Find the angle θ | A Nice Geometry Problem | 2 Different Methods

Find the angle θ | A Nice Geometry Problem | 2 Different Methods
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[@user-sw9lb2zs6e]

Third method: divide AC with perpendicular to B into two segments c and d.Euklid: c*d=64 and c +d=16. Hence c=d=8 , perpendicular also, Theta =45 degrees

[@margaretcorfield9891]

This is what I did. In my head, took less than a minute.

[@mauroFsc1972]

Ese método es el mejor y más corto...Teorema de Euclides..!

[@sumanbasak3507]

Required angle is 45°....I solved it by the 1st method😊

[@imetroangola4943]

*Solução Elegante:*
Trace um segmento DB ( no sentido da figura para o lado esquerdo) tal que DB=BC e depois ligue o ponto D ao ponto A.
Note que os triângulos ABC e ADB são congruentes pelo caso LAL, pois:
AD=BC (Por construção);
AB é comum nos triângulos;
Os ângulos ABD=ABC=90°.
Logo, AD=16 e DAB= BAC=£.
[ADC]= (AD×AC×sin 2£)/2=
2×[ABC]. Assim,
[ADC]= (16×16×sin 2£)/2=128
256×sin 2£= 256 =>
sin 2£=1=> sin 2£=90° => *£=45°.*

[@Mediterranean81]

Let h be the perpendicular to AC from B
64=1/2*16*h
h= 8
Construct D such as
AB=CD
AD=BC
To complete a rectangle
One of the properties of rectangles is that the diagonals bisect each other at the centroid (let's call it O)
So BO=AO=CO=DO=16/2=8
tangent theta = BO/AO=8/8=1
So theta is 45°
Li

[@gerardmasson7654]

calculer et surface(l'aire)et l'on obtient HB= BA/2 d'où le tiangle ABC est isocèle rectangle ....cqfd.

[@Mediterranean81]

@@gerardmasson7654 ouais

[@denisrenaldo3506]

If you draw the height H from B, you can compute S = 1/2*H*AC = 64, so H = 8 cm = AC/2. Then, this height is also a mediator, being equal to the radius of the circumscribed circle (the hypotenuse being its diameter). So, the right triangle is also isosceles and θ = 45°. Pure geometric demonstration

[@devondevon4366]

45
since a = b . Hence, an isosceles triangle
A = 1/2 a*b
64 = 1/2 a*b
128 = a*b
Hence, a = 128 /b Equation A
a^2 + b^2 = 16^2 (Pythagorean)
a^2 + b^2 = 256 Equation B
(128/b)^2 + b^2 = 256 [Substitute Equation A into Equation B]
128^2/b^2 + b^2 = 256
128^2 + b^4 = 256b^2 [ Multiply both sides by b^2]
Let n= b^2 [ Introduce a new variable n)
Hence, n^2 = b^4
Hence, 128^2 + n^2 = 256 n
n^2 -256n + 128^2 = 0 [Quadratic form]
(n-128)(n-128) = 0 [ Factor]
n = 128 and n=128
Hence, b^2 = 128 since b^2 = n
b=sqrt 128, and a = sqrt 128
Since a = b, the triangle is an isosceles right triangle. Hene, theta = 45 degrees

[@harikatragadda]

Draw a circle around ∆ABC and draw a perpendicular BE on AC.
BE= 2*Area/16=8
AC is the diameter, Radius = 8, which is also equal to BE. Hence ∆ABE is an Isosceles Right triangle and θ=45°

[@guyhoghton399]

Once you have shown that _|AB| = |BC|_ it follows immediately that _θ = 45°_ : angles in a right angled isosceles triangle.

[@juanalfaro7522]

I did it by a variant of method 2: Let a=AB - 16*cos (Zeta) and b=BC = 16*sin (Zeta). Then [ABC] = ab/2 = 16^2 * cos (Zeta) * sin (Zeta) / 2 = 16^2 / 2^2 * [2*cos (Zeta) * sin (Zeta)] = 8^2 * sin(2*Zeta) = 64 * sin(2*Zeta) = 64 --> sin (2*Zeta) = 1 --> 2*Zeta = 90 --> Zeta=45

[@imetroangola4943]

*Solução 2: Simples*
Cos θ= AB/16 e sin θ=BC/16, logo,
AB×BC= 16×16 ×cos θsin θ
AB×BC= 16×8 sin 2θ
AB×BC/2= 8×8 sin 2θ
[ABC]= 64 sin 2θ
64 sin 2θ = 64 => sin 2θ= 1
sin 2θ= sin 90° =>
2θ=90° => *θ=45°.*

[@sergeyvinns931]

Angle Theta = 45 degrees. AB=BC=8\/2.

[@marioalb9726]

A = 64 cm² = ½ b.h
h = 2 A / b = 2 . 64 / 16
h = 8 cm
m + n = 16
m. n = h² = 8² = 64
m + 64/m = 16
m² -16m + 64 = 0
m = 8 cm. ; n = 8 cm
It is an isosceles right triangle, m=n
θ = 45° ( Solved √ )

[@denisrenaldo3506]

Starting the same way, but when you obtain h=8cm, you can easily demonstrate that the height is also a mediator, being equal to the radius of the circumscribed circle (the hypotenuse being its diameter). So, the right triangle is also isosceles and θ = 45° (as you finish rightly)

[@marioalb9726]

​@@denisrenaldo3506
Thanks

[@Antony_V]

In a generic right triangle once Area and hypotenuse "i" are known we can extrapolate the general formula to find the lenght of the legs a, b (in a way similar to the 1st method):
a, b = (1/2)*[√(i^2+4A)+-√(i^2-4A)]. In this case a=b=8√2 thus ABC is an isosceles right triangle and θ=45°

[@AmirgabYT2185]

45°

[@marcelowanderleycorreia8876]

Nice question!

[@michaeldoerr5810]

I think that this problem shows is that there are inter related implications involved in solving theta. And while the first method is more intuitive, the second method shows another result that can be checked but definitely is a direct result of theta being 45 degrees, if not a little counter-intuitive.

[@devondevon4366]

45

[@haiduy7627]

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