Construct a $Δ A B C$ in which $A B = 5.8 c m, ∠ B = 60^{\circ}$ and $B C + C A = 8.4 c m$ . Justify your constructions.

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Solution

Steps of construction:
1. Draw a line segment $A B = 5.8 c m .$
2. Construct $∠ A B X = 60^{\circ}$
3. Cut an arc $B P = 8.4 c m$ taking B as the center.
4. Join $P A .$
5. Draw the right bisector of $P A,$ meeting $B P$ at $C .$
6. Join $A C .$
Thus, $Δ A B C$ is the required triangle.

Justification:
In $Δ A P C,$ we have
$∠ C A P$ $= ∠ C P A$
[By construction,(Triangles on the two sides of perpendicular bisector) the two triangles are congruent by SAS congruence Rule]
$\Rightarrow C P = A C$ (Sides opposite to equal angles are equal)
Now, $B C = P B - P C = P B - A C$
$\Rightarrow B C + A C = P B = 8.4 c m$

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