A satisfying geometry question - circle exterior to a triangle side

Thanks to Amit from India for the suggestion! A circle is tangent on the exterior of a side of a triangle and tangent to the lines through the other two sides. The tangent side length 7, and the other side lengths are 8 and 9. What is the radius of the circle?
0:00 Problem
1:22 Geometry
5:35 Trigonometry
Twitter thread (additional solutions)
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Excircle formula
artofproblemsolving.com/wiki/...
Excircle formula Wikipedia
en.wikipedia.org/wiki/Incircl...
Tangent half-angle
en.wikipedia.org/wiki/Tangent...
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[@zzstoner]

Dark Mode: Activated. Nice. :)

[@fishcanroll0]

now its just organic chemistry channel

[@ArunaVishwajith123]

I am in Sri Lanka .... this is one of my favorite channels .... your service is very valuable .... I learned a lot from this channel ...Thank you PRESH TALWALKAR 😍😍

[@Spartans-community]

Gammac mageth favourite channel ekak dana hama ganama hadnw 🖤❤️

[@user-ci5kh7pl8y]

Hi. I'm from your neighbouring country.😁😁

[@Spartans-community]

@@user-ci5kh7pl8y India 🇮🇳 ✋

[@IS-py3dk]

@@user-ci5kh7pl8y Me too neighbouring country 😃 India

[@xxdxx3776]

@DoTheMath lmaooo

[@pwmiles56]

To avoid the cosine rule: make angle NOD=theta, angle DOM=phi. Then angle NOM = theta+phi By right triangle properties 3 = r tan(theta/2), tan(theta/2) = 3/r 4 = r tan(phi/2), tan(phi/2) = 4/r 12 = r tan((theta+phi)/2) Use tan sum formula 12 = r (3/r + 4/r)/(1 - 12/r^2) 12 = 7/(1-12/r^2) 12 - 144/r^2 = 7 144/r^2 = 5 r^2 = 144/5 r = 12/sqrt(5)

[@Tiqerboy]

That looks far more complicated than avoiding the law of cosines.

[@pwmiles56]

@@Tiqerboy Oh, I thought it brought out the symmetries in the problem. Thnx for comment

[@dickson3725]

No, Just use similiarity to avoid complexity

[@philosophical_aspect.]

@@pwmiles56 man i have an ezz solution Just imagine like its a half rectangle Formula for diagonal of a rectangle = √l²+b²find that then half it and you get your answer !!!!

[@pwmiles56]

@@philosophical_aspect. What are l and b?

[@spdigital7752]

I'm from India, 17 years old and I solved this question by my own made Formula within 5 seconds.

[@nirmalaverma8127]

Its taught in india in class 10/11 th. So it was easy for me. Its basically ∆/s-a

[@mevadavraj4178]

Stop boasting please it's very cringey, ofcourse question is easy but u don't have to tell this is taught in india std 10 or 11

[@dickson3725]

But obviously you didn't know how to derive it hahaha

[@pravegrajput]

@@mevadavraj4178 yes bro you are right , it is not taught in class 11 in India but in class 10 I am learning it right now ,

[@pravegrajput]

If you want just check class 10 cbse circle previous year questions , you will find it.

[@mevadavraj4178]

@@pravegrajput i am not saying it is not taught in std 10 i am saying stop boasting urself and also i am in std 10

[@SAM00780]

I got exact same question is my 10th prelims but only to derive until we find the length of tangent segments.. btw, u taught this much easier than my own teacher..

[@pat7594]

you may also know that the radius of the escribed circle is equal to : 2S/(8 + 9 - 7), where S is the area of the triangle (found with Heron's formula)

[@paulgets2737]

Welcome back, Presh.

[@gaurav7582]

My maths olympiad exam is on 6th march love from india

[@nutmegninja23]

Who knew thanos was Indian

[@qwertyuiop-pc2xw]

Thanoswaram iyer

[@notananimenerd1333]

Ioqm

[@Anonymous-kw7ls]

@@qwertyuiop-pc2xw Thanoswamy muttuswammy iyyer.

