Question

(a) For any triangle, ABC, draw: [4 MARKS]
(i) All possible medians.
(ii) All possible altitudes.


(b) Find the unknown x and y.

Answer 1

(a) Correct diagram : 2 Marks
(b) Solution: 1 Mark
Properties: 1 Mark

(a)

(i) (ii)


(b)



$\angle BAC={92}^{\circ }$
$\angle ACB=x^{\circ }$ (Equal sides have equal angles opposite to them.)
$\angle ACB+\angle ABC+{92}^{\circ }={180}^{\circ }$
$2x+{92}^{\circ }={180}^{\circ }$
$x+{46}^{\circ }={90}^{\circ }$
$x={90}^{\circ }-{46}^{\circ }$
$x={44}^{\circ }$

$\angle ACB+y={180}^{\circ }\Rightarrow x+y={180}^{\circ }$
$\Rightarrow {44}^{\circ }+y={180}^{\circ }$
$\Rightarrow y={180}^{\circ }-{44}^{\circ }$
$\Rightarrow y={136}^{\circ }$