Question

ɪғ ᴀ 1.5m ᴛᴀʟʟ ɢɪʀʟ sᴛᴀɴᴅs ᴀᴛ ᴀ ᴅɪsᴛᴀɴᴄᴇ ᴏғ 3m ғʀᴏᴍ ᴀ ʟᴀᴍᴘ ᴘᴏsᴛ ᴀɴᴅ sʜᴀᴅᴏᴡ ᴏғ ʟᴇɴɢᴛʜ 4.5m ᴏɴ ᴛʜᴇ ɢʀᴏᴡ ᴛʜᴇɴ ғɪɴᴅ ᴛʜᴇ ʜᴇɪɢʜᴛ ᴏғ ᴛʜᴇ ʟᴀᴍᴘ ᴘᴏsᴛ.
sᴏʟᴠᴇ .


Dᴏɴᴛ sᴘᴀᴍ​

Answer 1

Step-by-step explanation:

HEY MATE......

$let \: ab \: be \: the \: lamp \: post \: : \\ \\ cd \: be \: the \: girl \: \\ \\ and \: de \: be \: her \: shadow \\ \\ we \: have \: \\ \\ cd = 1.5 |m| \\ \\ ad = 3 |m| \\ \\ de = 4.5 |m| \\ \\ let \: < e = thetha \\ \\ in \: triangle \: cde \\ \\ tan \: thetha = \frac{cd}{de} \\ \\ \implies \: tan \: thetha = \frac{1.5}{4.5} \\ \\ \implies \: tan \: thetha = \frac{1}{3} ...(1) \\ \\ now \: in \: triangle \: abe \\ \\ tan \: thetha = \frac{ab}{ae} \\ \\ \implies \: \frac{1}{3} = \frac{ab}{ad + de} (using \: \\ \\ (1)eqn) \\ \\ \implies \: \frac{1}{3} = \frac{ab}{3 + 4.5} \\ \\ \implies \: ab = \frac{7.5}{3} \\ \\ ab = 2.5 |m|$

Answer 2

Let AB be boy = 1.5m

CD be lamp = Let it be x m.

OB be shadow cast by boy = 4.5m

BD be distance between boy and lamp = 3m

Therefore

In ΔAOB,

=> tan θ = AB/OB

In ΔCOD,

=> tan θ = CD/OD = CD/(OB+BD)

∴ AB/OB = CD/(OB+BD)

=> 1.5/4.5 = CD/(4.5+3)

=> CD = 7.5/3 = 2.5m

need brainliest :)