[@akhandanand_tripathi]

@@Anonymous-kw7ls thanoswamy mutthuswamy venugopal iyyer

[@Bugrick92]

In Darkmode now, nice! Interessting puzzle :)

[@SQRTime]

Hi Bugrick. If you are interested in math competitions, please consider kzbin.info/www/bejne/gaiadJaCmKiIm5Y and other videos in the Olympiad playlist. Hope it will be useful 👍

[@kashishgautam7040]

R= area of traingle/ semiperimeter - side touched by Circle

[@falldamage2008]

Hello, there is a simpler way to do this. Area of triangle ABC = r1(s-a) r1 is the radius we need to find, s is the semiperimeter and a is length of side BC. By herons formula we can find the area of triangle and then equate it to the RHS

[@TheMemesofDestruction]

Thank you!

[@helpmefortheloveofshrek6623]

I actually solved this! I’m so happy :) I ended up creating a square, made up of two right triangles. Using the fact that a square is made up of 4 right angles, I split up the square into two right triangles. I then did trig to solve for the bottom of the triangle, using sin(45)=x/7 and got around 5 ( I just rounded up a lot to make my life easier). Since it is a square, all of its sides must be the same, therefore the side that I made have the same length as the radius is also 5, making the radius 5. I suck at explaining, so I hope any of it make sense

[@devphalswal419]

Sorry to tell you but there is no way you can create a square in this fig and all sides of square are equal which is not possible, you might have ended up making a kite

[@helpmefortheloveofshrek6623]

@@devphalswal419 maybe. I’m not an expert in math, and I just did this for fun, so I don’t what can and can’t be done. I just did something that somehow worked, so make of that what you will🤷‍♀️

[@helpmefortheloveofshrek6623]

@@devphalswal419 Also, I would like to know your reasoning. Maybe I can learn something.

[@animezoneamv9116]

@@helpmefortheloveofshrek6623 But please tell me what square you formed

[@spencer1980]

The geometry portion was elegant. Good work.

[@vashon100]

2:44 From the 7-8-9 triangle (ie ABC), solving angle C and its complement are 58.41 deg and 121.59 deg. Then segment ND can be obtained from law of cosines (triangle NCD). Last step, triangle NOD with sides of r solved with law of cosines again.

[@MattColler]

Nice approach - without a calculator, cos(angle ACB) = 11/21, so cosine of the supplementary angle NCB is -11/21. This gives segment ND² = 192/7, and with angle NOD being supplementary again, you can get the correct radius of 12/√5.

[@rahulpaul5539]

I so much like your trigonometry method .......it's so easier than geometry

[@justnowi8967]

Loved the new dark theme ❤️

[@Vishalkr12]

Your way of explanation is great. Love from INDIA. 01-03-2022..1:58 PM

[@shashwatvats7786]

Nice question . Please give these types of question countinously. It refrains our mind. It took me more than five minutes to solve.

[@WoodyC-fv9hz]

~5.3665. I worked out 3 angles of the triangle given and side-shifted its side (labeled 7 units) so far to the left, until tangential with the circle, thus inscribing the circle within a trapezoid of 2 parallel sides. "Similar triangles" finds all inner angles of the trapezoid. Connecting the incentre with point C and D, finds another triangle (to the left of triangle ABC), sharing it's base with line BC. Two of its angles are bisectors of the complementary angles (with respect to 180°) of BCA and ABC. Looking at the new triangle, having its apex at the incentre, 2 of its angles are already known and also known is the length of the base (7). I constructed a line from the incentre (O) perpendicular to BC (point D), which resembles the altitude of the new triangle, being equivalent to radius r.

[@kimjennie7937]

Nice question👌🏼

[@HungNguyen-ey9wm]

Love these solutions ❤

[@neysantos2147]

The two ways are amazing. Thank you!

[@eroraf8637]

Sweet, merciful dark theme.

[@pilu1966]

thanks for boosting my love for maths!

