In the given circle with points A, B, C, and D on the circumference, we are given that $m(\angle BCD) = 90^\circ$, $m(\angle BAC) = 49^\circ$, and $m(\angle ADB) = 61^\circ$. We need to find: a) $m(\angle ACB)$ b) $m(\angle ABC)$ c) $m(\angle CAD)$ d) $m(\angle BEC)$

1. Problem Description

In the given circle with points A, B, C, and D on the circumference, we are given that $m(\angle BCD) = 90^\circ$, $m(\angle BAC) = 49^\circ$, and $m(\angle ADB) = 61^\circ$. We need to find:

a) $m(\angle ACB)$

b) $m(\angle ABC)$

c) $m(\angle CAD)$

d) $m(\angle BEC)$

2. Solution Steps

a) To find $m(\angle ACB)$, we know that $\angle ADB$ and $\angle ACB$ subtend the same arc $AB$. Therefore, they are equal.

$m(\angle ACB) = m(\angle ADB) = 61^\circ$

b) To find $m(\angle ABC)$, we consider triangle $ABC$. The sum of the angles in a triangle is $180^\circ$. We have $m(\angle BAC) = 49^\circ$ and $m(\angle ACB) = 61^\circ$. Therefore,

$m(\angle ABC) = 180^\circ - m(\angle BAC) - m(\angle ACB)$

$m(\angle ABC) = 180^\circ - 49^\circ - 61^\circ = 180^\circ - 110^\circ = 70^\circ$

c) To find $m(\angle CAD)$, we know that $\angle CBD$ and $\angle CAD$ subtend the same arc $CD$. First, we need to find $m(\angle CBD)$.

Since $\angle BCD = 90^\circ$, triangle $BCD$ is a right triangle. We can find $m(\angle CBD)$ using the fact that the angles in triangle $BCD$ add up to $180^\circ$.

$m(\angle CBD) = 180^\circ - m(\angle BCD) - m(\angle BDC)$

We know that $\angle BAC$ and $\angle BDC$ subtend the same arc $BC$. Thus, $m(\angle BDC) = m(\angle BAC) = 49^\circ$.

$m(\angle CBD) = 180^\circ - 90^\circ - 49^\circ = 180^\circ - 139^\circ = 41^\circ$

Therefore, $m(\angle CAD) = m(\angle CBD) = 41^\circ$

d) To find $m(\angle BEC)$, we know that $m(\angle BEC)$ is an exterior angle of triangle $ABE$. Therefore, it is equal to the sum of the two opposite interior angles.

$m(\angle BEC) = m(\angle EAB) + m(\angle EBA)$

$m(\angle EAB) = m(\angle BAC) = 49^\circ$

$m(\angle EBA) = m(\angle ABC) = 70^\circ$

$m(\angle BEC) = 49^\circ + 70^\circ = 119^\circ$

3. Final Answer

a) $m(\angle ACB) = 61^\circ$

b) $m(\angle ABC) = 70^\circ$

c) $m(\angle CAD) = 41^\circ$

d) $m(\angle BEC) = 119^\circ$