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Sagot :
Given:-
- ABC is a triangle.
- [tex]\angle[/tex]BAC = 40°.
- [tex]\angle[/tex]ABC = 70°.
To Find:-
- Measure of [tex]\angle[/tex]ACD.
Solution:-
➾Firstly, we will find the measure of [tex]\angle[/tex]ACB.
We know that,
Sum of all interior angles of a triangle = 180°.
[tex]\rightarrow[/tex][tex]\angle[/tex]BAC + [tex]\angle[/tex]ABC + [tex]\angle[/tex]ACB = 180°
[tex]\rightarrow[/tex] 40°+70°+[tex]\angle[/tex]ACB = 180°
[tex]\rightarrow[/tex] 110°+[tex]\angle[/tex]ACB = 180°
[tex]\rightarrow[/tex] [tex]\angle[/tex]ACB = 180°-110°
[tex]\rightarrow[/tex][tex]\angle[/tex]ACB = 70°
So, the measure of [tex]\angle[/tex]ACB = 70°.
➾ Side BC of the triangle is extended till D. So, we have to find the measure of [tex]\angle[/tex] ACD.
As we know that,
Sum of any two or more angles that lie on a straight line = 180°.
So, [tex]\sf\angle{ACB}[/tex]+[tex]\sf\angle{ACD}[/tex] = 180°
[tex]\rightarrow[/tex] 70° + [tex]\sf\angle{ACD}[/tex] = 180°
[tex]\rightarrow[/tex][tex]\sf\angle{ACD}[/tex] = 180°-70°
[tex]\rightarrow[/tex][tex]\sf\angle{ACD}[/tex] = 110°.
Therefore, measure of [tex]\sf\angle{ACD}[/tex] = 110°.
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