Find the Distance X in this Triangle | Quick & Easy Tutorial

[@pranavamali05]

Everyday i just egerly Waite for the notification of pre math well done

[@PreMath]

So nice of you Pranav dear Thank you! Cheers! You are awesome. Keep it up😀 Love and prayers from the USA!

[@SupratimBhowmick]

I have been following your math videos for a few days. I always pause the video for 5-10 minutes before resuming for the solution. Today for the first time I could solve a problem unassisted. I used the same methodology as yours. And I did the calculations mentally without using paper or calculator. It's all because of your teaching. Thank you.

[@silviatotaro9372]

I am an Italian mathematician. Now I am retired, you make my brain awake! My English is not so good, but my compliments are.

[@flyingspirit3549]

Your English is just fine! Had you not said that you were not a native speaker of English, I would not have guessed!

[@philipkudrna5643]

Before watching: by a^2+b^2=c^2 and if we divide the isosceles triangle into two right triangles, the height of the triangle ABC is sqrt(48). If we create a right triangle CED (E being the middle between A and B), it follows that (1+x)^2+48=64 or x=3. (No quadratic formulas or law of sines or cosines needed…) After watching: 😀 Exactly what I did!

[@PreMath]

Super Philip! Thank you for sharing! Cheers! You are awesome. Keep it up😀

[@iZAPMath]

Best seller! It's not easy to come up with new math problems every day. You are pretty much writing an interactive math book!

[@PreMath]

Wow, thanks! So nice of you dear You are awesome.😀

[@ABhaim]

In triangle ABC, according to the law of cosines: 7^2 = 7^2 + 2^2 - 2*7*2*cos(alpha), so cos(alpha) = 1/7 in triangle BCD, angle B and angle alpha are complimentary to 180 deg. so cos(B) = -cos(alpha) = -1/7. applying that yet again to the law of cosines, we get: 8^2 = 7^2 + x^2 + 2*7*x/7 x^2 + 2x - 15 = 0 taking only a non-negative solution, x=3

[@PreMath]

Super job Ohad! Thank you! Cheers! You are awesome. Keep it up😀

[@calspace]

The same way I did it.

[@guilhermelessadeazeredo7223]

I used the Law of Cosines and got the same result. First, I used it to calculate the cosine of the angle A. I had: 7^2=7^2+2^2- (2)(2)(7)(coisa). At the end, I found cosA to be 1/7. After this, I applied the same Law to find the length of (2+x). So, I had: 8^2=7^2+(2+x)^2- 2(2+x)(7)(cosA). I had a quadratic formula to solve but it was all good at the end. :)

[@PreMath]

Excellent Guilherme! Thank you! Cheers! You are awesome. Keep it up😀

[@holyshit922]

I used the same approach

[@mahatmapodge]

Because the base was 2, I calculated the height of the isosceles triangle by calculating its area S = (a+b+c)/2 A= sqrt((s(s-a)(s-b)(s-c)) And the base is length 2, so area ==perpendicular height The rest I did the same way as you

[@PreMath]

Super job! Thank you! Cheers! You are awesome. Keep it up😀

[@davidfromstow]

Did it slightly differently and ended up with (x+5)=0 or (x-3)=0; since x cannot be a negative number x must equal 3. Your maths lessons are working well!

[@PreMath]

Super job David! Thank you! Cheers! You are awesome. Keep it up😀

[@marcello3621]

We can use the Stewart's Relation too. 7²x + 8²2 - 7²(2 + x) = (2 + x)2x 49x + 128 - 98 - 49x = 4x + 2x² -2x² - 4x + 30 = 0 x' = -5 x" = 3 x = 3, because a length can't be negative.

