Name the side AD y Use the Pythagorean theorem on ABD to find AB Y^2-X^2=3 Use the algrebraic identity (x+y)(x-y)=X^2-Y^2 (x-y)(x+y)=3 We know x+y>x-y and 3 is a prime number So x+y=3 and x-y=1 Subtract the two equation to get 2x=2 => x=1
My solution uses more geometry, less calculation and no trigonometry. Let O be the circum-center of triangle ACD. Draw OA, OC, OD, then OA = OC = OD. Since angle at center is 2 times angle at circumference, angle COD = 2*angle CAD. Angle COD = 2*30 = 60 degrees. OC = OD and angle COD = 60 degrees --> triangle OCD is an equilateral. --> OC = OD = CD = 2 units. So OA = OC = OD = CD = 2. Let E be the midpoint of CD. Draw OE. CE = 2/2 = 1 and OE will be perpendicular to CD. By Pythagoras; OC = 2, CE = 1 then OE = sqrt (3). OE and AB are perpendicular to BC so OE // AB. OE = AB = sqrt (3) --> OABE is a rectangle. So BE = OA = 2. BE = x + CE --> 2 = x + 1 --> x = 1
If you cross-multiplied @ 13:18, then all the √3's will go away, and after bring up the 3 from the denominator of the denominator on the LHS, you get 3(x+1) = (3-x)(x+2) , and so on...
Posto BDA=α .teorema dei seni(scusate se uso sempre questo),con t=AD..t/sin90=√3/sinα..t/sin(α-30)=2/sin30..divido le due equazioni..sin(α-30)=(√3/4)/sinα..sinαsin(α-30)=√3/4...Werner..(1-√3)/2=cos(2α-30)...α=70,73535.. ovviamente risulta √3/x=tgα..x=√3/tgα=0,605356..mah?!?..ovviamente i calcoli sono sbagliati..risulta 2α-30=90..α=60(infatti sin60sin30=√3/4)...perciò x=√3/tgα=1..fare i calcoli velocemente,senza controllarli,ha delle controindicazioni...ah ah,scusate
With these data, are infinites triangles. Slide the "2 units" to the right or left, and mantain angle of 30 degrees, and "x" made shorter ir longer... ¿Ok?
Name the side AD y
Use the Pythagorean theorem on ABD to find AB
Y^2-X^2=3
Use the algrebraic identity (x+y)(x-y)=X^2-Y^2
(x-y)(x+y)=3
We know x+y>x-y and 3 is a prime number
So x+y=3 and x-y=1
Subtract the two equation to get 2x=2 => x=1
6:43
x,y are not to be sure belong to N.
x+y=3
x-y=1
is wrong
AD^2=AB^2+x^2
AC^2=AB^2+(x+2)^2
area ACD=1/2AB x CD=1/2 AD x AC x sin30
we can get equation
x^4+4x^3+10x^2+12x-27=0
(x-1)(x+3)(x^2+2x+9)=0
x=1
Nicely done! I got stuck with a 4th degree equation.
φ = 30°; ∆ ABC → AB = √3; BC = a + 2; AC = AT + CT → sin(DTA) = 1 →
AT = AB → DAB = TAD = φ → BD = DT = a = 1 → TDC = 2φ → DCT = φ → cos(φ) = √3/2 = CT/2 → CT = √3
My solution uses more geometry, less calculation and no trigonometry.
Let O be the circum-center of triangle ACD. Draw OA, OC, OD, then OA = OC = OD.
Since angle at center is 2 times angle at circumference, angle COD = 2*angle CAD.
Angle COD = 2*30 = 60 degrees. OC = OD and angle COD = 60 degrees --> triangle OCD is an equilateral.
--> OC = OD = CD = 2 units. So OA = OC = OD = CD = 2.
Let E be the midpoint of CD. Draw OE. CE = 2/2 = 1 and OE will be perpendicular to CD. By Pythagoras; OC = 2, CE = 1 then OE = sqrt (3).
OE and AB are perpendicular to BC so OE // AB. OE = AB = sqrt (3) --> OABE is a rectangle. So BE = OA = 2.
BE = x + CE --> 2 = x + 1 --> x = 1
If you cross-multiplied @ 13:18, then all the √3's will go away, and after bring up the 3 from the denominator of the denominator on the LHS, you get 3(x+1) = (3-x)(x+2) , and so on...
Method 1 very interesting. Thank you.
Posto BDA=α .teorema dei seni(scusate se uso sempre questo),con t=AD..t/sin90=√3/sinα..t/sin(α-30)=2/sin30..divido le due equazioni..sin(α-30)=(√3/4)/sinα..sinαsin(α-30)=√3/4...Werner..(1-√3)/2=cos(2α-30)...α=70,73535.. ovviamente risulta √3/x=tgα..x=√3/tgα=0,605356..mah?!?..ovviamente i calcoli sono sbagliati..risulta 2α-30=90..α=60(infatti sin60sin30=√3/4)...perciò x=√3/tgα=1..fare i calcoli velocemente,senza controllarli,ha delle controindicazioni...ah ah,scusate
With these data, are infinites triangles.
Slide the "2 units" to the right or left, and mantain angle of 30 degrees, and "x" made shorter ir longer... ¿Ok?
X=1
This can be even more easily with my method.
In the original question it was not stated that B = 90° you just wasted my fucking time