Question

In Fig 3, ABCD is a cyclic quadrilateral in which AB is a diameter of the circle passing through A, B, C and D. If ∠ADC = 130°, then find ∠BAC.

Answer 1

Given: ABCD is a cyclic quadrilateral, AB is a diameter and ∠ADC = 130°

In cyclic quadrilateral ABCD,

$$\begin{gathered} \angle \mathrm{ABC}+\angle \mathrm{ADC}=180°\left(\mathrm{Opposite}\mathrm{angles}\mathrm{of}\mathrm{cyclic}\mathrm{quadrilateral}\mathrm{are}\mathrm{supplimentary}\right) \\ \Rightarrow \angle \mathrm{ABC}=180°-\angle \mathrm{ADC} \\ \Rightarrow \angle \mathrm{ABC}=180°-130° \\ \Rightarrow \angle \mathrm{ABC}=50° \end{gathered}$$

Also,

As AB is the diameter, it subtends right angle at the circumference (property of circles)

So, ∠ACB = 90°

Now, in $∆$ABC,

$$\begin{gathered} \angle \mathrm{ABC}+\angle \mathrm{BAC}+\angle \mathrm{ACB}=180° \\ \Rightarrow 50°+\angle \mathrm{BAC}+90°=180° \\ \Rightarrow \angle \mathrm{BAC}+140°=180° \\ \Rightarrow \angle \mathrm{BAC}=180°-140° \\ \therefore \angle \mathrm{BAC}=40° \end{gathered}$$