In the diagram, $\mathrm{AB}$ is a circle chord and $\mathrm{AOC}$ is its diameter, so $50°$ If the tangent to the circle at point A is $\mathrm{AT}$then $\angle \mathrm{BAT}$is equal to

Remember to join point O to C and use Angle CAT is equal to 90 by property of diameter in this question; following these instructions will help you get closer to the solution.

The line$\mathrm{AC}$ passes through $\mathrm{O}.$ So it is a diameter of the circle.

$\therefore \angle \mathrm{ABC}$ is an angle inscribed in a semicircle.

$\therefore \angle \mathrm{ABC}=90°\text{. }$

Now $\angle ABC+\angle ACB+\angle BAC=180°$... (sum of the  angles of a $\left.\Delta =180°\right)$

$\Rightarrow {50}^{°}+{90}^{°}+\angle \text{BAC}={180}^{°}$

$\Rightarrow {50}^{°}+{90}^{°}+\angle \text{BAC}={180}^{°}$

$\Rightarrow \angle \text{BAC}={40}^{°}$

Again, AT is a tangent to the circle at $\mathrm{A}$.

$\therefore \angle \mathrm{CAT}=90°$

$\Rightarrow \angle \mathrm{BAT}+\angle \mathrm{CAB}=90°$

$\Rightarrow \angle \mathrm{BAT}+40°=90°$

$\Rightarrow \angle \mathrm{BAT}=50°$