The horizontal distance between two towers is 140m.The angle of elevation of top of the 1st tower.When seen from the top 2nd tower is 30°.If the height of the 2nd tower is 60m.Find the height of the 1st tower
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Let AB be the height of second tower and CD be the height of first tower.
Given, BD = AE = 140 m
& AB = DE = 60 m
In ΔAEC,
tan 30°= Perpendicular /Base= CE/AE
tan 30 ° = CE/140
1/√3= CE /140
CE= 140 / √3
CE= 140 ×√3/ (√3×√3)
[ Rationalising the denominator]
CE= 140√3 /3
Height of the first tower CD= CE+DE
= 140√3/3 + 60
=( 140 × 1.73) /3 + 60
[ √3=1.73]
= 242.2/3 + 60
= 80.73 + 60
= 140.73 m
Height of the first tower (CD)=140.73 m
==================================================================
Hope this will help you...
Given, BD = AE = 140 m
& AB = DE = 60 m
In ΔAEC,
tan 30°= Perpendicular /Base= CE/AE
tan 30 ° = CE/140
1/√3= CE /140
CE= 140 / √3
CE= 140 ×√3/ (√3×√3)
[ Rationalising the denominator]
CE= 140√3 /3
Height of the first tower CD= CE+DE
= 140√3/3 + 60
=( 140 × 1.73) /3 + 60
[ √3=1.73]
= 242.2/3 + 60
= 80.73 + 60
= 140.73 m
Height of the first tower (CD)=140.73 m
==================================================================
Hope this will help you...
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140.73ᴄᴍ
ʟᴇᴛ ᴀʙ ʙᴇ ᴛʜᴇ ʜᴇɪɢʜᴛ ᴏғ sᴇᴄᴏɴᴅ ᴛᴏᴡᴇʀ ᴀɴᴅ ᴄᴅ ʙᴇ ᴛʜᴇ ʜᴇɪɢʜᴛ ᴏғ ғɪʀsᴛ ᴛᴏᴡᴇʀ.
ɢɪᴠᴇɴ, ʙᴅ = ᴀᴇ = 140 ᴍ
& ᴀʙ = ᴅᴇ = 60 ᴍ
ɪɴ Δᴀᴇᴄ,
ᴛᴀɴ 30°= ᴘᴇʀᴘᴇɴᴅɪᴄᴜʟᴀʀ /ʙᴀsᴇ= ᴄᴇ/ᴀᴇ
ᴛᴀɴ 30 ° = ᴄᴇ/140
1/√3= ᴄᴇ /140
ᴄᴇ= 140 / √3
ᴄᴇ= 140 ×√3/ (√3×√3)
[ ʀᴀᴛɪᴏɴᴀʟɪsɪɴɢ ᴛʜᴇ ᴅᴇɴᴏᴍɪɴᴀᴛᴏʀ]
ᴄᴇ= 140√3 /3
ʜᴇɪɢʜᴛ ᴏғ ᴛʜᴇ ғɪʀsᴛ ᴛᴏᴡᴇʀ ᴄᴅ= ᴄᴇ+ᴅᴇ
= 140√3/3 + 60
=( 140 × 1.73) /3 + 60
[ √3=1.73]
= 242.2/3 + 60
= 80.73 + 60
= 140.73 ᴍ
ʜᴇɪɢʜᴛ ᴏғ ᴛʜᴇ ғɪʀsᴛ ᴛᴏᴡᴇʀ (ᴄᴅ)=140.73ᴄᴍ
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