[@lowdaolei615]

From tri ABC, we can easily obtain Cos A = 1/7. From tri ABD, using the Cisine formula, 8^2=7^2+(1+x)^2-2.7.(1+x)cos A 64 = 49 + 1 + 2x + x^2 - 2 . 7 (1+x) . 1/7 64 = 48 + x^2 x^2 = 16 x = +4 or - 4

[@darkomarkovic3373]

Very elegant solution of this not too hard problem! 👍

[@kennethstevenson976]

It was great to solve a problem and have the solution agree with yours step-for-step, very good feeling.

[@PreMath]

Super job Kenneth! Thank you! Cheers! You are awesome. Keep it up😀 Love and prayers from the USA!

[@RealSlopeDude]

Interesting problem. I used the Law of Cosines to find angle B of the isosceles triangle and worked it from there. Your solution was definitely easier than mine (and didn't require a calculator). Thanks.

[@dhrubajyotidaityari9240]

Angle ABC=θ, Angle CBD= 180 -θ. cosθ=1/7 & cos(180-θ)=-1/7 -1/7={7²+x²-8²)/2.7.x x=3.

[@PreMath]

Super! Thank you Dhrubaijyoti! Cheers! You are awesome. Keep it up😀

[@satyapalsingh4429]

Your method of solving the problem is interesting , dear professor ! Keep it up!

[@kafaichan3371]

1. Draw a perpendicular line from pt C to line AB (height = h), since ABC is an isos triangle -> this line bisects AB. Apply Pythagorean Theorem to find h (sqrt(48)) first, then (1+x)^2 + h^2 = 8^2 -> x^2+2x-15 = 0 -> (x-3)(x+5) = 0 -> x=3 or x=-5 (rejected) 2. Use cosine law, 7^2 = 7^2+2^2 -2(2)(7)cosB -> cosB = 1/7, 8^2 = 7^2+x^2-2(7)(x)cos(180-B) -> x^2+2x-15 = 0 -> (x-3)(x+5) = 0 -> x=3 or x=-5 (rejected)

[@PreMath]

Great job Chan! Thank you! Cheers! You are awesome. Keep it up😀

[@temyraverdana6421]

After your explanation, everything it's so easy

[@PreMath]

Glad to hear that! Thank you Temyra! Cheers! You are awesome. Keep it up😀

[@TinkeringJohn]

It is neat that these problems can be solved without using trig. When I see a problem with triangles, I always use trig to try to solve them not knowing the solution can be reached through just using algebra.

[@davidr1781]

Circle radius 7 at centre c. Sectors intersect extermally at D. Sector sector ratio rule x(2+x) = 1(7+7+1) x=3

[@govindashit6524]

Simple problem,but I love your explanation very much. Thanks sir .

[@PreMath]

You are most welcome Govinda So nice of you dear Thank you! Cheers! You are awesome. Keep it up😀

[@albertomarin5264]

the solution of the quadratic equation (x = -5) means another side CE to the left of AC with distance 3 from segment EA

[@sandanadurair5862]

In triangle ABC Using c^2=a^2+b^2-2abCos(c) we get 49=4+49-28.cosB CosB = 1/7 In triangle BCD 64=X^2+49-14X.Cos(180-B)....eqn1 Since Cos(180-B)= -CosB Cos(180-B) = -1/7 Hence Eqn 1 becomes 64 = X^2+49+14.X/7 X^2+2X-15 = 0 Positive root of X = 3 : Ans

[@mcorruptofficial6579]

Thanks, solved with the same steps you performed. Regards

[@PreMath]

Super job! Thank you! Cheers! You are awesome. Keep it up😀

[@devondevon3416]

x=3 And this is proven by Heron's formula as the triangle with sides 7, 7, 2 plus the triangle with sides 7,8,3 = the largest triangle with sides 7, 8, 5

[@PreMath]

Super! Thank you my friend! Cheers! You are awesome. Keep it up😀

[@spacepotato3677]

I thought of using heron's formula then substitute with base*height/2 and then use Pythagoras theorem. Nonetheless, i think your method is more efficient than mine. Great video. Very thorough explanation. Easy to interpret 🤓

[@PreMath]

Great suggestion! Thank you Vince! Cheers! You are awesome. Keep it up😀

[@krislegends]

I got 3 by using laws of cosine to find angle b, subtracting it from 180 to get angle cbd, then use laws of sines to find x

[@PreMath]

Excellent Kris! Thank you Dhrubaijyoti! Cheers! You are awesome. Keep it up😀

[@Gargaroolala]

I used cosine rule to find angle CBA. Then I can find angle CBD. I then used sine rule to find angle CDB. Then I can find angle DCB. I used sine rule again to find x.