[@alaaalkhafaji2884]

Excellent for you to this 👏question

[@JohnJones-pu4gi]

Just to add the missing bracket: r_n = TriangleArea / (s - side_n) is the radius of the excircle on side n this is easy to see by looking at triangles OAB, OAC, and OBC and r_0, the incircle radius is TriangleArea/s which is even easier to see using the same technique. Of interest then by multiplying these together Pi(r_i) [i=0..3] = TriangleArea^2

[@TheDigiWorld]

There is another way to go at the problem: We start with the Heron's formula to calculate the area, that (as explained in the video) is √720 As the question stated that one side of the triangle is tangent to the circle while the other two sides also extend as tangents to the circle, such a circle is called an Escribed circle and there is a formula for the radius of such circle r = Area of triangle/(s - a), where a is the length of the side of the triangle that touches the circle (in this case, BC). Simply putting in the value we get r = √720/(12-7) which simplifies to 5.366 or 5.37

[@mirfayozmirgiyosov9594]

Where did u get the s=12? U should've explained that, too.

[@user-ly5bc4xd2s]

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

[@kennethchristianguiwo9407]

You can use Area of ∆ = r(s-a) where s is the semi perimeter and a is the tangent line to the circle and is also a side of the triangle. This is the shortcut.

[@luismuller6505]

And you then use heron's formula to get s = (7 + 8 + 9) / 2 = 12 r = A / (s-a) = sqrt(s * (s-b) * (s-c) / (s-a)) = sqrt(12 * 3 * 4 / 5) = 12 / sqrt(5). Multiply both the numerator and denominator by sqrt(5) to get r = 12*sqrt(5)/5 which is the correct result.

[@dee5559]

Thank you, can't believe I went my whole life never having heard of that so-easy Heron formula oO

[@xz1891]

If u know so-called excircles 旁切圆, the 3 rads are, 2A/(a+b-c), 2A/(a-b+c), 2A/(-a+b+c), respectively

[@newzero1000]

escribed circles : r1 = A/( s-a) where A = rt { s(s-a)(s-b)(s-c)} and s = (a+b+c)/2

[@harinandnk123]

This is the creative logic

[@honsthebronze]

Yes exactly that's what I am saying 😂 We also had a problem like this in our geometry book in Iran He made suck a mess didn't he😂

[@honsthebronze]

@@newzero1000 exactly That's right

[@jeyakumarleveenth6528]

Yesterday we got a similar question in Sri Lankan mathematics competition.But there triangle ABC was a right angle triangle.I solved the question,as I knew the formula [A=(s-a)r] where A-Area of triangle,s-triangle perimeter/2,a-length of side opposite to

[@natashok4346]

Thank you! The black screen is best for my eyes.

[@242math]

great job, thanks for sharing

[@donasaloum9154]

Presh😍😍❤️‍🔥

[@AzaabcG]

amazing!

[@hinsontarrayo5822]

in short to get the radius of the circle just use the radius formula of the TRIANGLE ESCRIBED TO CIRCLE: *first find the "S" by using Semi-Perimeter of the Heron's formula, S=(a+b+c)/2 *let "a" the side of the triangle near to the circle or tangent to the circle so our a is 7, b is 8 and c is 9 *S=(7+8+9)/2 S=12 *now use the radius formula of the TRIANGLE ESCRIBED TO CIRCLE, r=A/(S-a) "A" is the Area of the triangle and also equal to the Heron's formula, √S(S-a)(S-b)(S-c) *r=(√S(S-a)(S-b)(S-c))/(S-a) r=(√12(12-7)(12-8)(12-9))/(12-7) r=5.36656

[@markstahl1464]

This is a really awesome question! Thanks for sharing! I was thinking though, that it would be cool knowing the level of math required to solve the puzzle going in, that way I wouldn’t have to beat my head against the wall, not realizing that I needed something that I vaguely remember seeing maybe one time, 25 years ago. Even still, I did get a fair way through the process before getting stuck enough that I realized I didn’t have the necessary knowledge to progress. It’s kind of like playing an old school adventure game where the game state has become unwinnable without you realizing it.

[@powersulca3033]

I solved the exercise in the second way, it had not occurred to me to use areas, the first way was very good, greetings from Peru

[@rakesh01778]

Nice bro keep making.

