Math Olympiad | Find height h in the triangle | [Important Geometry skills explained]

Learn how to find the height h in the triangle. Important Geometry and Algebra skills are also explained: Pythagorean theorem; Triangle sum theorem. Step-by-step tutorial by PreMath.com
Today I will teach you tips and tricks to solve the given olympiad math question in a simple and easy way. Learn how to prepare for Math Olympiad fast!
Step-by-step tutorial by PreMath.com
• Math Olympiad | Find h...
Math Olympiad | Find height h in the triangle | [Important Geometry skills explained] #math #maths
Olympiad Mathematical Question! | Learn Tips how to solve Olympiad Question without hassle and anxiety!
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[@bigm383]

Professor Premath is a godsend!👍🥂❤

[@PreMath]

Wow, thanks! You are very generous, my dear friend. Keep it up 👍 Stay blessed.

[@bigm383]

@@PreMath 😀🍺

[@engralsaffar]

I did it in a weird way! I named the right hypotenuse c and the left a I used sin(2x)=2sinxcosx h/c =2* 21/a*h/a Or 42 c = a^2 Then I used pythagoras twice h^2=c^2-784 h^2=a^2-441=42c-441 Then subtract them c^2-42c-343=0 (c-49)(c+7)=0 Giving c=49 Substitute back h^2=49^2-28^2=1617 h=7*sqrt(33)

[@misterenter-iz7rz]

28tan2x=21cotx, 28tanxtan2x=21, 56tan^2 x=21(1-tanx^2),77tan^2 x=21, tanx=sqrt(3/11), h=21cotx=21sqrt(11/3)=40.212 approximately. 😊

[@PreMath]

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

[@abdulmajidmohieldin1194]

Also I solve it by your method 👍

[@abdulmajidmohieldin1194]

Sorry solved!!!

[@markkinnard796]

Very nice.

[@williamwingo4740]

There's also a trigonometric solution: tan x = 21/h; and tan 2x = h/28; but tan 2x = (2 tan x/(1 -- tan^2 x); so h/28 = (42/h)/(1 -- (21/h)^2) = (42/h)/((h^2 -- 1)/(441/h^2)); a purely algebraic expression. After some manipulation (tedious details omitted) we get the same answer: h^2 = 1617 = (49)(33); so h = sqrt((49)(33)) = 7sqrt(33) = 40.2367, approximately. Cheers. 🤠

[@umerhussain3855]

I actually got the right answer, but I had to use A-level Maths tricks like the Double Angle Identity for Tan(2x) which was a lot longer than your method, But I do believe that the way you solved it is actually a really good method for people who haven’t gotten to higher level mathematics yet. Appreciate your videos, as always.

[@covenslayer]

That's exactly what I did. Glad we got showed another approach.

[@117hippo3]

Good!! I also think tan2X

[@mohanramachandran4550]

Very smart solution

[@harikatragadda]

By reflecting triangle CDA about AD to form another triangle ADC', a circle of radius (21+ 28 = 49) can be drawn passing through A, B, C such that the chord AC subtends 2x at the center and x at C' and by symmetry also x at C. By extending AB to D on the circle, h is the perpendicular distance to the base of the Right Triangle ADC. h² = 21*(28+49) h = 7√33

[@PreMath]

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

[@markkinnard796]

Your solution is beautiful! Thank you, I really learned something. I tried to find a synthetic solution like yours, but was unsuccessful.

[@TNKFF]

I am unable to understand

[@TNKFF]

Now , I understand. You did it so nicely

[@KAvi_YA666]

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

[@ghhdcdvv5069]

تمرين جيد . رسم واضح مرتب. شرح واضح مرتب. شكرا جزيلا لكم والله يحفظكم ويحميكم ويرعاكم جميعا .تحياتنا لكم من غزة فلسطين

[@KookyPiranha]

if ur doing math olympiad, it's highly reccomended to learn ahead of your grade level so u can just use basic methods

[@JLvatron]

Yes, I got it!

[@devondevon4366]

Answer 40.21 tan x = 21/h tan 2x =h/28 equation 2 but tan 2x = 2 tanx { double angle formulae for tangent) -----------' 1 - tan ^2x Hence 2 tan x = 2 (21/h) = 42/h tan ^2x = (21/h)^2 = 441/h^2 hence 1 -tan 2x= 1 - 441/h^2 = ( h^2 -441)/h^2 hence tan 2x = 42/h * h^2/(h^2 -441) = 42 * h/(h^2 -441) [ h cancels out] tan 2x = h/28 [ see equation 2 ] hence h/28 = 42 * h/(h^2-441) 1/28 = 42 * 1(h^2 -441) [ multiply both sides by 1/h] 1/28 * 1/42 = 1/(h^2 -441) [ multiply both sides by 1/42] 42 * 28 = h^2 -441 [ cross multiply 1176 = h^2 -441 1176 + 441 =h^2 1617 =h^2 40.21 = h (square root of both sides)

[@misterenter-iz7rz]

Note that BAC is an isosceles triangle, thus h=sqrt(49^2-28^2)=7sqrt(49-16)=7sqrt(33)=40.212 approximately 😅

[@NaceerAlassady-x2l]

Good job......we can solve it in other way by using trigonometric functions

[@MultiYUUHI]

tanx=21/h tan2x=h/28 tan2x=2tanx/(1-(tanx)^2)=h/28 42/(h^2-21^2)=1/28 h=7√33

[@PreMath]

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

[@ShivrajTattapure-kf4sc]

my method was same

[@gelbkehlchen]

Solution: angle DAC = 90°-x = 90°-2x+x = angle DCB + angle ACD = angle ACB ⟹ CB = AB = AD+DB = 21+28 = 49 Pythagoras in the triangle DBC: h²+28² = 49² |-28² ⟹ h² = 49²-28² |√() ⟹ h = √(49²-28²) = √(7²*7²-7²*4²) = 7*√(7²-4²) = 7*√33 ≈ 40,2119

