# the angle of elevation of the top q of a vertical tower pq from a point x on the ground is 60 degree

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## The angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is 60^∘ . From a point Y, 40 m vertically above X, the angle of elevation of the top Q of tower is 45^∘ . Find the height of the tower PQ and the distance PX. (Use √(3) = 1.73 )

Click here👆to get an answer to your question ✍️ The angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is 60^∘ . From a point Y, 40 m vertically above X, the angle of elevation of the top Q of tower is 45^∘ . Find the height of the tower PQ and the distance PX. (Use √(3) = 1.73 )

The angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is 60Question ∘

. From a point Y, 40 m vertically above X, the angle of elevation of the top Q of tower is 45

∘

. Find the height of the tower PQ and the distance PX. (Use

3 =1.73) Hard Open in App

Updated on : 2022-09-05

Solution Verified by Toppr In ΔYRQ, we have tan45 o = YR QR 1= YR x

YR=x or XP=x [because YR=XP] ---- (1)

Now In ΔXPQ we have tan60 o = PX PQ 3 = x x+40 (from equation 1) x( 3 −1)=40 x= 3 −1 40 x= 1.73−1 40 =54.79 m

So, height of the tower, PQ=x+40=54.79+40=94.79 m.

Distance PX=54.79 m.

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## The angle of elevation of the top Q of a vertical tower PQ, from a point X on the ground is 60∘ . From a point Y , 40 m vertically above X, the angle of elevation of the top Q of tower is 45∘. Find the height of the tower PQ and the distance PX. Use √3=1.73

The angle of elevation of the top Q of a vertical tower PQ, from a point X on the ground is 60∘ . From a point Y , 40 m vertically above X, the angle of elevation of the top Q of tower is 45∘. Find the height of the tower PQ and the distance PX. Use √3=1.73

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The angle of elevation of the top Q of a vertical tower PQ, from a point X on the ground is 60∘ . From a point Y , 40 m vertically above X, the angle of elevation of the top Q of tower is 45∘. Find the height of the tower PQ and the distance PX. Use √3=1.73

Question

The angle of elevation of the top Q of a vertical tower PQ, from a point X on the ground is

60 ∘

. From a point Y, 40 m vertically above X, the angle of elevation of the top Q of tower is

45 ∘

. Find the height of the tower PQ and the distance PX. (Use

√ 3 = 1.73 ) Open in App Solution

We have PQ as a vertical tower

In Δ Y Z Q t a n 45 ∘ = Q Z Y Z Q Z Y Z = 1 Q Z = Y Z . . . ( i ) And, in Δ X P Q t a n 60 ∘ = Q P X P √ 3 = Q Z + 40 X P √ 3 = Q Z + 40 Y Z ( ∵ X P = Y Z ) √ 3 Q Z = Q Z + 40 [Using (i)] √ 3 Q Z − Q Z = 40 Q Z ( √ 3 − 1 ) = 40 Q Z = 40 √ 3 − 1 = 40 √ 3 − 1 × √ 3 + 1 √ 3 + 1 = 20 ( √ 3 + 1 ) = 20 ( 2.73 ) = 54.60 m ∴ Q Z = 54.6 m And P Q = X Y + Q Z = ( 54.6 + 40 ) m = 94.6 m. Suggest Corrections 47

## The Angle of Elevation of the Top Q of a Vertical Tower Pq from a Point X on the Ground is 60° . at a Point Y, 40m Vertically Above X, the Angle of Elevation is 45° . Find the Height of Tower Pq.

The Angle of Elevation of the Top Q of a Vertical Tower Pq from a Point X on the Ground is 60° . at a Point Y, 40m Vertically Above X, the Angle of Elevation is 45° . Find the Height of Tower Pq.

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The angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is 60° . At a point Y, 40m vertically above X, the angle of elevation is 45° . Find the height of tower PQ.

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### SOLUTION

We have

XY = 40m,∠PXQ = 60° and ∠MYQ = 45°

Let PQ = h

Also, MP = XY = 40m, MQ = PQ - MP = h - 40

In ΔMYQ, tan45°=MQMY ⇒1=h-40MY ⇒ MY = H - 40

⇒ PX = MY = h - 40 ................(1)

Now , in ΔMXQ, tan60°=PQPX ⇒3=hh-40 [From (i)] ⇒h3-403=h ⇒h3-h=403 ⇒h (3-1)=403 ⇒h=403(3-1)

⇒h=403(3-1)×(3+1)(3+1)

⇒h=403(3+1)(3-1) ⇒h=403(3+1)2 ⇒h=203(3+1) ⇒h=60+203 ⇒h=60+20×1.73 ⇒h=60+34.6 ∴ h = 94.6m

So, the height of the tower PQ is 94. 6 m.

Concept: Heights and Distances

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Chapter 14: Height and Distance - Exercises

Q 24 Q 23 Q 25

### APPEARS IN

RS Aggarwal Secondary School Class 10 Maths

Chapter 14 Height and Distance

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