Answer:
Part 1) Option 2. 180°
Part 2) Option 1. 55°
Part 3) Option 4. 30°
Part 4) Option 3. 45°
Explanation:
Part 1) Determine
m ∠ABC + m∠BAC + m∠ACB
Remember that
The sum of the interior angles in any triangle must be equal to 180 degrees
so
In this problem
The sum of the three interior angles of triangle ABC must be equal to 180 degrees
so
m ∠ABC + m∠BAC + m∠ACB=180°
Part 2) Determine m∠ABC
when m∠BAC = 70° and ΔABC is an isosceles triangle with AB=AC
In this problem the vertex angle is m∠BAC = 70°
m∠ABC=m∠ACB ---> because AB=AC
Remember that
The sum of the interior angles in any triangle must be equal to 180 degrees
so
2m∠ABC+m∠BAC=180°
substitute the given values
2m∠ABC+70°=180°
2m∠ABC=180°-70°
2m∠ABC=110°
m∠ABC=55°
Part 3) Determine m∠QPR
when m∠QRP = 30° and ΔPQR is an isosceles triangle with PQ=QR
In this problem
m∠QPR=m∠QRP ---> because PQ=QR
we have
m∠QRP = 30° ---> given problem
therefore
m∠QPR = 30°
Part 4) Determine m∠BDE
when m∠BAC = 45° and points D and E are the midpoints of AB and BC in ΔABC
we know that
Segment DE is parallel to segment AC (because the points D and E are the midpoints of segment AB and BC
so
m∠BDE=m∠BAC ----> by corresponding angles
we have
m∠BAC = 45° ---> given problem
therefore
m∠BDE = 45°