Triangle in circle

Let $AB=b$ and $BC=a$ .

Since AC is the diameter, then $\triangle ABC$ is right-angled at $B$ .

By the Pythagorean Theorem, $AC=\sqrt{a^{2}+b^{2}}$

Area of the circle $=\dfrac{\pi (\sqrt{a^{2}+b^{2}})^{2}}{4}$

Area of $\triangle ABC=\dfrac{ab}{2}$

And $\dfrac{\pi (\sqrt{a^{2}+b^{2}})^{2}}{2ab}=2\pi$

Giving us $\dfrac{a^{2}+b^{2}}{2ab}=2$

By Sine Law,

$\dfrac{a}{\sin A}=\dfrac{b}{\sin C}=\sqrt{a^{2}+b^{2}}$

Substituting to the above equation yields

$\dfrac{a^{2}}{2(\sin^{2} A)(a)(\dfrac{\sin C}{\sin A})(a)}=2$

$\dfrac{1}{2\sin A \sin C}=2$

$\dfrac{1}{\cos (A-C)}=2$

$\cos (A-C)=\dfrac{1}{2}$

And finally,

$A-C=\boxed{60^\circ}$

As triangle ABC is a right triangle, angle ABC is a right angle.

Let angle ACB be theta.

AC = 2r.

AB = 2r sin theta

BC = 2r cos theta

Area of triangle = 4r^2( sin theta) ( cos theta) /2 = 2r^2 (sin theta) (cos theta)

Using double identity, sin (2 theta)= 2 (sin theta)( cos theta)

Area of triangle = r^2 (sin 2 theta)

Area of circle / area of triangle = pi r^2 / r^2 sin (2 theta)= pi/sin (2 theta) = 2pi

sin (2 theta) = 1/2

2 theta = 30 degrees.

theta = 15 degrees

The complementary angle BAC = 75 degrees

Difference in angles = 60 degrees

Let $AC=2r$ where $r$ is the radius. Then we have $\text{Area of Triangle}=\dfrac{\text{Area of Circle}}{2\pi}=\dfrac{\pi r^2}{2\pi}=\dfrac{r^2}{2}.$ Let $\angle ACB=\theta$ . Hence $AB=AC\sin\theta=2r\sin\theta,$ $BC=AC\cos\theta=2r\cos\theta.$ Clearly $\angle ABC=90^\circ$ . Hence $\text{Area of Triangle}=\dfrac{1}{2}\times AB\times BC=2r^2\sin\theta\cos\theta=\dfrac{r^2}{2}.$ $\therefore ~ 2\sin\theta\cos\theta=\sin 2\theta=\dfrac{1}{2}=\sin 30^\circ.$ Therefore $2\theta = 30^\circ\implies \theta =15^\circ$ . So the other angle is $90^\circ-15^\circ=75^\circ$ . Hence their difference is $75^\circ-15^\circ=\boxed{60^\circ}$ .

A = PI/2 - C ; Area(Circle)/Area(Triangle) = 2PI -> sin(2C) = 1/2; 2C = PI/6; C = PI/12 -> 15°; A = PI/2 - PI/12 = (5/12)PI -> 75°; A - C = 60°

Since one of the side of the triangle is the diameter of the circle, it is a right triangle. Let $AC=2r$ and $\angle ACB=\theta$ , then $AB=2r \sin \theta$ and $BC=2r \cos \theta$ .So the area of the triangle is $\dfrac{1}{2}(2r\sin \theta)(2r \cos \theta)=2r^2(\sin \theta)(\cos \theta)$ . Using the double angle identity, $\sin~2x=2\sin~x\cos~x$ , we have

$A_T=r^2 \sin 2\theta$

Given in the problem that $\dfrac{A_C}{A_T}=2 \pi$ , so

$\dfrac{\pi r^2}{r^2 \sin~2\theta}=2 \pi$

$\dfrac{1}{2}=\sin 2\theta$

$2\theta=\sin^{-1}\left(\dfrac{1}{2}\right)$

$2 \theta = 30$

$\theta =15$

It follows that $\angle CAB=90-15=75$ .

So the desired answer is $70-15=$ $\boxed{60^\circ}$

Let the circle centre be O to which join B. Now, 2∏=∏R²/(r h) where h is the altitude from B to AC. h = R/2=R sin(30). This implies that / BOA=2*/ BCA=30° or / BCA=15° & / BAC=75° giving us 60° as the required difference.

$a^2+c^2=b^2=4r^2..~~But~\dfrac{\pi*r^2}{(1/2*a*c)}=2*\pi~.. ~\therefore~~ a *c=r^2. \\ SinA=\dfrac{a} {2r}=CosC,~~~~~~~~~CosA=\dfrac{c}{2r}=SinC\\~Cos(A-C)=\dfrac{c}{2r}*\dfrac{a}{2r}+\dfrac{a}{2r}*\dfrac{c}{2r}=2*\dfrac{a*c}{4r^2}= ~2*\dfrac{r^2}{4r^2}\\= 1/2=Cos(60)\\ \boxed{60}$

This is same as NaveenKmuar Venkataraj, just used latex.

Let the angles be x and y and AB=b,BC=a. Given that (area of circle)/area of triangle=2 * pi or pi * r^2/{(1/2)ab }=2 * pi.......................(1), Now sin x=a/2r or a=2r * sin x.........(2) And cos x=b/2r or b=2r * cos x....(3) Now putting those into (1) pi * r^2/{(1/2) * (2r * cos x) * (2r * sin x)}=2 * pi or pi * r^2/(2r^2 * sin x * cos x)=2 * pi or pi/(2 * sin x *cos x)=2 * pi or 1/(sin x * cos x)=(2 * 2 * pi)/pi or (sin x * cos x)^2=(1/4)^2 or sin^2 x * cos^2 x=(1/4)^2 or sin^2 x(1-sin^2 x)=1/16 or sin^2 x-sin^4 x=1/16 or 16sin^2 x-16sin^4 x=1 or 0=16sin^4 x-16sin^2 x+1 either sin x=sqrt{(2+sqrt 3)/4} or sinx=sqrt {(2+sqrt 3)/4} so x=(75,15). Now 90(semi circle angle is always 90)+75+y=180 or 90+15+y=180 that makes y=(15,75) So either way its [x-y=75-15=60 or y-x=75-15=60] 60. SO ANSWER IS 60

Ratio of the areas $\frac{2\pi R^2}{AB\cdot BC}=2\pi$ gives $R^2= AB\cdot BC$, while by Pythagorean Theorem we get $AB^2 + BC^2 = 4R^2$. Substituting, $AB^2 + BC^2 = 4AB\cdot BC$. By letting $\tan C=\frac{AB}{BC}=x$, the previous equation can be written as $x^2 - 4x + 1=0$, which can be solved by the quadratic formula to obtain $x= 2 \pm \sqrt{3}\Rightarrow \tan C=2\pm\sqrt{3}$. Thus, $C=15^\circ$ (which gives $A=75^\circ$) or $C=75^\circ$ (which gives $A=15^\circ$). In either case, the answer is $60^\circ$.

Say a,b,c are angles at A,B,C and AB = w, BC = h and AC = 2r Easy to get r.r = w.h from what is given. cos(a-c) = cos(a)cos(c) + sin(a)sin(c) =(h/2r)(w/2r) + (w/2r)(h/2r) substitute rr = wh we get cos(a-c) = 1/2 , a-c = 60deg