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Side condition for a 120 degree angle

Side condition for a 120 degree angle

IP 2
7 marks

In triangle \(ABC\), the sides opposite \(A\), \(B\) and \(C\) are \(a\), \(b\) and \(c\).

  1. Show that if \(c^2=a^2+ab+b^2\), then \(C=120^\circ\).[3]
  2. Verify that the triangle with \(a=3\), \(b=5\), \(c=7\) has this property, and find its exact area.[2]
  3. A triangle with \(C=120^\circ\) has \(a=5\) and \(b=16\). Show that \(c\) is a whole number.[2]

Solution:

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Answer:(a) \(\cos C=-\dfrac{1}{2}\Rightarrow C=120^\circ\) (b) \(49=7^2\); area \(\dfrac{15\sqrt{3}}{4}\) (c) \(c=19\)

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