[@davspa6]

AM = AN, because the circle will touch the opposite lines at the same distance from A. BM = BD, and CN = CD, same reason. Also BD + CD = 7, so BM + CN = 7. So then AM + AN = 9 + 8 + 7 = 24. So AM = AN = 12. Use the law of cosines to find that cosine of angle A is 96/144 = 2/3. So then tan of half that angle = r/12. r = 12 tan(arccos (2/3)/2) ≈ 5.367

[@pratapjadhao388]

Excellent for me , Thanks sir.

[@SQRTime]

Hi Pratap. If you are interested in math competitions, please consider kzbin.info/www/bejne/gaiadJaCmKiIm5Y and other videos in the Olympiad playlist. Hope it will be useful 👍

[@michellethaler1832]

The trig solution is so elegant… but I’m a geometry fan- hard work, step by step .. old school . Beautiful nonetheless!

[@dhruvsingla6533]

This question comes in board examinations of class 10 and trigonometric method is very useful and time saving

[@lindseyd.2707]

trig really is overpowered

[@jimmykitty]

Thank you boss 🥺🥰 I love your videos a lot ❤🧡

[@YTN137]

I loved the first way using to solve this problem .

[@jacklogansmith7646]

This was fun. I had most of the building blocks for the Geometry solution, but despite knowing that AM=AN I did not connect the dots to realize that BM=BD and CN=CD as well.

[@falldamage2008]

Do you now a topic called Incentre-Excentre configuration? It covers many patterns in this diagram

[@masterbullets3105]

Just confusing problem but awesome explaination 🔥

[@Hanyamanusiabiasa]

That's it, my weakness is at geometry.

[@JLvatron]

I used the Law of Cos. Great vid-jo!

[@mutsisloll]

I got an A from my geometry exam, thanks to you

[@rome8726]

Nice

[@user-xt8ky2nu4u]

I respect you !!!! I LOVE YOU !!!!

[@nevyns9285]

i would not have thought to use the areas to figure that out, I prefer trig. very interesting way of doing it.

[@kolappan6517]

Why can't we do like this Join OM, Formed MONA is a quadrilateral Angle ONC and ANGLE OMB = 90° ( radius are perpendicular to the tangent at point of contact ) Angle O + angle A + Angle ONC + ANGLE OMB = 360° Hence, Angle O + angle A + 90° + 90° = 360° Angle O + angle A = 180 Now let us construct a perpendicular bisector OA, which divides the angle equally Hence, angle NOA = angle OAN = 45° Using trigonometry functions, Tan(theta) = opposite/adjacent Tan 45°= x/9 1 = x/9 × = 9cm That is radius = 9cm I am just a class 10 underling, so please correct me if I am wrong 😊

[@williamweatherall8333]

I'm back after trying this for a week. lol you got me good with this one. I'm interested in doing this without brute forcing it using trig.

[@anwardiggs8748]

Where was this during my sophomore year? 😭

[@zdrastvutye]

two lines that are tangents have an angle together, whose bisector "leads" towards the circle's center point, there you are. after this thoughts, the whole calculation leads to 2 linear equations with 2 unknown dimensions xm and ym. einfach die winkelhalbierenden der tangenten bilden und schneiden, dann erhält man die koordinaten des Mittelpunkts

[@Qermaq]

5:00 or so - a far easier way for me to uses Heron's Law is to work with the prime factors. In this case, S is 2^2*3, S-7 is 5, S-8 is 2^2 and s-9 is 3. So we need the square root of the product 2^4*3^2*5. All the prime factors to an even power will form the integer portion of the answer just by dividing all the exponents by 2. The one to an odd power, 5, is left under the radical. So 2^2*3*sqrt(5) or 12sqrt(5) is the answer. (BTW if you have a prime to the third power, just split it. So if you found 7^3, the &^2 part comes out as a 7, and one 7 remains under the radical.)