[@PreMath]

Nice tip Thank you Garrick! Cheers! You are awesome. Keep it up😀

[@skwbusaidi]

I used the cosine law on trinagle to ABC to find the cos A And used again on triangle ACD

[@debdasmukhopadhyay4692]

Very good indeed. Yes now I eagerly wait for your videos. Thanks a lot.

[@PreMath]

So nice of you Debdas dear Thank you! Cheers! You are awesome. Keep it up😀 Love and prayers from the USA!

[@DB-lg5sq]

شكرا لكم يمكن استعمال sin في ABCنجد cos BAC =1/7 مبرهنة الكاشي فيBCD نجد X=3

[@arshadalam8550]

Sir we also can apply Stewart theorem

[@PreMath]

Here is the video link for the Stewart Theorem that I've already uploaded: kzbin.info/www/bejne/haqvnoOgo6Z8hqs

[@PreMath]

Great tip Thank you Arshad! Cheers! You are awesome. Keep it up😀

[@SonuKumar-ml5yz]

I think CE will act as a median only if it is angle bisector of ACE and it's not given.

[@susennath6035]

Pls make some parts in book format for our future collection

[@PreMath]

Great suggestion! Thank you Susen! Cheers! You are awesome. Keep it up😀

[@predator1702]

Awsome, thank you teacher 🙏.

[@PreMath]

You are welcome! Thank you dear! Cheers! You are awesome. Keep it up😀

[@242math]

very well explained, thanks for sharing bro

[@PreMath]

So nice of you dear friend Thank you! Cheers! You are awesome.😀 Love and prayers from the USA!

[@mikelm222]

Does the solution really have to be that long winded ? for me once we get the square of perpendicular as 48, Then it's just x + 1 = root ( 64 - 48 )

[@theophonchana5025]

CE = sqrt (48)

[@mahalakshmiganapathy6455]

Great explanation thank you

[@PreMath]

So nice of you Mahalakshmi dear Thank you! Cheers! You are awesome. Keep it up😀 Love and prayers from the USA!

[@lailafernandes3649]

Muito bonito 🇧🇷

[@PreMath]

Dear Laila, Tão legal da sua parte, querida. Obrigado! Saúde! Você é demais. Continue assistindo😀 Amor e orações dos EUA!

[@kalyanbasak6494]

Namaskar sir,from area of triangle I can get height✓48unit H²-perpendicuka²=✓16=4unit, X=4-1=3unit answer sharing

[@PreMath]

Super! Thank you Kalyan! Cheers! You are awesome. Keep it up😀 Love and prayers from the USA!

[@xellosblackforest1685]

*interesting approach* ..

[@PreMath]

Glad you think so! Thank you Xellos! Cheers! You are awesome. Keep it up😀

[@alexramzes8021]

good, but i used theorem of sines, theorem of cosines and equality tgA+tgB+tgC=tgA*tgB*tgC

[@nicholascage3699]

Go for easy question....your channel will be a popular!

[@PreMath]

Thanks Nicholas for the suggestion. Our efforts is to give a mixture of problems- easy and challenging both! You are awesome. Keep it up😀

[@avijitdas2683]

This is very simple to solve . Hight is the main point of this question

[@PreMath]

Great tip! Thank you Avijit! Cheers! You are awesome. Keep it up😀

[@eldrixace8902]

Very nice.

