In Figure 6.35, we are given $m\angle BCD = 90^\circ$, $m\angle BAC = 49^\circ$, 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 Figure 6.35, we are given $m\angle BCD = 90^\circ$, $m\angle BAC = 49^\circ$, 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 are 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 $180^\circ$.

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

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

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

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

$m\angle ABC = 70^\circ$

c) Finding $m\angle CAD$:

In quadrilateral $ABCD$, we have $m\angle BCD = 90^\circ$, $m\angle ABC = 70^\circ$, $m\angle ADB = 61^\circ$. Since $\angle BAC=49$, $\angle BAD = \angle BAC+\angle CAD = 49+\angle CAD$.

The sum of the angles in a quadrilateral is $360^\circ$.

$m\angle ABC + m\angle BCD + m\angle CDA + m\angle DAB = 360^\circ$

$70^\circ + 90^\circ + 61^\circ + m\angle CAD + 49^\circ = 360^\circ$

$270^\circ + m\angle CAD=360^\circ-10^\circ = 260^\circ$

$70^\circ + 90^\circ + 61^\circ + m\angle BAD=360^\circ$

$221^\circ + m\angle BAD=360^\circ$

$m\angle BAD = 360^\circ -221^\circ$

$m\angle BAD = 139^\circ$

$m\angle CAD=m\angle BAD -m\angle BAC$

$m\angle CAD = 139^\circ - 49^\circ = 90^\circ$.

Since $\angle BCD=90^\circ$, $BD$ is a diameter. Therefore, $\angle BAD=90^\circ$

In $\triangle ABD$, $m\angle ABD=180^\circ - m\angle BAD -m\angle ADB = 180^\circ - 90^\circ -61^\circ=29^\circ$

Also $m\angle BDA = 61^\circ$, since arc BA has a degree measure of $2\times61^\circ = 122^\circ$, the central angle made by AB has degree measure $122^\circ$, so since the circle has degree measure $360$, arc BAD has degree measure $360-122=238^\circ$. This makes $\angle BCD=1/2\times238 =119^\circ$, contradiction to having $\angle BCD=90^\circ$

$\angle CAD = \angle BAD - \angle BAC$. $\angle BAD=\angle BAC+\angle CAD$, so $\angle BAD = 90$. So $m\angle CAD = 90^\circ - 49^\circ=41^\circ$

d) Finding $m\angle BEC$:

$m\angle BEC$ is an exterior angle of $\triangle ABE$.

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

$m\angle BEC = m\angle CAB + m\angle CBA$

$m\angle BEC = 49^\circ + 70^\circ$

$m\angle BEC = 119^\circ$

3. Final Answer

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

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

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

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