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

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given 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 \(20\sqrt { 3 } m=34.64m \)

**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

**More Resources**

**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 = \(\frac { { 200 } }{ { \sqrt { 3 } } } \times \frac { { \sqrt { 3 } } }{ { \sqrt { 3 } } } \) = 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 \(60(1+\sqrt { 3 } ) \)

**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),

\(20\sqrt { 3 } =\sqrt { 3 } h \)

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°

Hope given RS Aggarwal Solutions Class 10 Chapter 14 Height and Distance are helpful to complete your math homework.

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