[@gaelik10]

Solving it through a coordinate plane is a little bit more tedious but it can be done. We need 6 functions and 6 variables, at least in the way I did it. These are the equations: y=(1÷1.626)(x-z)+h y=-1.626 (x-9) h=sqrt((z-x)^2+(h-y)^2) o=1.118a o=(-1/1.1188)(a-z)+h h=sqrt((a-z)^2+(h-o)^2) Where: ## h and z determine the coordinates (x,y) from the center of the circle ## -1.626 is the slope of line BC (simplified to 3 decimals), x and y the coordinates where it intersects with the circle ## 1.1188 is the slope of line AB and I renamed the x and y axis of this plane a and o to differenciate it from the other ones, since here x and y (thus a and o) will determinate the intersection point of the circle with line AB and will therefore have different values from the other x and y. ## the equations involving h and z state that the segment line between the center of the circle and the intersection with both AB and BC have to equal h. Why h? Because these segments are 2 of the radius and h is the y of the coordinates of the center of the circle, which is therefore the lenght of its 3rd radius. Therefore I am assigning to the equations that all three radius must be equal ## the answer to the problem is therefore the value of h, which is of course the same one as in the video

[@mathewpv681]

Very interesting problem. Let the sides of the triangle ABC be a, b, c as in standard notation. Let the circle be such that c is tangent and extended b, and extended a be also tangents. Then it can be shown that radius r=[ABC] /(S-c) where S is the semi perimeter of triangle ABC. That is r=sqrt[S×(S-a)×(S-b)÷(S-c)]

[@shrijagadish]

Very nice observation. This can be symmetrically used to find radii of circles touching one of the other sides a or b also. Further an interesting thing is that the point where the circle touches the extended side is always at a distance of S (half perimeter ) from the opposite vertex. This is true for all the three possible circles.

[@mathewpv681]

@@shrijagadish This also means that all triangles drawn by joining the two tangents from an external point to a circle such that it is also tangent to the circle will have the same perimeter, provided this tangent is on the same side of the circle as the external point.

[@siddhantkumar1935]

U can also use AM = 1/2(AB+AC+BC) for finding BD and DC

[@fangtooth-1125]

There is a really easy way to do this. Circle theorems suggest that triangle ONC and ODC are congruent right angled triangles. This is the same for triangle OMB and triangle ODB. Due to this fact, we can say that NC = CD and MB = DB. Let NC = x and MB = y Then we can say that x + y = 7 Triangle ONA and OMA are congruent due to circle theorems, therefore 9 + x = 8 + y. Rearrange to get y = x+1, and solve simultaneously with x + y = 7 to get x = 3. Therefore, NA = 9+3=12. (…) means that i rounded. Also use cosine rule on triangle ABC to get angle BAC = 48.2(…). Angle OAN = 0.5angleBAC due to the two triangles being congruent as mentioned before. Using SOHCAHTOA, sin(24.1(…))=r/12 Hence, r = 12sin(24.1(…)) = 5.367

[@kshitijjain9302]

Actually this is a excircle, and we have well known formulae for exradius, r=∆/(s-a)

[@eugene7518]

China had no known tradition of building lifelike sculptures before Qin's reign. According to Li Xiu Zhen a senior archeologist at the Terra Cotta army site this departure of scale and style likely occurred when when influences arrived in China specifically from ancient Greece, i.e. chariots, hairstyles. They believe the appearance may have been modeled/inspired by Greek sculptures. Hence Chinese and Greeks had contact then and they exchanged knowledge such as the Pythagorean theorem.

[@anujsangwan6718]

Dude that trigo solution ❤️

[@SQRTime]

Hi Anuj. If you are interested in math competitions, please consider kzbin.info/www/bejne/gaiadJaCmKiIm5Y and other videos in the Olympiad playlist. Hope it will be useful 👍

[@SuperYoonHo]

thanks

[@RenatoSilva-sy9tj]

You could use the easy relation for the radius of the circunferente related to the side 7 as r1, radius to side 8 as r2, radius to side 9 as r3: S=r1(p-7)=r2(p-8)=r3(p-9)=sqrt((p(p-7)(p-8)(p-9))

