Thanks for giving us two examples to reinforce the methodological approach for dealing with such problems. Keep up the good work! Have a great new year!
The remote angles or opposite interior angles and exterior angles are Supplementary! Along with the Triangle Sum Theorem (all interior angles add up to 180⁰) and the Linear Pair that is formed by the Exterior Angle and Adjacent Angle, setting the Triangle Sum Theorem and Linear Pair Equations Equal to each other makes it easy to solve for unknown angles of a Triangle that doesn't share a Side or Vertex with Exterior Angle. Real life applications are used in many fields. I absolutely love this stuff. Thanks for posting puzzles! 🙂
Now I attempt to do part 2, adopting the same skill, CE to make an isosceles triangle with angle 32,32, 116, then EAC is 16,16, 148, join BE to get ABE with 60,60,60, and EBC is isosceles triangle with 180-60-32=88, 46,46, x=46-16=30.😊
Please, I would like someone to clarify two questions for me: 1) In excerpt 05:13 of the video, how was the value of 55⁰ of the angle of the 📐 EDC obtained? 2)In excerpt 09:23, when the narrator says "...If we add all these three angles that adds up 280⁰...", How did that number come about?
Hello dear, in triangle EDC, one of the angle is 70. => Two other angles in this triangle are 110. This triangle is isosceles as well => so each angle is going to be the half of 110. That is 55 deg each. Hope I explained well 😀 At 09:23, I said 180 degrees! My apologies if I said it otherwise. Take care and stay blessed 🌹
In the first problem, I don't understand why we may draw the line BE. How do we now both angles D and B (from triangle DBE) are 25° when we draw the line BE?
Well: There was NOT A SINGLE THING relating to "exterior angle theorem". The whole solution comes down to the introduction of a useful helper point (E) and helper line (EC) - but not at all to something even remotely related to "exterior angle theorem". I'm somewhat disappointed.
Iniziamo il 2024 facile,molto facile..il1)..teorema dei seni a/sin25=t/sin50...a/sin(x+35)=t/sinx...divido le equazioni risulta..sin(x+35)/sin25=sinx/sin50.,ctgx=1/2cos25sin35-ctg35..x=-65...x=-65+180=115....il 2) è lo stesso procedimento...ctgx=2cos16/sin44-ctg44=√3...x=30
Happy new year to all fans of this channel. And thanks a lot for the variety of very nice challenges in the past year.
Thanks ❤️
Happy, safe, and prosperous New Year!
Thanks for giving us two examples to reinforce the methodological approach for dealing with such problems. Keep up the good work! Have a great new year!
Thanks ❤️
Happy, safe, and prosperous New Year!
The remote angles or opposite interior angles and exterior angles are Supplementary! Along with the Triangle Sum Theorem (all interior angles add up to 180⁰) and the Linear Pair that is formed by the Exterior Angle and Adjacent Angle, setting the Triangle Sum Theorem and Linear Pair Equations Equal to each other makes it easy to solve for unknown angles of a Triangle that doesn't share a Side or Vertex with Exterior Angle. Real life applications are used in many fields. I absolutely love this stuff. Thanks for posting puzzles! 🙂
Thanks ❤️
Happy, safe, and prosperous New Year!
👍🏻❣️
Thanks ❤️
Happy, safe, and prosperous New Year!
Thank you
Thanks ❤️
Happy, safe, and prosperous New Year!
Now I attempt to do part 2, adopting the same skill, CE to make an isosceles triangle with angle 32,32, 116, then EAC is 16,16, 148, join BE to get ABE with 60,60,60, and EBC is isosceles triangle with 180-60-32=88, 46,46, x=46-16=30.😊
Thanks ❤️
Happy, safe, and prosperous New Year!
Nice shrieng❤❤❤❤
Thanks ❤️
Happy, safe, and prosperous New Year!
Happy New Year 2024 Premath friends
Triangle AEB is equilateral.
Good evening, Pre Math, how's it going? "May hope bloom in your heart like a garden of infinite possibilities." Happy New Year!❤❤❤❤❤❤
Thanks ❤️
Happy, safe, and prosperous New Year!
Please, I would like someone to clarify two questions for me:
1) In excerpt 05:13 of the video, how was the value of 55⁰ of the angle of the 📐 EDC obtained?
2)In excerpt 09:23, when the narrator says "...If we add all these three angles that adds up 280⁰...", How did that number come about?
Hello dear, in triangle EDC, one of the angle is 70.
=> Two other angles in this triangle are 110. This triangle is isosceles as well => so each angle is going to be the half of 110. That is 55 deg each.
Hope I explained well 😀
At 09:23, I said 180 degrees! My apologies if I said it otherwise.
Take care and stay blessed 🌹
Good evening, Pre Math. The two angles of the EDC triangle are 70⁰ and 25⁰ and the sum of both is 85.
In the first problem, I don't understand why we may draw the line BE. How do we now both angles D and B (from triangle DBE) are 25° when we draw the line BE?
You make the angle 25, and the rest follows.
@@RAG981I see now, thanks for your response!
Just to clarify...
AB = BE ⇒ ∠BAE = ∠EAB = 50° which is an exterior ∠ of ΔDEB.
∠EAB = ∠EDB + ∠DBE
50° = 25° + ∠DBE ⇒ ∠DBE = 25°.
Happy, safe, and prosperous New Year!
@@gulshanjoshi7682 Thank you, this explanation is very clear. I appriciate it a lot
Nice bro
Thanks ❤️
Happy, safe, and prosperous New Year!
Well: There was NOT A SINGLE THING relating to "exterior angle theorem". The whole solution comes down to the introduction of a useful helper point (E) and helper line (EC) - but not at all to something even remotely related to "exterior angle theorem". I'm somewhat disappointed.
1)BDC=30
Thanks ❤️
Happy, safe, and prosperous New Year!
Iniziamo il 2024 facile,molto facile..il1)..teorema dei seni a/sin25=t/sin50...a/sin(x+35)=t/sinx...divido le equazioni risulta..sin(x+35)/sin25=sinx/sin50.,ctgx=1/2cos25sin35-ctg35..x=-65...x=-65+180=115....il 2) è lo stesso procedimento...ctgx=2cos16/sin44-ctg44=√3...x=30
Thanks ❤️
Happy, safe, and prosperous New Year!