Question

(a) What is the maximum possible number of obtuse angles in a triangle? Justify your answer.
(b) In a $△ABC$ the side BC is extended to D. Given that $\angle ABC={30}^{\circ }and\angle ACD={160}^{\circ }$. Find $\angle BAC$. [3 MARKS]

Answer 1

(a) Answer: 1 Mark
Reason: 1 Mark
(b) Solution: 1 Mark

(a) In a triangle, only one obtuse angle can exist.

If we have more than one obtuse angle in a triangle, then the sum of two (obtuse) angles will exceed ${180}^{\circ }$.

But the sum of three angles of a triangle is ${180}^{\circ }$.

$\Rightarrow$ A triangle cannot have more than one obtuse angle.



(b)

$\angle BCD={180}^{\circ }$ (Straight angle)
$\therefore \angle ACB=\angle BCD-\angle ACD=(180-160{)}^{\circ }={20}^{\circ }$
$\Rightarrow \angle BAC={180}^{\circ }-(\angle ABC+\angle ACB)=180-(30+20)=180-50=130$
$\therefore \angle BAC={30}^{\circ }$