Question

ᴛʜᴇ ᴀɴɢʟᴇs ᴏғ ᴇʟᴇᴠᴀᴛɪᴏɴ ᴀɴᴅ ᴅᴇᴘʀᴇssɪᴏɴ ᴏғ ᴛʜᴇ ᴛᴏᴘ ᴀɴᴅ ʙᴏᴛᴛᴏᴍ ᴏғ ᴀ ʟᴀᴍᴘ ᴘᴏsᴛ ғʀᴏᴍ ᴛʜᴇ ᴛᴏᴘ ᴏғ ᴀ 66ᴍ ʜɪɢʜ ᴀᴘᴀʀᴛᴍᴇɴᴛ ᴀʀᴇ 60° ᴀɴᴅ 30° ʀᴇsᴘᴇᴄᴛɪᴠᴇʟʏ.ғɪɴᴅ

1) ᴛʜᴇ ʜᴇɪɢʜᴛ ᴏғ ᴛʜᴇ ʟᴀᴍᴘ ᴘᴏsᴛ
2)ᴛʜᴇ ᴅɪғғᴇʀᴇɴᴄᴇ ʙᴇᴛᴡᴇᴇɴ ʜᴇɪɢʜᴛ ᴏғ ᴛʜᴇ ʟᴀᴍᴘ ᴘᴏsᴛ ᴀɴᴅ ᴛʜᴇ ᴀᴘᴀʀᴛᴍᴇɴᴛ
3)ᴛʜᴇ ᴅɪsᴛᴀɴᴄᴇ ʙᴇᴛᴡᴇᴇɴ ᴛʜᴇ ʟᴀᴍᴘ ᴘᴏsᴛ ᴀɴᴅ ᴛʜᴇ ᴀᴘᴀʀᴛᴍᴇɴᴛ (√3=1.732)​

Answer 1

Your Answer:

Given:-

• Height of apartment = 66m
• Angle of Elevation = 60°
• Angle of Depression = 30°

To Find:-

• The height of lamp post
• The difference between height of the lamp post and the apartment
• The distance between the lamp post and the apartment

Solution:-

Let CD be the height of apartment

Let AB be the height of Lamp post

Let the distance between lamp post and apartment is BD

$\tt In \ \ \triangle BDC \\\\ \tt \angle ECB = \angle CBD = 30^o \ \ \ (\because alternate \ angles) \\\\ \tan 30^o = \dfrac{CD}{BD} \\\\ \tt \Rightarrow \dfrac{1}{\sqrt3} = \dfrac{66}{BD} \\\\ \tt \Rightarrow BD = 66\sqrt3m$

$\tt BD = EC = 66\sqrt3m$

$\tt In \ \ \triangle AEC \\\\ \tt \tan 60^o = \dfrac{AE}{EC} \\\\ \tt \Rightarrow \sqrt3 = \dfrac{AE}{66\sqrt3} \\\\ \tt \Rightarrow (\sqrt3)(66\sqrt3) = AE \\\\ \tt \Rightarrow 198m = AE$

i)

$\tt AB = AE + EB \\\\\tt AB = AE + CD (\because EB = CD) \\\\ \tt AB = 198 + 66 = 264m$

So, height of lamp post is 264m

ii) Difference between height of the lamp post and the apartment

= 264 - 66 = 198m

iii) Distance between the lamp post and apartment is BD = $\tt 66\sqrt3 = 66 \times 1.732 = 114.312m$