Question

In the figure $ \mathrm{AB}$ is a chord of the circle and $ \mathrm{AOC}$ is its diameter such that $ \angle \mathrm{ACB}=50°.$If $ \mathrm{AT}$ is the tangent to the circle at the point $ \mathrm{A}$, then $ \mathrm{BAT}$ is equal to

• A. $65°$
• B. $60°$
• C. $50°$
• D. $40°$

Answer 1

The correct option is C

$50°$

The explanation for the correct option:

Step 1: Find the angle $BAC$.

Given: $\angle \mathrm{ACB}=50°.$

Since the angle in the semicircle is right-angled, therefore $\angle \mathrm{CBA}=90°.$

$\mathrm{In}∆\mathrm{ABC}$, $\angle \mathrm{ACB}+\angle \mathrm{CBA}+\angle \mathrm{BAC}=180°$ (Angle sum property of triangle)

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

Step 2: Find the angle $BAT$.

Since the tangent at any point on the circle is perpendicular to the radius through the point of contact.

Therefore angle $CAT$ is $90°$

$$\begin{gathered} \Rightarrow \angle \mathrm{CAB}+\angle \mathrm{BAT}=\angle \mathrm{CAT} \\ \Rightarrow 40+\angle \mathrm{BAT}=90 \\ \Rightarrow \angle \mathrm{BAT}=90-40 \\ \Rightarrow \angle \mathrm{BAT}=50° \end{gathered}$$

Hence option $C$ is the correct answer.