📐thanks again - everyday is a premath day 📏

@PreMath

9 ай бұрын

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Angle DEC = 30 degrees, angle CED = 30 degrees as well. Then CD and DE are congruents. Both segmentes have size of 15 units. Then BC = 45. The side "x" is half of BC, then x = 22.5.

@salimahmad7414

9 ай бұрын

Exactly, This is how it can be done easily.

@PreMath

9 ай бұрын

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍

@aiboljusupov9116

8 ай бұрын

I found tha same way

@PreMath

9 ай бұрын

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Third method; starts out like the first, but takes a different direction in the middle. All angles are in degrees. EDB = 60, so EDC = (180 -- 60) = 120. CED = 30, so ECB = (180 -- 120 -- 30) = 30. Therefore triangle DEC is isosceles, as you said, even though it doesn't look it; but hey, the diagram is "not necessarily to scale." Anyway, since the triangle is isosceles, side DE = side DC. In a 30-60-90 triangle like EDB, the side opposite the 30-degree angle is one-half the hypotenuse; so DC = DE = (DB/2) = 30/2 = 15; and BC = 30 + 15 = 45. And of course the big triangle ABC is also 30-60-90 and CA is opposite the 30-degree angle; so x = 45/2 = 22.5. Alea jacta est! 🤠

@PreMath

9 ай бұрын

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍

@jhill4874

9 ай бұрын

Me, too!

@williamwingo4740

8 ай бұрын

@@PreMath I've always thought so but I'm usually to modest to mention it.

Because angle BDE is and exterior angle to triangle CDE, we have angle BDE = angle DCE + angle DEC, so that 60 = 30 + angle DEC => angle DEC = 30, which makes triangle CDE isosceles => CD = DE = 15, so that segment BC = BD + CD = 30 + 15 = 45. Because triangle is a 30-60-90 right triangle, AC = (1/2)(BC), So, x = (1/2)(45) = 45/2 = 22.5. Done!

@PreMath

9 ай бұрын

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍

Thanks for video.Good luck sir!!!!!!!!!!

Well worked out.👍

@PreMath

9 ай бұрын

Thanks for your feedback! Cheers! 😀 You are awesome, Ramani. Keep it up 👍

Thank you

@PreMath

9 ай бұрын

You are very welcome! You are awesome. Keep it up 👍

Great problem! Good example to illustrate the importance of discovering all relationships and recording relationships on the diagram. (30,60,90 triangle) and isosceles triangle relationships.

@PreMath

8 ай бұрын

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Thank you, we always enjoy your teachings. We created a 3rd method by using twice the 30°-60°-90° triangle properties: AE = a (shortest leg), AC= a√3 = x, and CE = 2a = 15√3, therefore x = a√3 = ((15√3)/(2)) (√3) = 15(3)/2 = 45/2 =22.5 units

@PreMath

9 ай бұрын

Very good!❤️ Thanks for your feedback! Cheers! 😀 You are awesome. Keep it up 👍

Sir, which whiteboard do u use ?

Yay! I solved it. I used the trigonometry functions. I used law of sines. I used the proportions for the similar triangles. I also used the Pythagorean theorem. All the methods returned the value of x at 22.5 units. I suspect the answer is 22.5.

@PreMath

9 ай бұрын

Excellent! Thanks for sharing! Cheers! You are awesome. Keep rocking 👍

@shadowwarrior5697

8 ай бұрын

I do the same first i find DE by sin theta and then prove triangles DEB~CAB and make sides proportional

YEEESSS! I finally got one of your videos correct; its probably one of the easiest but am still happy.

@PreMath

9 ай бұрын

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Thanks Sir Thanks PreMath While geometry exercises are hard , we are enjoying with your explain . With my respects .

@PreMath

8 ай бұрын

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DE = 1/2 DB = 15 If angle DEA = 90° and angle CEA = 60° then angle DEC = 30°. If angle CBA = 30° then angle ACB = 60° and angle ECD = 30° => CD = DE = 15. CB = CD+DB = 15+30 = 45. If angle CBD = 30° => CA = 1/2 CB = 1/2•45 = 22.5

@PreMath

9 ай бұрын

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For the last part in the 1st method, I used the basic geometry i.e. 30-60-90 special rt∆EAC Since c= 15√3 and c= 2b b= 15√3/2 And a= b√3 a= (15√3/2)(√3) a= (15)(3)/2 a= 45/2 or 22.5 units And this is technically the 3rd method. On a side note, I always use mnemonics for the rt∆s legs... a for altitude and b for base.

