Math Olympiad Geometry Problem | You should know this trick!

[@giuseppemalaguti435]

(2/sin10)/sin20=x/sin100...x=2sin100/sin10sin20=2cos10/(sin10*2sin10cos10)=1/(sin10)^2

[@dost77]

Use tan(80) in triangle BAD to get AB = 11.34. Use sin (20) in triangle ABC to get x = AB/sin(20) i.e 11.34/.34 = 33.35 units

[@pinjalabhagiradha4897]

Angle BDC = 180°-(60°+20°) =80= Angle BDA = 180°-80° = 100° Angle ABD = 180° - (100°+80°) = 10° In the Triangle ABD Tan 10° = 2/AB AB = 2 / Tan 10° In Triangle ABC Sin 20° = AB/BC = (2 / Tan 10°)/X X= 2 / (Tan 10° . Sin 20°) X= 2/ (( Sin 10°/Cos 10°).(2 Sin 10°.Cos 10°)) X=1/Sin 10° . Sin 10° = 1/(Sin 10°)^2

[@ludmilaivanova1603]

after @ 7:10 where we found AC= x-2, we can draw a circle around triangle ABC with the centre O, connect O with A, Angle AOB =40degrees. Extend BD to meet circumfernce in (.) F, connect F with C. Triangle AFC has a 30 degrees angle C. Therefore, BF(against 30 degrees angle) =1/2 BC. Let's name BO, OC, OF l- R(radius). Then, x=2R, BF=R. Angle FOB=60 degrees. Ratios of chrds BA/BF=40/60=2/3. BA=2/3R. (2/3R)^2 + (2R-2)^2 = (2R)^2 R= 9 -+ 6 times sqrt 2. fFinal answerr is x=2R

[@jimlocke9320]

Let AB have length y.

[@lasalleman6792]

Used the sine rule on triangle ABD. Angle ADB is 80 degrees. Angle ABD is 10 degrees. Under Sine rule, line AB works out to 11.3456. Which is also the sine of angle ACB. Line BC works out to hypotenuse of 33.1632. For me, anyway.

[@user-lk5kh7we2p]

The Theorem Exent Angle

[@52soccerstar]

I think it can be done with minimal trigonometry and no need for a trigonometric value: the long way

[@imetroangola4943]

Solução: No triângulo retângulo ∆ABD: sin 20°= AB/x → *x=AB/sin 20°* Por outro lado, ∠ADB=80°, logo no triângulo retângulo ∆ABD, temos: Tg 80°= AB/AD →AB=2tg 80°. Daí, x= 2tg 80°/ sin 20° tg 80°= ctg (90°- 80°)= ctg 10°, por sua vez ctg 10°= cos 10°/sin10° e sin20°= 2 sin 10°cos10° x=2cos 10°/2(sin10°)²cos 10° *x=1/(sin10°)².* lembrando que: cos 20°=1-2(sin 10°)², ou seja, (sin 10°)²=(1- cos 20°)/2 Podemos reescrever x, da seguinte forma: *x=2/(1 - cos 20°)*

[@jairoeveliogordillomarin5780]

Sen 10°= 2/BD. Entonces: BD= 11.52 u. (AB)²= BD²-AD²= (11.62)²-(2) ²= 11.35 u. Tan 20°= 11.35/AC. Entonces: AC= 31.18 u. X²= (11.35)²+(31.18)²= 1101.01 u². X= 33.18 u.

[@Emmashar]

2/Sin10/Sin20*Sin100≈33.16.

[@appybane8481]

x=2/2sin^2(10)=csc^2(10)

[@jarikosonen4079]

x~33.16

[@RealQinnMalloryu4]

{60°B+20°C+20°D}=100°BCD {20°B+20°D20°A}=60°BDA {100°BCD+60°BDA}=160°BCDBDA {160°BCDBDA ➖ 180}=20°BCDBDA 2^2^10 1^1^2^5 2^1(BCDBDA ➖ 2BCDBDA+1).

[@sumankundu93]

Bro post some problems on circles.

[@harikatragadda]

5:15 ∆EBD is Similar to ∆BEC. CD = X-4, and if BD = EB = a, then X/a= a/4 a²=4X Applying Cosine rule in ∆CBD, (X-4)²= X²+ a² -2aXCos60 The solution is cubic a³-12a²+64=0 With X=a²/4

[@user-lk5kh7we2p]

X=1/sin10

[@oguzhanbenli]

x ~ 33.163

[@clapptschon]

x = (2 * sin(100)) / (cos(80) * sin(20)) 😀

[@ald6980]

Too hard. Answer x=sqrt(4/tan^2(10)+4/(tan^2(10)tan^2(20))) can be obtained much more easily. 2/a = tan(10); a/(2+b)=tan(20); x^2=a^2+(2+b)^2. If somebody wants he can prove that this tan-x is equal to 2/(1-cos(20)). Hint: "both" x is a root of x^3 - 36 x^2 + 96 x - 64=0 near 33. But it is possible to express cos(20) and tan(20) using tan(10) and universal trigonometric substituton.

[@jelalatinovic5195]

Mi piace ragionamento