TRIANGLE
[tex]\normalsize{\blue{\overline{\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad}}}[/tex]
[tex]\rm{\underline{QUESTION:}}[/tex]
50.) In the adjoining figure, side BC of triangle ABC is extended to D. What is value of aº?
[tex]\\[/tex]
[tex]\rm{\underline{SOLUTION:}}[/tex]
First of all, we know that, <ACB + <ACD = 180º
- [tex]\rm{\rightarrow 6y^{\circ} + 3y^{\circ}= 180^{\circ}}[/tex]
- [tex]\rm{\rightarrow 9y^{\circ}= 180^{\circ}}[/tex]
- [tex]\rm{\rightarrow \dfrac{9y^{\circ}}{9} = \dfrac{180^{\circ}}{9} }[/tex]
- [tex]\rm{\rightarrow y^{\circ} = 20^{\circ}}[/tex]
[tex]\\[/tex]
Therefore:
- [tex]\rm{ < ACB = 6y^{\circ} = 120^{\circ}}[/tex]
- [tex]\rm{ < ABC = 2y^{\circ} = 40^{\circ}}[/tex]
- [tex]\rm{ < ACD = 3y^{\circ} = 60^{\circ}}[/tex]
Now that we have the values, we can solve for aº:
- [tex]\rm{ a^{\circ} = 180^{\circ} -(40^{\circ} + 120^{\circ} )}[/tex]
- [tex]\rm{ a^{\circ} = 180^{\circ} - 160^{\circ} }[/tex]
- [tex]\rm{ a^{\circ} = \boxed{\green{20^{\circ} }}}[/tex]
[tex]\\[/tex]
Hence, the answer to your question would be option A. 20 or D. 20. You can answer any of them since they are both 20.
[tex]\normalsize{\blue{\overline{\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad}}}[/tex]
[tex]\large{\boxed{\tt{\blue{06/12/2024}}}}[/tex]
