$\angle ACD$ is an exterior angle of $\Delta ABC$. If $\angle A = 50^{\circ}$ and $\angle B = 65^{\circ}$,then $\angle ACD = \dots$ (in $^{\circ}$)

  • A
    $110$
  • B
    $120$
  • C
    $115$
  • D
    $105$

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Similar Questions

State whether each of the following statements is true or false:
$(1)$ In $\Delta XYZ$,if $XY > XZ$,then $\angle Z > \angle Y$.
$(2)$ In $\Delta ABC$ and $\Delta PQR$,if $\frac{AB}{PR} = \frac{BC}{QP} = \frac{CA}{RQ} = 1$,then $\Delta ABC \cong \Delta RPQ$.

In the given figure,$PN$ and $QM$ are both perpendicular to line segment $PQ$. Also,$X$ is the midpoint of $PQ$ as well as $MN$. Prove that $\triangle PNX \cong \triangle QMX$.

If $\Delta ABC \cong \Delta YZX,$ then in $\Delta XYZ,$ which side is equal to side $AB$?

In $\Delta ABC$,$AB = 8 \, cm$ and $BC = 5 \, cm$,then $AC > \ldots \ldots \ldots cm$.

$Q$ is a point on the side $SR$ of a $\triangle PSR$ such that $PQ = PR$. Prove that $PS > PQ$.

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