Question

An aeroplane is flying at a height of 300m above the ground. Flying at this height, the angle of depression from the aeroplane of two points on both banks of a river in opposite directions are 45 and 60 respectively..Find the width of the river . ( Use √3= 1.732)

Answer 1

||✪✪ QUESTION ✪✪||

An aeroplane is flying at a height of 300m above the ground. Flying at this height, the angle of depression from the aeroplane of two points on both banks of a river in opposite directions are 45 and 60 respectively..Find the width of the river . ( Use √3= 1.732)

|| ✰✰ ANSWER ✰✰ ||

❁❁ Refer To Image First ..❁❁

From image we have :-

→ P = Position of Plane ..

→ ∠PBM = 45°

→ ∠PAM = 60°

→ AM = Let x m.

→ BM = Let y m.

→ AB = width of River .

→ PM = Height of Aeroplane From Ground = 300m.

So, in Rt.∆PMB we have now :-

➼ Tan45° = Perpendicular / Base

➼ 1 = PM / BM

➼ PM = BM

➼ BM = y = 300 m ------------------ Equation ❶

______________________

In Rt.∆PMA now :-

➻ Tan60° = PM / MA

➻ √3 = (300/x)

➻ X = (300/√3) .

Rationalising RHS part Now,

➻ X = (300/√3) * (√3/√3)

➻ X = (300√3)/3

➻ X = 100√3

Putting √3 = 1.732 (Given) now,

→ X = 1.732 * 100

→ X = 173.2 m --------------------------- Equation ❷

___________________

Now,

➺ width of River = AB

➺ AB = AM + MB

Putting Values From both Equations now, we get,

➺ AB = 173.2 + 300

➺ AB = 473.2 m .

∴ width of River is 473.2 m.

Answer 2

ॐ $\sf\red{R}$$\sf\orange{a}$$\sf\green{d}$$\sf\blue{h}$$\sf\purple{e}$ $\sf\red{R}$$\sf\orange{a}$$\sf\green{d}$$\sf\blue{h}$$\sf\purple{e}$ ॐ

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$\huge{\boxed{\bold \orange{\fcolorbox{red}</h3><h3>{black}{QUESTION}}}}$

Aɴ ᴀᴇʀᴏᴘʟᴀɴᴇ ɪs ғʟʏɪɴɢ ᴀᴛ ᴀ ʜᴇɪɢʜᴛ ᴏғ 300ᴍ ᴀʙᴏᴠᴇ ᴛʜᴇ ɢʀᴏᴜɴᴅ. Fʟʏɪɴɢ ᴀᴛ ᴛʜɪs ʜᴇɪɢʜᴛ, ᴛʜᴇ ᴀɴɢʟᴇ ᴏғ ᴅᴇᴘʀᴇssɪᴏɴ ғʀᴏᴍ ᴛʜᴇ ᴀᴇʀᴏᴘʟᴀɴᴇ ᴏғ ᴛᴡᴏ ᴘᴏɪɴᴛs ᴏɴ ʙᴏᴛʜ ʙᴀɴᴋs ᴏғ ᴀ ʀɪᴠᴇʀ ɪɴ ᴏᴘᴘᴏsɪᴛᴇ ᴅɪʀᴇᴄᴛɪᴏɴs ᴀʀᴇ 45° ᴀɴᴅ 60° ʀᴇsᴘᴇᴄᴛɪᴠᴇʟʏ.Fɪɴᴅ ᴛʜᴇ ᴡɪᴅᴛʜ ᴏғ ᴛʜᴇ ʀɪᴠᴇʀ . ( Usᴇ √3= 1.732).

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$\huge{\boxed{\bold \orange{\fcolorbox{red}</h3><h3>{black}{SOLUTION}}}}$

Lᴇᴛ BD ʙᴇ ᴛʜᴇ ʟᴇɴɢᴛʜ ᴏғ ᴛʜᴇ ʀɪᴠᴇʀ ᴀɴᴅ A ʙᴇ ᴛʜᴇ ᴀʀᴇᴏᴘʟᴀɴᴇ.

❝ ★ ʀᴇғᴇʀ ᴛʜᴇ ɢɪᴠᴇɴ ɪᴍᴀɢᴇ ★ ❞

☞ tan 45° = AC / BC

☞ 1 = 300 / BC

☞ BC = 300 m

☛ tan 60° = AC / CD

☛ √3 = 300 / CD

☛ CD = 300/√3 × √3 / √3

☛ 300√3 / 3

☛ 100√3

☛ 1.732 × 100

☛ 173.2

ᴛʜᴇ ᴡɪᴅᴛʜ ᴏғ ᴛʜᴇ ʀɪᴠᴇʀ = ʙᴄ + ᴄᴅ

➠ 300 + 173.2

➠ 473.2 m

∴ The width of the river = $\huge\pink{473.2\:m}$

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