Triangles [NEW] — Maths STD 9 — Question

3. 70º

Solution:

AX is bisector of $\angle\text{DAC}.$

$\Rightarrow\angle\text{DAX}=\angle\text{XAC}=70^\circ$

$\Rightarrow\angle\text{DAC}=2\times70^\circ=140^\circ$

Now, $\angle\text{A}=180^\circ-\angle\text{DAC}=180^\circ-140^\circ=40^\circ$

In $\triangle\text{ABC},$

$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$

$\Rightarrow40^\circ+\angle\text{B}+\angle\text{C}=180^\circ$

$\Rightarrow40^\circ+\angle\text{C}+\angle\text{C}=180^\circ\dots(\angle\text{B}=\angle\text{C})$

$\Rightarrow2\angle\text{C}=140^\circ$

$\Rightarrow\angle\text{C}=70^\circ$

$\text{i.e}\angle\text{ACB}=70^\circ$