[@HeribertoEspinosa-t7o]

I see a lot of work with tangents, but I don't see anyone taking advantage of the Law of Sines by combining both triangles into one larger triangle. Here's my work: sin(∠ACD+∠BCD)/(AD+BD) = sin(∠CAD)/(BC) sin(x+(90-2x))/49 = sin(90-x)/sqrt(h^2+28^2) sin(90-x)/49 = sin(90-x)/sqrt(h^2+28^2) cosx/49 = cosx/sqrt(h^2+28^2) cosx/49 - cosx/sqrt(h^2+28^2) = 0 cosx(1/49 - 1/sqrt(h^2+28^2)) = 0 cosx = 0 (solutions irrelevant), 1/49 - 1/sqrt(h^2+28^2)) = 0 . . . h^2 +28^2 = 49^2 h^2 + (7^2)(4^2) - (7^2)(7^2) = 0 h^2 + (7^2)((4^2) - 7^2)) = 0 h^2 - (33)7^2 = 0 h = ±7sqrt(33) h = 7 sqrt(33)

[@marioalb9726]

tan x = 21/ h tan 2x = h / 28 h = 21/tan x = 28 . tan 2x tan 2x . tan x = 21/28 x = 27,57° h = 40.2 cm ( Solved √ )

[@117hippo3]

thank you for the good solution!! I think another method to use tanX ^^ first, angle A = 90-x, tan(90-x)= h/21, and tan2x = h/28 tan(90-x)=cotx = h/21, so cotx=1/tanx=h/21, tanx= 21/h and tan2x = 2tanx/1-(tanx)^2 , then (2*21/h)/1-(21/h)^2 = h/28 , If I summarize it in h, h^2= 7^2*33 at last h=7Root(33) ^^

[@wuchinren]

Your method can be more simplified. tan(2x)=CD/BD=h/28 tan x=AD/CD=21/h tan(2x)=(2*tan x)/[1-(tan x)^2] so h/28=(2*21/h)/[1-(21/h)^2] The following part is the same with yours.

[@NahidMiah-e6m]

The first viewer, please prined me shake a leg. Love from my deepest heart ❤️ for your sharing your super knowledge. Congratulations from Bangladesh 🇧🇩.

[@PreMath]

You're the best! Thank you, dear! Cheers! 😀 You are awesome. Keep it up 👍

[@gersantru]

Another way is: tanx=21/h, togheter with tan2x=h/28 and solve for h.

[@ybodoN]

To express h in terms of AD and DB, today's formula is: h = √(AD² + 2 ⋅ AD ⋅ DB) 💡

[@ShivrajTattapure-kf4sc]

sir we can solve it by another method In tringle ADC angle CDA=90 degree by appying tan ratio tan (x)=21/h now in triangle CDB angle CDB =90 tan (2x)= h /28 we know that tan 2x =2tan x /(1 -- tan x^2 ) h/28= (2. 21/h)/ (1-441/h^2 ) after solving we get same answer

[@murdock5537]

Nice! Great way to solve it, many thanks, Sir! tan⁡(x ) = 21/h → 2tan⁡(x) = 42/h → tan^2(x) = (21/h)^2 → 1 - tan^2(x) = (1/h^2)(h^2 - 21^2) → tan⁡(2x) = h/28 = (2tan⁡(x))/(1 - tan^2(x)) = (42/h)/((1/h^2 )(h^2 - 21^2)) → h^2 = 28(42) + 21(21) = 1617 = 49(33) → h = 7√33 ≈ 40,21 → x ≈ 27,575°

[@gregoryeze9329]

You can use the tan2x formula

[@wackojacko3962]

@ 7:53 ,... Approximations over very long distances have been resolved with Fresnal equations. Just sayin . 🙂

[@PreMath]

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

[@NaceerAlassady-x2l]

‏‪0:02‬‏ ‏‪0:02‬‏ ‏‪0:02‬‏

[@Copernicusfreud]

Yay! I solved it.

[@aboubachir7286]

This exercise has been a struggle for me for a while. Finding competent people to help me is always a pleasure for me. ABC is a rectangular triangle on A. Proves that: (AB - AC)² (BC² + 4 * AB * AC°)² = < 2 * BC ^6

[@aboubachir7286]

@@Grizzly01-vr4pnThe triangle is right-angled at A. English is not my native language. I would like to be corrected.

[@mibsaamahmed]

I used a rather complicated method. I calculated the height using tan in both triangles and set them equal as they were both for the height then I simplified and got something along the lines of 37.789 or something so I rounded it of to 40 units anyways nicee

[@markkinnard796]

I did that too and used numerical methods to get 40.212

[@giuseppemalaguti435]

Per il teorema dei seni...sinx=sqrt(3/14)... Quindi h=7sqrt33

[@mudangkano5267]

I used trigonometry to solve this.

[@DOGLAPAN3015]

tanx=21/h and tan2x=h/28 tan2x = 2tanx/1-tan*2x by solving these we get h=root 1617 = 40.21 Sir comment on my comment please

[@ShivrajTattapure-kf4sc]

my method was same

[@vidyadharjoshi5714]

AB = BC = 49. 2X = 55.15, h = 40.21

[@PreMath]

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

[@JSSTyger]

h = sqrt(49^2-28^2) h = sqrt(7^2(49-16)) h = 7sqrt(33)

[@bryananthony2192]

7 (sqrt 33)

[@racquelsabesaje4562]

42

[@racquelsabesaje4562]

21 28 2x

[@racquelsabesaje4562]

28+14= 42

[@devondevon4366]

40.2119