EBD is a 30-60-90 therefore ABC is the same as is ACE we know BD is 30 units, therefore DE must be 15 units A line drawn from E perpendicular to a point P on BC bisects BC and creates another 30-60-90 Given DE ==15 , DP must then be 15/2 X==BC/2 == CP==PB DB == 30 DP == 15/2 X == DB - DP ≈= 22.5units

@PreMath

8 ай бұрын

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At a quick glance, The intersection of two lines, BC and EA at C give the horizontal distance x. The equation of a line is y= m * x + c. Where m is the slope and c is the intersection of the Y axis at x = 0. The two slopes are tan(60) = sqrt(3) and tan(30) = 1/sqrt(3). Moving our origin to E , the line CE has intersection of Y at 0 and line DB at D. Then the two line equations are Y1 = sqrt(3) and Y2 = 1/sqrt(3) + DE. The intersection when Y1 = Y2 = ( 'X' in diagram) . Then DE = sin(30) * 30 = 15. Then sqrt(3) *x = x/sqrt(3) + 15= - x*sqrt(3)( 1 - 1/ 3) = 15 . Then x = - 13. This is the distance AE Then X, the vertical height = tan (60) * 13 = 22.5

I like the first method only because it seems too give more insight too sciences related too geometry like navigation, solid and celestial mechanics, especially periodic functions sine cosine and tangent for analysis of equally spaced data. 🙂

@PreMath

9 ай бұрын

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x= 22.5 DE= 15 30-60-90 in which the height is 1/2 the hypotenuse CD= 15 since angle CE is 30 degrees CB =45 ( 15 + 30) Hence x= 22.5 or 1/2 the hypotenuse Second method DE: sine 30 degree= 1/2 * 30 = 15 Hence, CD= 15 isosceles triangle CB= 45 x/15 =45/30 since triangle ABC and triangle BDE are similarlar triangle. x/15 = 1.5 x= 15 * 1.5 = 22.5

@PreMath

9 ай бұрын

Thanks for sharing! Cheers! You are awesome. Keep it up 👍

Thanks for your english easy to understand by a French

@PreMath

8 ай бұрын

You are very welcome! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

DBE is a 30°60°90° right triangle so its minor cathetus DE = 15 and DE = CD because CDE is isosceles being angles at the base 30° CB = CD + DB = 15+30 = 45 then we can work on the similarity between ABC and DBE CB : DB = AC : DE 45 : 30 = X : 15 X = 45/2

@PreMath

9 ай бұрын

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method 3 DE=15 The triangle CDE is isosceles =>CD=15 BC=45 => AC=22,5

@PreMath

9 ай бұрын

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I used similar triangles. CAB and DEB are similar, so (AE+15sqrt3)/15sqrt3 = x/15. Crossmultiplication simplifies this to 15AE+225sqrt3=15sqrt3*x. Since the 30-60-90 relations stipulate that x=AEsqrt3, the previous equation can be written as 15AE + 225sqrt3 = 45AE. Solving for AE produces AE= (15sqrt3)/2. Again using the 30-60-90 relations means x = 45/2 or 22.5. Is this a third way, a fourth way?

@PreMath

8 ай бұрын

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Triangle ∆DEB: A/H = cos 30° EB/30 = (√3)/2 EB = 15√3 As ∠AEC = 30° and ∠CAE = 90°, ∠ECA = 30°. As ∠ABC = 30° and ∠CAB = 90°, ∠BCA = 60°. As ∠BCA = 60° and ∠ECA = 30°, ∠BCE = 30°. As ∠BCE and ∠EBC are both 30°, ∆CEB is isosceles and EC = EB = 15√3. Triangle ∆CAE: O/H = sin 60° x/15√3 = (√3)/2 x = 15(3)/2 = 45/2

x=22,5

This one I worked out mentally using ratios of right triangle sides. How useful they are! I’m sure many of you could have done or actually did the same.

