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

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

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

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

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°

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

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

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 √ )

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

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°

*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°.*

*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°.*

Using the 1st method, there is no need to compute tg theta as long as you get a=b. The triangle is then a right isoceles triangle and with to equal angles at 45 dg.

45°

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.

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

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

Nice question!

45

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