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26 Nov 2022, 00:34
Kudos
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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:
65%
(hard)
Question Stats:
50% (02:47) correct
50%
(01:37)
wrong
based on 18
sessions
History
Date
Time
Result
Not Attempted Yet
An aeroplane is flying at a height of 300 m above the ground. Flying at this height, the angles of depression from the aeroplane of two points on both banks of a river in opposite directions are 45° and 60° respectively. Find the width of the river.
\( A) 300\sqrt{3}\)
\( B) 100 (\sqrt{3}+1)\)
\( C) 100 (\sqrt{3} +3)\)
\( D) 100 (3\sqrt{3}) \)
\( E) 300 (\sqrt{3}+1) \)
\( A) 300\sqrt{3}\)
\( B) 100 (\sqrt{3}+1)\)
\( C) 100 (\sqrt{3} +3)\)
\( D) 100 (3\sqrt{3}) \)
\( E) 300 (\sqrt{3}+1) \)
26 Nov 2022, 01:15
Kudos
Bookmarks
Angle of depression is the angle between horizontal line and the line of sight to the bank of the river.
The problem can be visually represented as shown. B and C represent two points on the bank of the river. AD is the perpendicular distance between the aero plane and the ground and is equal to 300m.
BC = Width of the river
BC = AD + DC
In \(\triangle ADC\), \(\angle DAC = 45^{\circ}\)
In \(\triangle ADB\), \(\angle DAB = 30^{\circ}\) & \(\angle DBA = 60^{\circ}\)
\(\triangle ADC\)
DC = 300 m (using 45 - 45 - 90 property)
\(\triangle ADB\)
BD = \(\frac{300}{\sqrt{3}}\) m (using 30 - 60 - 90 property)
BC = \(\frac{300}{\sqrt{3}} + 300\)
= \(\frac{300}{\sqrt{3}}(1+\sqrt{3})\)
Multiplying numerator and denominator by \(\sqrt{3}\)
= \(\frac{300\sqrt{3}}{3}(1+\sqrt{3})\)
=\( 100(\sqrt{3}+3)\)
Option C
P.S. - Doesn't mimic a GMAT like question as the definition of angles of depression is not given up-front. IMO GMAT generally provides such information in the question stem.
The problem can be visually represented as shown. B and C represent two points on the bank of the river. AD is the perpendicular distance between the aero plane and the ground and is equal to 300m.
BC = Width of the river
BC = AD + DC
In \(\triangle ADC\), \(\angle DAC = 45^{\circ}\)
In \(\triangle ADB\), \(\angle DAB = 30^{\circ}\) & \(\angle DBA = 60^{\circ}\)
\(\triangle ADC\)
DC = 300 m (using 45 - 45 - 90 property)
\(\triangle ADB\)
BD = \(\frac{300}{\sqrt{3}}\) m (using 30 - 60 - 90 property)
BC = \(\frac{300}{\sqrt{3}} + 300\)
= \(\frac{300}{\sqrt{3}}(1+\sqrt{3})\)
Multiplying numerator and denominator by \(\sqrt{3}\)
= \(\frac{300\sqrt{3}}{3}(1+\sqrt{3})\)
=\( 100(\sqrt{3}+3)\)
Option C
P.S. - Doesn't mimic a GMAT like question as the definition of angles of depression is not given up-front. IMO GMAT generally provides such information in the question stem.
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