Question

ᴀ ᴛᴠ ᴛᴏᴡᴇʀ sᴛᴀɴᴅs ᴠᴇʀᴛɪᴄᴀʟʟʏ ᴏɴ ᴀ ʙᴀɴᴋ ᴏғ ᴀ ᴄᴀɴᴀʟ. ғʀᴏᴍ ᴀ ᴘᴏɪɴᴛ ᴏɴ ᴛʜᴇ ᴏᴛʜᴇʀ ʙᴀɴᴋ ᴅɪʀᴇᴄᴛʟʏ ᴏᴘᴘᴏsɪᴛᴇ ᴛʜᴇ ᴛᴏᴡᴇʀ, ᴛʜᴇ ᴀɴɢʟᴇ ᴏғ ᴇʟᴇᴠᴀᴛɪᴏɴ ᴏғ ᴛʜᴇ ᴛᴏᴘ ᴏғ ᴛʜᴇ ᴛᴏᴡᴇʀ ɪs 60∘. ғʀᴏᴍ ᴀɴᴏᴛʜᴇʀ ᴘᴏɪɴᴛ 10 ᴍ ᴀᴡᴀʏ ғʀᴏᴍ ᴛʜɪs ᴘᴏɪɴᴛ ᴏɴ ᴛʜᴇ ʟɪɴᴇ ᴊᴏɪɴɢ ᴛʜɪs ᴘᴏɪɴᴛ ᴛᴏ ᴛʜᴇ ғᴏᴏᴛ ᴏғ ᴛʜᴇ ᴛᴏᴡᴇʀ, ᴛʜᴇ ᴀɴɢʟᴇ ᴏғ ᴇʟᴇᴠᴀᴛɪᴏɴ ᴏғ ᴛʜᴇ ᴛᴏᴘ ᴏғ ᴛʜᴇ ᴛᴏᴡᴇʀ ɪs 30∘ .. ғɪɴᴅ ᴛʜᴇ ʜᴇɪɢʜᴛ ᴏғ ᴛʜᴇ ᴛᴏᴡᴇʀ ᴀɴᴅ ᴛʜᴇ ᴡɪᴅᴛʜ ᴏғ ᴛʜᴇ ʀᴏᴀᴅ..

don't spam ans correcly​

Answer 1

Given:

AB is the tower

C is a point 10 m away from B

Find:

DB and AB

Solution:

Let DC be x.

In ΔABC,

$Tan30^{0} = \frac{AB}{CB} \\\\\frac{1}{\sqrt{3} } = \frac{AB}{10} \\\\\frac{10}{\sqrt{3} } = AB\\\\\frac{10\sqrt{3} }{3} = AB$

In ΔABD,

$Tan60^{0} = \frac{AB}{DB} \\\\\sqrt{3} = \frac{\frac{10\sqrt{3} }{3} }{x + 10} \\\\x + 10 = \frac{10\sqrt{3} }{3\sqrt{3} } \\\\x = \frac{10}{3}-10 = \frac{-20}{3} \\\\DB = \frac{-20}{3} + 10 = \frac{30-20}{3} = \frac{10}{3} m$

So ,

Height = $\frac{10\sqrt{3} }{3} m$

Width = $\frac{10}{3} m$

I hope it will help you.

Regards.

Answer 2

Answer:

refer the image.....

Step-by-step explanation: