We are given a circle with tangents $AB$ and $AC$ drawn from point $A$. Point $D$ is on the circle. We are given that $m\angle BDC = (6x+43)^\circ$ and $m\angle A = (3x+11)^\circ$. We are asked to find the value of $x$ and the measure of angle $A$.

1. Problem Description

We are given a circle with tangents $AB$ and $AC$ drawn from point $A$. Point $D$ is on the circle. We are given that $m\angle BDC = (6x+43)^\circ$ and $m\angle A = (3x+11)^\circ$. We are asked to find the value of $x$ and the measure of angle $A$.

2. Solution Steps

Since $AB$ and $AC$ are tangent to the circle, $\angle ABC$ and $\angle ACB$ subtend the same arc, thus $\angle ABC = \angle ACB$. Therefore, triangle $ABC$ is isosceles.

Since $AB$ is tangent to the circle at $B$, the inscribed angle $\angle BDC$ is equal to the angle between the tangent and the chord $BC$. So, $\angle BDC = \angle CBA$.

Therefore, $\angle CBA = (6x+43)^\circ$.

Also, $\angle BCA = (6x+43)^\circ$.

The sum of angles in triangle $ABC$ is 180 degrees. So,

$\angle CBA + \angle BCA + \angle BAC = 180^\circ$

$(6x+43)^\circ + (6x+43)^\circ + (3x+11)^\circ = 180^\circ$

$6x + 43 + 6x + 43 + 3x + 11 = 180$

$15x + 97 = 180$

$15x = 180 - 97$

$15x = 83$

$x = \frac{83}{15}$

However, we are given that $\angle A = (3x+11)^\circ$.

Consider the quadrilateral formed by $O$ as the center of the circle.

Since tangents are perpendicular to the radius, $\angle OBC = 90^\circ$ and $\angle OCB = 90^\circ$.

$\angle BOC = 2 \angle BDC = 2 (6x+43)^\circ = (12x + 86)^\circ$

In quadrilateral $OBAC$, $\angle BOC + \angle OBC + \angle OCB + \angle BAC = 360^\circ$

$12x+86 + 90 + 90 + 3x+11 = 360$

$15x + 277 = 360$

$15x = 360 - 277$

$15x = 83$

$x = \frac{83}{15}$

$m\angle A = 3x+11 = 3(\frac{83}{15}) + 11 = \frac{83}{5} + \frac{55}{5} = \frac{138}{5} = 27.6^\circ$

$2m\angle BDC + m\angle A = 180^\circ$ since the angle at the center is twice the angle at the circumference.

$2(6x+43) + 3x+11 = 180$

$12x + 86 + 3x + 11 = 180$

$15x + 97 = 180$

$15x = 83$

$x = \frac{83}{15}$

$m\angle A = 3(\frac{83}{15}) + 11 = \frac{83}{5} + 11 = \frac{83+55}{5} = \frac{138}{5} = 27.6$

3. Final Answer

$x = \frac{83}{15}$

$m\angle A = 27.6^\circ$