$\triangle ABC$ is inscribed in a circle, Find $\angle{BAC}$

Question

An acute angled $\triangle ABC$ is inscribed in a circle with its center at the origin. If $B$ is $(3,4)$ and $C$ is $(-4,3)$, then $\angle{BAC}$ equals?

My attempt:

All the points will lie on the circle, $x^2+y^2=25$.

No idea how to proceed. Any hints would be helpful.

Answer 1

Hint: Calculate the angle of the arc that starts at $B$ and ends at $C$. Use your trig functions to do this.

Having this number in hand, then what is the (simple) relation between an angle of arc and an angle inscribed in the circle that subtends that arc?

Answer 2

The radius $r$ of the circle is $5$ because the equation of a circle centered at the origin is $x^2+y^2=r^2$.

Also, $\angle BAC$ will always remain constant. From this we can use any point in the circle as $A$ to calculate the angle. I will choose $(0,-5)$.

Using trigonometry, $\angle BAC$ can be calculated: $$\angle BAC=\pi-\arctan\left(\frac{8}{4}\right)-\arctan\left(\frac{9}{3}\right)$$ $$\Rightarrow \angle BAC=\pi-\arctan(2)-\arctan(3)=\arctan(1)=\frac{\pi}{4}$$ $$\boxed{\angle BAC=\frac{\pi}{4}=45^{\circ}}$$

Answer 3

Thanks to @ReinhardMeier, I was able to solve this. I will just add the solution here for future references.

Calculating, BC we have,

$$BC=\sqrt{(4-3)^2+(3-(-4))^2}$$

$$BC=\sqrt{50}$$

Also we get that, $$R^2 + R^2=25+25=50=BC^2$$

Therefore, triangle $BOC$ is right angled.

Using inscribed angle theorem, $\angle{BAC}$ will be $45$ degrees.

Answer 4

As per inscribed angle theorem Ψ =1/2 θ here, Ψ = ∠BAC As the triangle ABC is inscribed in a circle with its centre at the origin, hence θ = 90∘ - (Using Pythagoras theorem) Hence ∠BAC = 45∘

Answer 5

To find ( tan of) angle of triangle at C

$$ \tan^{-1}\frac{3}{4}+ \tan^{-1}\frac{4}{3} = \tan^{-1}\frac{..}{1-1}= \infty \to C= 90^0$$

BCA is a right angled isosceles triangle of sides $(5, 5 \sqrt2, 5)$ . So the angles at A,B are each $45^0.$