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01 Jul 2020, 23:02
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
63% (02:26) correct
37%
(02:21)
wrong
based on 84
sessions
History
Date
Time
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Not Attempted Yet
The top of a 15 metre high tower makes an angle of elevation of 60° with the bottom of an electronic pole and angle of elevation of 30° with the top of the pole. What is the height of the electric pole?
A. 5 metres
B. 8 metres
C. 10 metres
D. 12 metres
E. 13 metres
Let AB be the tower and CD be the electric pole.
\(∠ACB=60∘ , ∠EDB=30∘ , AB=15\)
\(Let BE=AB−AE=15−h\)
\(\frac{AB}{AC}=tan 60∘= \sqrt{3}\)
=>\(AC=\frac{AB}{\sqrt{3}}\)
=>\(AC=\frac{15}{\sqrt{3}}\)
Also,
\(\frac{BE}{DE}=tan 30∘=\frac{1}{\sqrt{3}}\)
=>\(DE=BE\sqrt{3}=\sqrt{3}(15−h)\)
AC = DE
=> \(\frac{15}{\sqrt{3}} = \sqrt{3}(15−h)\)
=> \(15 = 45 - 3h\)
=> \(3h = 30\)
=> \(h = 10\)
Hence, OA is (C).
\(∠ACB=60∘ , ∠EDB=30∘ , AB=15\)
\(Let BE=AB−AE=15−h\)
\(\frac{AB}{AC}=tan 60∘= \sqrt{3}\)
=>\(AC=\frac{AB}{\sqrt{3}}\)
=>\(AC=\frac{15}{\sqrt{3}}\)
Also,
\(\frac{BE}{DE}=tan 30∘=\frac{1}{\sqrt{3}}\)
=>\(DE=BE\sqrt{3}=\sqrt{3}(15−h)\)
AC = DE
=> \(\frac{15}{\sqrt{3}} = \sqrt{3}(15−h)\)
=> \(15 = 45 - 3h\)
=> \(3h = 30\)
=> \(h = 10\)
Hence, OA is (C).
Attachments
answer.jpg [ 7.15 KiB | Viewed 8818 times ]
01 Jul 2020, 23:27
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Explanation:
Check Figure below for better understanding:
∠ACB = 60 , ∠EDB = 30 , AB=15 , AC=DE
BE = AB − AE =15−h
AB/AC=tan60 = √3
AC=15/√3
BE/DE= tan30 = 1/√3
DE=√3*(15−h)
As AC = DE
15/√3 = √3*(15−h)
15 = 45 - 3h
h = 30/3
h = 10
IMO-C
Check Figure below for better understanding:
∠ACB = 60 , ∠EDB = 30 , AB=15 , AC=DE
BE = AB − AE =15−h
AB/AC=tan60 = √3
AC=15/√3
BE/DE= tan30 = 1/√3
DE=√3*(15−h)
As AC = DE
15/√3 = √3*(15−h)
15 = 45 - 3h
h = 30/3
h = 10
IMO-C
Attachments
File comment: Attachement
height.jpg [ 7.15 KiB | Viewed 8539 times ]
