Question

application of trignometry​

Answer 1

Let, ABP is a triangle in which PD is the height=3m and AD is the width of the river.

QR is the line joining at P

<PAD= 30°

<PBD=45°

PAD and PBD are right angle triangles.

In right angle triangle PAD_

tan A = side opposite to <A/side adjacent to <A

         = PD/AD

∴ tan 30°=3/ad

⇒1/√3=3/AD

⇒AD=3√3 m

In triangle PDB_

tan B= side opposite to <B/side adjacent to <B

        = PD/BD

∴ tan 45°= PD/BD

⇒ 1 = PD/BD

⇒PD=BD

So, BD = 3 m

∴ AB=AD+BD

       =3√3+3

      = 3(√3+1) m

∴ Width of the river = AD = 3(√3+1) m

Answer 2

Answer:

$\huge{\red{\boxed{\boxed{\green{\sf{width=3(1+\sqrt{3})m}}}}}}$

Step-by-step explanation:

$\huge{\red{\boxed{\boxed{\green{\underline{\red{\sf{SOLUTION-}}}}}}}}$

In above figure(2):

□DP=3m (height of the bridge)

□∠PAD=30°

□∠PBD=45°

To find:

○Width of river=?

$IN \: RIGHT\triangle \: APD \: \\\\ ∠A = 30 \degree \\ \tan30 \degree = \frac{PD}{ AD} \\ \frac{1}{ \sqrt{3} } = \frac{3}{ad} \\ ad = 3 \sqrt{3} m \\\\ again \\ IN\: RIGHT\triangle \: PBD \: \\ ∠B= 45 \degree \\ so \: BD = PD = 3m \\ AD = BD + AD \\ = )3 + 3 \sqrt{3} \\ = )3(1 + \sqrt{3} )m \\ \\ \therefore \: width = 3( 1 + \sqrt{3} )m$

$\huge{\red{\boxed{\boxed{\green{\sf{width=3(1+\sqrt{3})m}}}}}}$

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