[@fsyi8395]

it doesn't necessary to find MB and NC, the three triangles ONC, OCB, and OBM can combine together and form a 7×r rectangle, while triangle OMA and ONA can form a 12×r rectangle

[@AmanKumar-vd1jc]

Take A as Origin(0,0),Line AC as x axis Then, Coordinates of C is (-9,0) Using Some Trigonometry Coordinates of B is (-16/3,8√5/3) Now Radius of Circle is The Y cordinate of excentre Y cordinate of Excentre which is [(-7×0+ 9×8√5/3 +8×0)/(-7+8+9)] =24√5/10=12√5/5

[@jomertomale]

Circumcircle R = abc/4A Incircle r = A/s Excircles r [a/b/c] = A/(s - [a/b/c])

[@prabhathchathuranga2392]

❤️🔥

[@sudoheckbegula]

Direct formula=area/(s-a)

[@SQRTime]

Hi Nipun. If you are interested in math competitions, please consider kzbin.info/www/bejne/gaiadJaCmKiIm5Y and other videos in the Olympiad playlist. Hope it will be useful 👍

[@hans-rudigerdrzimmermann]

Nice g ood problem. I solved with the trigonometry concept.

[@davidseed2939]

If the triangle hss tangent legs a,b and cross tangent c. Then the whole area is sr where s = (a+b+c)/2 The area of the triangles in the circle is cr. The area of the exterior triangle is K=sqrt(s(s-a)(s-b)(s-c)) r=K/(s-c) r²=s(s-a)(s-b)/(s-c) Here s=12 a,b,c = 9,8,7 r²=12.3.4/5 r=12¥5/5 As an exercise choose a,b,c so that r is an integer

[@Kakarot_Karthi]

This makes you simple..... Ex radius r = (tri area)/(semi perimeter - a) r = (12√5) / (12-7) = 12√5/5 = 12/√5 Find traingle area using Heron's formula a = represents the side touches the circle s = (7+8+9)/2 = 12

[@math2693]

This is just omg

[@karryy01]

This is so... easy...

[@My_IRyS_is_You]

Great channel always come with black screen

[@ravitejakakarala7858]

Excircle opposite to A is r1 = delta /(s-a)

[@rterentius]

I constructed it on AutoCAD and got the same answer. Tanzania

[@RohitSingh-jx7ml]

Area of triangle /s - 7 =radius of circle

[@WahranRai]

Let S area of triangle ABC and half perimeter p = 0.5(a+b+c) with a = 7, b=9 and c= 8 ---> we have the classical formula : Heron formula : *S = sqrt(p(p-a)(p-b)(p-c))* the external circle is the circle exscribed of the triangle ABC ---> *radius = S/(p-a)* I let substitue and find the solution

[@shadrana1]

At 0:29, MB=BD=x say................................(1) tangent rule. DC=(7-x)=NC.................................(2) tangent rule. MA=x+8=NA=(7-x)+9...................(3) tangent rule. x+8=(7-x)+9 >>>>>>x=4..............(4) We need angle BAC, Take Cosine Rule about triangle BAC, 7^2=8^2+9^2-2*8*9*cosBAC 49 =64+81-144cosBAC cosBAC =2/3 angle BAC=arcos(2/3)..................(5) We need length MN through circle, Take Cosine Rule about triangle MAN, MA=4+8=12 and AN=3+9=12 MN^2=12^2+12^2-2*12*12*(2/3) =288/3=96....................................(6) Take Cosine Rule about OMN, Cosine angle NOM= -cos BAC, MN^2=r^2+r^2-2*r^2(-cosBAC)=2r^2(1+2/3) 96=2*r^2*(5/3) 48=(5/3)*r^2 r^2=(144/5) r=12/(sqrt5) units in length. we need the +ve root. That is our answer.

[@jamessanchez3032]

What I did was extend the lines that start at point A a bit and then connect the two lines, such that you've formed a larger triangle and the circle fits in a trapezoid, with BC on "top," and the new third side of the triangle as the parallel base. I'll call the new vertices of the triangle P and Q. You have enough information to solve for PQ, PB, and QC, because 1) the big triangle is proportional to the small triangle, and 2) you have two new pairs of external tangents. PQ is 16.8, PB is 11.2, and QC is 12.6. Then, you can do some algebra recognizing that the trapezoid consists of a rectangle that is 7 x 2r, and two triangles also with height of 2r. (2r)^2 + x^2 = 11.2^2; (2r)^2 + ((16.8-7) - x)^2 = 12.6^2.

