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
@pinjalabhagiradha489723 күн бұрын
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
@ludmilaivanova160320 күн бұрын
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
@jimlocke932022 күн бұрын
Let AB have length y.
@lasalleman679222 күн бұрын
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-lk5kh7we2p22 күн бұрын
The Theorem Exent Angle
@52soccerstar22 күн бұрын
I think it can be done with minimal trigonometry and no need for a trigonometric value: the long way
@imetroangola494322 күн бұрын
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°)*
@jairoeveliogordillomarin578021 күн бұрын
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.
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-lk5kh7we2p22 күн бұрын
X=1/sin10
@oguzhanbenli23 күн бұрын
x ~ 33.163
@clapptschon23 күн бұрын
x = (2 * sin(100)) / (cos(80) * sin(20)) 😀
@ald698023 күн бұрын
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.