@PreMath

8 ай бұрын

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⇒x=22.5 but I used a third way. And I could only do this using lessons you've taught in your videos. So thank you. And sorry if I was too detailed in doing the proof. You tend to use "Recall that" and "Concentrating on" so I included RECALL and "For", as well as listing the rules before using them. GIVEN: ∠B°=30, line DB=30 units RULE: Sin is Opp/Hyp FOR △EDB, ∠B°=30, line DB=30, sin 30°=.5 ∴ line ED=15 RULE: Angles of △ sum to 180° For △EDB, ∠B°=30, ∠DEB=90° ∴ ∠EDB=60° RULE: Angles of any line sum to 180° RECALL: ∠EDB=60°, line BC=180° ∴ ∠EDC=120° For line AB: line AB=180°, ∠BED=90° ∴ AED=90° RECALL ∠AEC=60°, AED=90° ∴ ∠CED=30° For △EDB, RECALL ∠CED=30°, ∠EDC=120° ∴ ∠ECD=30° For △EDB, RECALL ∠CED=30°, ∠ECD=30° ∴ △EDB is isosceles RULE: Lines congruent to the unequal angle of an isosceles triangle are equal. RECALL line ED=15, △EDB is isosceles ∴ line CD=15 For line CB: RECALL line DB=30, line CD=15 ∴ line CB=45 RECALL: line CB=45, ∠B°=30 sin 30°=.5 x=45•.5 ∴ x=22.5 Solution: x=22.5

Yeah, I solved it using trigonometry ratios

@PreMath

9 ай бұрын

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍

As CE=EB=30 root 3/2, x=CEroot3/2=30× 3/4=22.5😊.

@PreMath

9 ай бұрын

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍

Angle ACB=60° and angle ACE =30°. Thus, angle ECD=30°.

@PreMath

8 ай бұрын

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30cos30+x/tan60=x/tan30 ∴(√3-1/√3)x=30(√3/2) ∴x=45/2

@PreMath

9 ай бұрын

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sir what is your name by the way ?

@PreMath

9 ай бұрын

Premath😀

@arnavkange1487

9 ай бұрын

What really .....but its your channel name na ...I was asking about your own real name

A math problem a day keeps the doctor away

CD=DE=30/2=15 30+15=45 　　　　　x = 45/2=22.5

@PreMath

8 ай бұрын

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这个图画的有问题，根本不用计算，根据题目条件可得，DE=15，∠BCA=60°，∠DEC=30°，∠DCE=30°，△DCE为等腰三角形，DE=DC=15，BC=45，AC=1╱2BC=22.5

@PreMath

8 ай бұрын

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍

Second method: (x/root 3+30 root 3/2)/x=root 3, thus root 3 x=x/ root 3+15 root 3, (root 3-1/root 3)x=15 root 3, 45=2x, x=22.5😅

@PreMath

8 ай бұрын

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍

I guess this is one of the two methods shown in the video, but we will see: . .. ... .... ..... Triangle ABC: ∠ACB = 180° − ∠ABC − ∠BAC = 180° − 30° − 90° = 60° Triangle ACE: ∠ACE = 180° − ∠AEC − ∠CAE = 180° − 60° − 90° = 30° ∠DCE = ∠ACB − ∠ACE = 60° − 30° = 30° ∠CED = 180° − ∠AEC − ∠BED = 180° − 60° − 90° = 30° Because of ∠CED = ∠DCE the triangle CDE is isosceles: ∠CED = ∠DCE ⇒ CD = DE Triangle BDE: DE / BD = sin(∠DBE) DE / 30 = sin(30°) DE / 30 = 1/2 ⇒ DE = 15 CD = DE = 15 ⇒ BC = BD + CD = 30 + 15 = 45 The triangles ABC and BDE are similar: AC / DE = BC / BD x / 15 = 45 / 30 ⇒ x = 45/2 Best regards from Germany

@PreMath

9 ай бұрын

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My name is Jose' Jimenez. 😉

@PreMath

9 ай бұрын

Thank you, Jose'! Cheers! 😀 Welcome aboard ❤️

x = 22.5

@PreMath

9 ай бұрын

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X=22.5

@PreMath

9 ай бұрын

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X=22,5

@PreMath

9 ай бұрын

Thanks for sharing! Cheers! You are awesome. Keep it up 👍

X=45/2..si tratta semplicemente di triangoli simili

@PreMath

9 ай бұрын

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍

This seems inefficient. We know, as you said, that DE is 15 and that angle ECD (or ECB) is 30 degrees. CED is also 30, so triangle CED is isoceles. Thus CD is 15 and CB is 45. By 30-60-90 we know that AC (x) is half of CB, 22.5.