A Nice Geometry Problem | Find the length X 

Why do the internal angles of ABD add up to 120 deg?

If you want to go non-trig route 🙂 Drop a perpendicular from D to AB. Let it meet AB at E. △ADE is a 45°-45°-90° triangle. Thus DE=8/√2 (because in such a triangle, sides maintain the ratio 1:1:√2). △DEB is a 30°-60°-90° triangle. Its sides maintain the ratio 1:√3:2. DE is opposite angle 60° and EB is opposite its 30° angle. Hence EB=DE/√3=8/(√2√3). Now drop a perpendicular from D to BC. Let it meet BC at F. △DFC is a 45°-45°-90° triangle. Its sides are in the ratio 1:1:√2. DF is equal to EB by construction. Hence DF=8/(√2√3). Hence DC=(DF)(√2)=[8/(√2√3)].√2=8/√3 units.

Находим угльі треугольника ABD, A=45°, B=90° => C=45°из точки D строим перпендикуляр к сторонам AB и BC. Получаем 2 равнобедренньіх треугольника с основаниями 8 и X. Отношение сторон tg30°=sqr3

... [ THÊTA = T = 15 deg. ] ... Triangle ABD ... SIN(60) / 8 = SIN(45) / BD ... BD = 8 * SIN(45) / SIN(60) = 8 * SQRT(2) / SQRT(3) ... Triangle BCD .... SIN(45) / BD = SIN(30) / X ... X = BD * SIN(30) / SIN(45) = [ 8 * SQRT(2) / SQRT(3) ] * [ 1 / 2 ] / [ SQRT(2) / 2 ] = 8 / SQRT(3) = 8/3 * SQRT(3) ...

Angle ABC wasn't a "given" as 90 degrees

X=8/tan(6o)

Calculate angles 3theta+5theta+4theta = 180 12theta = 180 theta =15 AB from sine rule in triangle ABD 8/sin(60)= AB/sin(75) AB = 8*sin(75)/sin(60) AB = 2*(sqrt(6)+sqrt(2))*2*sqrt(3)/3 AB = 4/3(sqrt(6)+3sqrt(2)) tan(45) = BC/AB 1 = BC/AB BC = AB In fact ABC is right, isosceles triangle 4/3(sqrt(6)+3sqrt(2))*sqrt(2)=8+x 4/3(2sqrt(3)+6) = 8+x 8/3sqrt(3) + 8 = 8+x x = 8/3sqrt(3)

I have to apologize: I've watched about 100 of this guy's videos and he doesn't make mistakes. So, I don't need to be so critical for him leaving out a "given" when his goal (I think) is to teach math... Now that I'm an old and cranky 73 year old math teacher... I really need to LIGHTEN UP!

4.619

Triangles corners add up to 180 degrees, not 120

Yes, there is something weird in the problem

3θ+4θ+5θ=180,θ=15...teorema dei seni..8/sin4θ=h/sin3θ..x/sin(90-4θ)=h/sin(90+3θ)..divido le equazioni,risulta x=8ctg4θ/ctg3θ=8ctg60=8/√3

In ∆ABD: 40+30+50=120°

asnwer=4.5 cm