We are given a cyclic quadrilateral ABCD in a circle. 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

We are given a cyclic quadrilateral ABCD in a circle. 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) Finding $m\angle ACB$:

Since $\angle ADB$ and $\angle ACB$ subtend the same arc $AB$, they must be equal.

$$m\angle ACB = m\angle ADB$$

$$m\angle ACB = 61^\circ$$

b) Finding $m\angle ABC$:

In triangle ABC, we know $m\angle BAC = 49^\circ$ and $m\angle ACB = 61^\circ$. The sum of angles in a triangle is $180^\circ$. Therefore:

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

$$m\angle ABC = 180^\circ - 49^\circ - 61^\circ$$

$$m\angle ABC = 180^\circ - 110^\circ$$

$$m\angle ABC = 70^\circ$$

c) Finding $m\angle CAD$:

Since $\angle BCD = 90^\circ$, we have a right angle at $C$. The angles $\angle BCA$ and $\angle ACD$ sum to $90^\circ$.

$$m\angle ACD = 90^\circ - m\angle ACB$$

$$m\angle ACD = 90^\circ - 61^\circ$$

$$m\angle ACD = 29^\circ$$

Also, $\angle CAD$ and $\angle CBD$ subtend the same arc CD, so

$m\angle CAD = m\angle CBD$.

Since ABCD is a cyclic quadrilateral, opposite angles sum to $180^\circ$.

$$m\angle BAD + m\angle BCD = 180^\circ$$

$$m\angle BAD + 90^\circ = 180^\circ$$

$$m\angle BAD = 90^\circ$$

Also,

$m\angle BAD = m\angle BAC + m\angle CAD$.

Thus, $m\angle CAD = m\angle BAD - m\angle BAC$.

$$m\angle CAD = 90^\circ - 49^\circ$$

$$m\angle CAD = 41^\circ$$

d) Finding $m\angle BEC$:

In triangle ABE, we have $m\angle BAC = 49^\circ$ and $m\angle ABC = 70^\circ$. Then

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

$$m\angle AEB = 180^\circ - (49^\circ + 70^\circ)$$

$$m\angle AEB = 180^\circ - 119^\circ$$

$$m\angle AEB = 61^\circ$$

Since $\angle AEB$ and $\angle BEC$ are supplementary angles, we have

$$m\angle BEC = 180^\circ - m\angle AEB$$

$$m\angle BEC = 180^\circ - 61^\circ$$

$$m\angle BEC = 119^\circ$$

3. Final Answer

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

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

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

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