

Let PB = x meter

In ∆ PBC,

∠CPB = 60°, ∠PBC = 90°

In ∆ APB,

∠APB = 30° and ∠ABP = 90°

Putting value of x in (1), we get

Thus, height of tower = 2.5m

**Question 12:**

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,

In right ∆ PAB,

Thus, height of the pedestal = 2m

**Question 13:**

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 150 m from C such that the angle of elevation of the top of tower at D is 60°.

Let h m be the height of the tower and AD = x m

In ∆ CAB, we have

Hence the height of tower is 129.9 m

**Question 14:**

Let AB be the tower and BC be flagpole, Let O be the point of observation.

Then, OA = 9 m, ∠AOB = 30° and ∠AOC = 60°

From right angled ∆ BOA

From right angled ∆ OAC

Thus

Hence, height of the tower= 5.196 m and the height of the flagpole = 10.392 m

**Question 15:**

Let AB be the hill and let CD be the pillar. Draw DE AB, then, ∠ACB = 60° and ∠EDB = 30° and AB = 200 m

Height of the pillar = CD = 133.33 m

Distance of the pillar from the hill = ED = [latex]frac { { 200 } }{ { sqrt { 3 } } } times frac { { sqrt { 3 } } }{ { sqrt { 3 } } } [/latex] = 115.33m

**Question 16:**

Let AB be the height of the window of house and CD be another house on the opposite side of the street AC

Then, AB = 60 m

Draw BE ⊥ CD and join BC

Then, ∠EBD = 60° and ∠ACB = ∠CBE = 45°

From right ∆ CAB, we have

From right ∆ BED, we have

Hence, the height of the opposite house is [latex]60(1+sqrt { 3 } ) [/latex]

**Question 17:**

Let O and B the two positions of the jet plane and let A be the point of observation.

Let AX be the horizontal ground.

Draw OC ⊥ AX and BD ⊥ AX.

Then, ∠CAO = 60°, ∠DAB = 30° and OC = BD = 1500√3 m

From right ∆ OCA, we have

From right ∆ ADB, we have

Thus, the aeroplane covers 3000 m in 15 seconds

Hence the speed of the aeroplane is

**Question 18:**

Let AB be the building and CD be the light house.

AE is drawn perpendicular to CD.

Now AB = 60 m

∠ADB = 60°, ∠CAE = 30°

Let BD = x m

AE = BD = x m

In right ∆ ACE, let CE = h

From (1) and (2),

[latex]20sqrt { 3 } =sqrt { 3 } h [/latex]

h = 20 m

Hence,

(i) Difference of heights of light house and building = 20m

(ii) The distance between light house and building = 34.64m

**Question 19:**

Let AB be the light house and let C and D be the positions of the ship.

Llet AD =x, CD = y

In ∆ BDA,

The distance travelled by the ship during the period of observation = 115.46 m

**Question 20:**

Let CD be the height of the building

Then, ∠CAB = 30°, ∠CBD = 45°, ∠ADC = 90° and AB = 30m

CD = h metres and BD = x metres.

From right ∆ CAD, we have

From right ∆ BCD, we have

Putting h = 40.98 in (2), we get x = 40.98 m

Hence height of building = 40.98 m and Distance of its base from the point

A = AB = (30 + x) m

= (30 + 40.98) m = 70.98 m

**Question 21:**

Let CD be a tree. Angle of elevation from A and B are 60° and 30° respectively.

Let AD = x m and CD = h m

In right ∆ ACD,

Height of the tree = 17.32 m

**Question 22:**

Let AB be the building 7 meters high. AE ⊥ CD, where CD is the cable tower.

In ∆ AED,

∠EAD = 30° = Angle of depression

Height of the tower = CD = CE + ED = (21 + 7) m = 28 m

**Question 23:**

Let AB be the tower and let C and D be the two positions of the observer. Then, AC = 9 meters, and AD = 4 meters.

Let ∆ ACB = θ

Then, ∠ADB = (90° – θ)

Let AB = h meters

From right ∆ CAB, we have

From right ∆ DAB, we have

Hence, the height of tower is 6 meters.

**Question 24:**

Let P be the point of observation RQ is the building and BR is the flag staff of height h, ∠BPQ = 45°, ∠RPQ = 30°

From (1) and (2), we have

Hence distance of building is and length of the flags staff is 7.3 m

**Question 25:**

Let AB be the 10 m high building and let CD be the multi – storey building. Draw BE ⊥ CD

Then, ∠DBE = 30° and ∠DAC = 45°

Let ED = x meters

Height of the Multi – storey building = (10 + 13.66)m = 23.66 m

Distance between two building = (10 + 13.66) m = 23.66 m

**Question 26:**

Let A and B be two points on the bank on opposite sides of the river. Let P be a point on the bridge at a height of 2.5 m

Thus, DP = 2.5 m

Then, ∠BAP = 30°, ∠ABP = 45° and PD = 2.5m

Height of the river = AB

**Question 27:**

Let AB be the tower. Let C and D be the positions of the two men.

Then, ∠ACB = 30°, ∠ADB = 45° and AB = 50 m

**Question 28:**

Let AB and CD be the first and second towers respectively.

Then, CD = 90 m and AC = 60 m.

Let DE be the horizontal line through D.

Draw BF ⊥ CD,

Then, BF = AC = 60 m

∠FBD = ∠EDB = 30°

**Complete RS Aggarwal Solutions Class 10**

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