[@ARN48411]

..

[@fried_ady6318]

Hey, you mentioned the tangent half angle formula, Heron's formula, and Al Kashi's law of cosines. I was wondering if you could make a book about these lesser known formulas (I'm an accelerating yr 8). Thanks

[@Gaurav_629]

Agreed

[@abirhossainshanto4900]

Best place to find quality math puzzle.

[@indianelectron5786]

If u consider radius is same the height of that triangle, then 4.5**2 + x**2 = 7**2, which solves to sqrt(115/4) = 5.36.

[@markstahl1464]

In what course do you learn Heron’s formula? I wasn’t familiar with it.

[@jumpman8282]

cos(2𝜃) = 2 ∕ 3 Using the double angle formula, we get 2 cos²(𝜃) − 1 = 2 ∕ 3 ⇒ cos²(𝜃) = 5 ∕ 6 From the trigonometric 1, we then get sin²(𝜃) = 1 − cos²(𝜃) = 1 ∕ 6 Thus, tan²(𝜃) = sin²(𝜃) ∕ cos²(𝜃) = 1 ∕ 5 0° < 𝜃 < 90° ⇒ tan(𝜃) = 1 ∕ √5 From the diagram we have tan(𝜃) = 𝑟 ∕ 12, which gives us 𝑟 = 12 tan(𝜃) = 12 ∕ √5 = 12√5 ∕ 5

[@anandk9220]

THIS IS ABSOLUTE CAKEWALK IN COMPARISON TO THE HARDEST EASY PROBLEM I SOLVED. Just TOO EASY !!! Why avoid cosine rule if we know it ???? I solved this by first finding 3 and 4 as parts of tangent segment of length 7 ( By assuming one part as x and other to be 7 - x and then using equal tangent property to solve 9 + x = 8 + 7 - x ) Then used cosine rule to find angle between sides 7 and 8, and hence found half of its exterior angle. And then simply applied trigonometric tangent ratio definition to get Radius = 5.36656312 units

[@smchoi9948]

For the trigo. way, the host applied cosine law to ∠BAC to find r. Applying the law to either of the other 2 ∠s of △ABC instead could also work: Let ∠NOC = x. tan x = 3/r sec² x = 1 + (3/r)² = (r² + 9)/r² cos² x = r²/(r² + 9) As ∠BCA = 2x, by cosine law, cos 2x = (7²+9²-8²)/[(2)(7)(9)] = 11/21. As cos 2x = 2 cos² x - 1, cos² x = 16/21. What to solve is r²/(r² + 9) = 16/21. Alternatively, let ∠MOB = y. tan y = 4/r sec² x = 1 + (4/r)² = (r² + 16)/r² cos² x = r²/(r² + 16) As ∠CBA = 2y, by cosine law, cos 2x = (7²+8²-9²)/[(2)(7)(8)] = 2/7. As cos 2x = 2 cos² x - 1, cos² x = 9/14. What to solve is r²/(r² + 16) = 9/14.

[@turel528]

Couldn't solve it in my mind. But it is a fascinating question!

[@d4django]

Thank you for switching to dark mode

[@jeffeloso]

Would be nice to create a general formula using the trig method.

[@efi3825]

If you follow along the video and replace the given sidelengths with general sidelengths a,b and c, you will end up with r = A/(s-c) (but this time, A is the area of the triangle.)

[@Tusharsharma-dr8qh]

This ques can also done by circle chapter theorem which is in class 10. Perimeter of triangle is equal to the sum of length of two tangent and then by trigometry various method are there this way is very complex in this video

[@amodbarnwal7189]

I am from India studied in class 9 and I solve this question in 5 minutes. Using Pytagorous theorem Some circle theorem Heron formula Area of triangle 😊