Question

From a point on a bridge across a river, the angles of depression of the river 30° and 45° respectively. If the bridge is at a height of 3m. from the bank, find the width of the river.​

Answer 1

GIVEN:

• The angles of depression of the river 30° and 45° respectively.
• The height of the bridge is 3m.

TO FIND:

• What is the width of the river ?

SOLUTION:

Let the width of the river be AB

and,

The height of the bridge (DP) is 3 m

∴ Angle of depression of the banks on the opposite sides of the river are 30° and 45° respectively.

⠀⠀⠀⠀⠀⠀∴ ∠QPA = 30°

⠀⠀⠀⠀⠀⠀∴ ∠RPB = 45°

We have to find the width of the river AB

Since,

⠀⠀⠀⠀PD ⊥ AB

So,

⠀⠀ ∠PDA = ∠PDB = 90°

Line QP || AB

and

AP is the transversal

So,

⠀⠀∠QPA = ∠PAD = 30° (Alternate angles)

Similarly,

Line PR || AB

and

PB is the transversal

So,

⠀⠀∠RPB = ∠PBD = 45° (Alternate angles)

In Right angled triangle PAD,

$\rm{\rightharpoonup tan \: A = \dfrac{Perpendicular}{Base}}$

$\rm{\rightharpoonup tan \: A = \dfrac{PD}{AD}}$

$\rm{\rightharpoonup tan \: 30\degree = \dfrac{3}{AD}}$

$\rm{\rightharpoonup \dfrac{1}{\sqrt{3}}\ =\ \dfrac{3}{AD}}$ [ tan 30° = 1/√3]

$\bf{\rightharpoonup \: \star AD = 3\sqrt{3}\:\star }$

In Right angled triangle PBD,

$\rm{\rightharpoonup tan \: A = \dfrac{Perpendicular}{Base}}$

$\rm{\rightharpoonup tan \: A = \dfrac{PD}{BD}}$

$\rm{\rightharpoonup tan \: 45\degree = \dfrac{3}{BD}}$

$\rm{\rightharpoonup 1 = \dfrac{3}{BD}}$[ tan 45° = 1]

$\bf{\rightharpoonup \: \star BD = 3\:\star }$

Now,

⠀⠀⠀AB = AD + DB

⠀⠀⠀AB = 3√3 + 3

★ Width of the river = 3(√3 +1) m ★

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Answer 2

GIVEN:

The angles of depression of the river 30° and 45° respectively.

The height of the bridge is 3m.

TO FIND:

What is the width of the river ?

SOLUTION:

Let the width of the river be AB

and,

The height of the bridge (DP) is 3 m

∴ Angle of depression of the banks on the opposite sides of the river are 30° and 45° respectively.

⠀⠀⠀⠀⠀⠀∴ ∠QPA = 30°

⠀⠀⠀⠀⠀⠀∴ ∠RPB = 45°

We have to find the width of the river AB

Since,

⠀⠀⠀⠀PD ⊥ AB

So,

⠀⠀ ∠PDA = ∠PDB = 90°

Line QP || AB

and

AP is the transversal

So,

⠀⠀∠QPA = ∠PAD = 30° (Alternate angles)

Similarly,

Line PR || AB

and

PB is the transversal

So,

⠀⠀∠RPB = ∠PBD = 45° (Alternate angles)

In Right angled triangle PAD,

\rm{\rightharpoonup tan \: A = \dfrac{Perpendicular}{Base}}⇀tanA=BasePerpendicular

\rm{\rightharpoonup tan \: A = \dfrac{PD}{AD}}⇀tanA=ADPD

\rm{\rightharpoonup tan \: 30\degree = \dfrac{3}{AD}}⇀tan30°=AD3

\rm{\rightharpoonup \dfrac{1}{\sqrt{3}}\ =\ \dfrac{3}{AD}}⇀31 = AD3 [ tan 30° = 1/√3]

\bf{\rightharpoonup \: \star AD = 3\sqrt{3}\:\star }⇀⋆AD=33⋆

In Right angled triangle PBD,

\rm{\rightharpoonup tan \: A = \dfrac{Perpendicular}{Base}}⇀tanA=BasePerpendicular

\rm{\rightharpoonup tan \: A = \dfrac{PD}{BD}}⇀tanA=BDPD

\rm{\rightharpoonup tan \: 45\degree = \dfrac{3}{BD}}⇀tan45°=BD3

\rm{\rightharpoonup 1 = \dfrac{3}{BD}}⇀1=BD3 [ tan 45° = 1]

\bf{\rightharpoonup \: \star BD = 3\:\star }⇀⋆BD=3⋆

Now,

⠀⠀⠀AB = AD + DB

⠀⠀⠀AB = 3√3 + 3

★ Width of the river = 3(√3 +1) m ★

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