Question

12- At two points on the opposite sides of a pole the angles of elevation of the top of the pole are 30' and 45. If the height of the pole is 20m, find the distance between the two points​

Answer 1

Step-by-step explanation:

Given :-

At two points on the opposite sides of a pole the angles of elevation of the top of the pole are 30° and 45°.

The height of the pole is 20m.

To find :-

The distance between the two points.

Solution :-

The height of the pole = 20 m

Angles of elevation = 30° and 45°

On converting the given data in the pictorial representation then we get a triangle,

Height of the pole = AD = 20 m

Angles of elevation = ∠ABD = 30° and

∠ ACD = 45°

Observation points are B and C

The distance between the two points = BC

In the right angled triangle ABD,

We know that

tan ∠ABD = Opposite side to ∠ABD / Adjacent side to ∠ABD

=> tan 30° = AD / BD

=> 1/√3 = 20/BD

=> BD × 1 = 20×√3

=> BD = 20√3 m

Therefore, BD = 20√3 m -----------(1)

and

In the right angled triangle ACD,

We know that

tan ∠ACD = Opposite side to ∠ACD / Adjacent side to ∠ACD

=> tan 45° = AD / CD

=> 1 = 20/CD

=> CD×1 = 20

=> CD = 20 m

Therefore, CD = 20 m -----------(2)

We have,

BC = BD + DC

=> BC = 20√3+20

=> BC = 20(√3+1) m

(or)

We know that √3 = 1.732

=> BC = 20(1.732+1)

=> BC = 20(2.732)

=> BC =54.64 m

Therefore, BC = 54.64 m

Answer :-

The distance between two points is 20(√3+1) m (or) 54.64 m

Used formulae:-

→ tan A = Opposite side to A/Adjacent side to A

→ tan 30° = 1/√3

→ tan 45° = 1

→ √3 = 1.732...

Answer 2

Step-by-step explanation:

Step-by-step explanation:

Given :-

At two points on the opposite sides of a pole the angles of elevation of the top of the pole are 30° and 45°.

The height of the pole is 20m.

To find :-

The distance between the two points.

Solution :-

The height of the pole = 20 m

Angles of elevation = 30° and 45°

On converting the given data in the pictorial representation then we get a triangle,

Height of the pole = AD = 20 m

Angles of elevation = ∠ABD = 30° and

∠ ACD = 45°

Observation points are B and C

The distance between the two points = BC

In the right angled triangle ABD,

We know that

tan ∠ABD = Opposite side to ∠ABD / Adjacent side to ∠ABD

=> tan 30° = AD / BD

=> 1/√3 = 20/BD

=> BD × 1 = 20×√3

=> BD = 20√3 m

Therefore, BD = 20√3 m -----------(1)

and

In the right angled triangle ACD,

We know that

tan ∠ACD = Opposite side to ∠ACD / Adjacent side to ∠ACD

=> tan 45° = AD / CD

=> 1 = 20/CD

=> CD×1 = 20

=> CD = 20 m

Therefore, CD = 20 m -----------(2)

We have,

BC = BD + DC

=> BC = 20√3+20

=> BC = 20(√3+1) m

(or)

We know that √3 = 1.732

=> BC = 20(1.732+1)

=> BC = 20(2.732)

=> BC =54.64 m

Therefore, BC = 54.64 m

Answer :-

The distance between two points is 20(√3+1) m (or) 54.64 m

Used formulae:-

→ tan A = Opposite side to A/Adjacent side to A

→ tan 30° = 1/√3

→ tan 45° = 1

→ √3 = 1.732...

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