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Kritisood
Attachment:
Screenshot 2020-04-30 at 11.31.59 PM.png

In the x-y plane, is angle BAC = 60 degrees?
(1) The length of AB = BC.
(2) The perpendicular from A on BC bisects BC.

Solution:
Pre Analysis:
  • In the triangle ABC, we are asked if \(\angle BAC=60^o\) or not

Statement 1: The length of AB = BC
  • According to this statement AB = BC which means \(\angle ABC=\angle ACB\)
  • However, this doesn't tell us if \(\angle BAC=60^o\) or not
  • 2 cases:
    • \(\angle ABC=\angle ACB=60\), then \(\angle BAC=60\)
    • \(\angle ABC=\angle ACB=45\), then \(\angle BAC=90\)
  • Thus, statement 1 alone is not sufficient and we can eliminate options A and D

Statement 2: The perpendicular from A on BC bisects BC
  • Accoring to this statement triangle ABC is either isosceles with AB = AC or equilateral with AB = AC = BC
  • Similar to statement 1, this is not sufficient to say if \(\angle BAC=60^o\) or not
  • Thus, statement 2 alone is also not sufficient

Combining:
  • From statement 1, we get AB = BC
  • From statement 2, we get either AB = AC or AB = AC = BC
  • Upon combining, we can be sure that ABC is an equilateral triangle and \(\angle BAC=60^o\)

Hence the right answer is Option C
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Kritisood
Attachment:
Screenshot 2020-04-30 at 11.31.59 PM.png

In the x-y plane, is angle BAC = 60 degrees?
(1) The length of AB = BC.
(2) The perpendicular from A on BC bisects BC.


The x-y plane doesn’t really affect the figure, so just concentrate on the triangle.

(1) The length of AB = BC.
This tells us that \(\angle C= \angle A\), but they could be anything between 0 and 90, both excluded.
Insufficient

(2) The perpendicular from A on BC bisects BC
Let it bisect BC at E. so, we have two triangles ABE and ACE.
AE is common, BE=CE and included angle is 90. Thus, the triangles are similar and \(\angle C= \angle B\)
But A could be again anything between 0 to 180, both excluded.
Insufficient

Combined
\(\angle C= \angle A\) \(= \angle B=60.\)
Sufficient



C
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