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Question 18
If in triangle ABC, if side AB $$= 6\sqrt{3}$$, AC $$= 12$$ and BC $$= 6$$, then angle B equals:
Solution
Given, AB $$= 6\sqrt{3}$$
AC $$= 12$$
BC $$= 6$$
$$AB^2+BC^2 = (6\sqrt{3})^2 + 6^2$$
$$= 108+36 = 144$$
$$AC^2 = 12^2 = 144$$
Therefore, Given triangle ABC is a right angled triangle right angled at B since the hypotenuse is AC.
Therefore, $$\angle B = 90^\circ$$
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