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Triangles ABC, ACD, and ABD are all isosceles triangles. Poi
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07 May 2020, 10:34
07 May 2020, 10:34
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#greprepclub Triangles ABC, ACD, and ABD are all isosceles triangles..jpg [ 17.03 KiB | Viewed 8060 times ]
Triangles ABC, ACD, and ABD are all isosceles triangles. Point E (not shown) is the midpoint of \(\overline{BD}\). If the ratio of the length of \(\overline{CE}\) to the length of \(\overline{BC}\) is equal to \(\sqrt{3} : 2\), then what is the measure, in degrees, of ∠CAD?
A. 10
B. 15
C. 30
D. 45
E. 60
Kudos for the right answer and explanation
Re: Triangles ABC, ACD, and ABD are all isosceles triangles. Poi
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07 May 2020, 20:11
07 May 2020, 20:11
1
△ABC, △ACD and △ABD are isosceles;
In △ABC: AC=BC, ∠BAC = ∠ABC = x
△ACD: AC=CD, ∠CAD = ∠CDA = x
△ABD: AB=AD,
Now, In △BCE, CE/BC = √3/2
sin ∠CBE = length of oppo site/ length of hypotenus
CE / BC = √3/2 = Sin 60°
hence, ∠CBE = 60°
as BC = CD; ∠CDE = 60°
In △ABD; ∠A + ∠B + ∠C = 180
2x + (x+60) + (x+60) = 180
4x = 60
x = 15
In △ABC: AC=BC, ∠BAC = ∠ABC = x
△ACD: AC=CD, ∠CAD = ∠CDA = x
△ABD: AB=AD,
Now, In △BCE, CE/BC = √3/2
sin ∠CBE = length of oppo site/ length of hypotenus
CE / BC = √3/2 = Sin 60°
hence, ∠CBE = 60°
as BC = CD; ∠CDE = 60°
In △ABD; ∠A + ∠B + ∠C = 180
2x + (x+60) + (x+60) = 180
4x = 60
x = 15
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#greprepclub Triangles ABC, ACD, and ABD are all isosceles triangles..jpg [ 12.99 KiB | Viewed 7999 times ]
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Joined: 30 Aug 2020
Posts: 23
Given Kudos: 0
GRE 1: Q165 V170
Re: Triangles ABC, ACD, and ABD are all isosceles triangles. Poi
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12 Oct 2020, 07:47
12 Oct 2020, 07:47
2
The correct answer is b) 15 degrees.
Now, let us work our way from the ratio CE/BC = √3:2 = > ECB is a 30-60-90 triangle.
=> Angle CBD = 60;
Angle CDB = 60
Now, notice that since two angles are 60 degree in Triangle CDB, CDB is equilateral
=> Angle DCB = 60 degree
Now notice that CE bisects the angle DCB
Hence, ACB + BCE = 180
=> ACB + 30 = 180
=> ACB = 150
Similarly,
ACD + DCE = 180
=> ACD + 30 = 180
=> ACD = 150
Now notice, that equal sides of the isosceles triangle DAC must be the sides AC and CD that are adjacent to angle ACD
(This is because if side opposite to ACD is the equal side, then 150 + 150 > 180 degrees. Hence not possible)
This ,means that CAD = CDA = x
=> x + x = 150 = 180
=> 2x = 30
=> x = 15
=> CAD = 15
Now, let us work our way from the ratio CE/BC = √3:2 = > ECB is a 30-60-90 triangle.
=> Angle CBD = 60;
Angle CDB = 60
Now, notice that since two angles are 60 degree in Triangle CDB, CDB is equilateral
=> Angle DCB = 60 degree
Now notice that CE bisects the angle DCB
Hence, ACB + BCE = 180
=> ACB + 30 = 180
=> ACB = 150
Similarly,
ACD + DCE = 180
=> ACD + 30 = 180
=> ACD = 150
Now notice, that equal sides of the isosceles triangle DAC must be the sides AC and CD that are adjacent to angle ACD
(This is because if side opposite to ACD is the equal side, then 150 + 150 > 180 degrees. Hence not possible)
This ,means that CAD = CDA = x
=> x + x = 150 = 180
=> 2x = 30
=> x = 15
=> CAD = 15