[@skwbusaidi]

On tringle ABC 7^2= 2^2+7^2 -2(2)(7) cos A Cos A = 1/7 On trinagle ACD 8^2=7^2+ (x+2)^2- 2(7)(x+2) × 1/7 (x+2)^2 -2(x+2) -15=0 Let u= x+2 u^2-2u-15 =0 u=5 or u=-3 Reject -ve x+2=5 x=3

[@eleall5295]

Actually, the best way is Stewart's Theorem

[@Yurchik55]

sin CDE=b/c=sqrt(3)/2, i.e. CDE=60. EAD=30/ Ed=8:2=4/ BD=4-1=3.

[@restablex]

Notice that √48 = 4(√3), 8 = 4(2)... So remaining side is 4(1) (triangle 1,√3,2)... Finally 4 - 1 = 3

[@PreMath]

Super! Thank you! Cheers! You are awesome. Keep it up😀

[@bentels5340]

What a coincidence. Solved it exactly the same way. 😉

[@PreMath]

Excellent Ben! Thank you! Cheers! You are awesome. Keep it up😀

[@theophonchana5025]

#triangle #isoscelestriangle #RightTriangle

[@mustafizrahman2822]

x=2.99998 mainly 3 units. I have solved it laws of cosine.

[@PreMath]

Super! Thank you Mustafiz! Cheers! You are awesome. Keep it up😀

[@jayquirk2297]

I did it that way, too, but I like the video's method better. Anytime you can get the answer without requiring a calculator is worth it. And then you don't have those nasty rounding errors.

[@theophonchana5025]

Right triangle CEB

[@DeepakKumar-nt2sd]

We can solve this question by Stewart theorem

[@PreMath]

Yes! Thanks Deepak 😀 Here is the video link for the Stewart Theorem that I've already uploaded: kzbin.info/www/bejne/haqvnoOgo6Z8hqs

[@AmirgabYT2185]

x=3

[@sameerqureshi-kh7cc]

Sir, Pakistan has sent flowers for you 🌹🌹🌹🌹

[@PreMath]

Wow! Thank you Sameer dear! You are awesome. Keep it up😀 Love and prayers from the USA!

[@nicholascage3699]

Understanding! Why 1+ 1= 2 !

[@Themiddlemanofnothing]

i got 3 using isoceles then pithagoras twic

[@PreMath]

Super Manuel! Thank you! Cheers! You are awesome. Keep it up😀

[@theophonchana5025]

#linearequation #binomial #polynomial

[@theophonchana5025]

b^(2) = 48 b = sqrt (48)

[@theophonchana5025]

a = 1 b = sqrt (48) c = 7

[@_josuke6034]

I got tis one :)

[@PreMath]

Super job! Thank you! Cheers! You are awesome. Keep it up😀

[@lawrencek1900]

used same method

[@theophonchana5025]

#Pythagoras #PythagoreanTheorem

[@billylowgroundvytaszukas5797]

true medizine to watch one video each morning by drinking coffee

[@holyshit922]

Law of cosines twice and we should get a result

[@PreMath]

Great tip Thank you Jacek! Cheers! You are awesome. Keep it up😀

[@raghavnegi232]

Alternative method

[@PreMath]

Dear Raghav, here is the alternative method: Stewart Theorem: Here is the video link for the Stewart Theorem that I've already uploaded: kzbin.info/www/bejne/haqvnoOgo6Z8hqs

[@susennath6035]

Nice

[@susennath6035]

Pls make a book for our future collection

[@PreMath]

So nice of you Susen Thank you! Cheers!😀

[@theophonchana5025]

Isosceles triangle ABC

[@adgf1x]

7/8=2/x=>7x=16=>x=16/7ans

[@shreyanshpatel9740]

stewart's theorem

[@theophonchana5025]

(x + 1)^(2) = 16 x + 1 = 4 x = 3

[@claireindongole8302]

27

[@oleg3701]

3 in less as 10s

[@adgf1x]

X=16/7

[@giuseppemalaguti435]

A me risulta 3 5

[@theophonchana5025]

Right triangle CED