MATH SOLVE

4 months ago

Q:
# GIVEN circle P with arc AE = 53 , arc BA =68 and arc CB =72match the following angles with their corresponding measurements

Accepted Solution

A:

the theory that is mainly used to find out the angles of the above diagram is that;

the sum of the internal angles of a triangle is 180°.

when the angles of arcs have been given(e.g: BA) that means from the centre of the circle to point B has been rotated by 53° to point A.

Next from point A to point E, the angle rotated is 68°

the addition of these 2 angles give the angle rotated from point B to point E,

that is angle no.7

m∠7 = 53° + 68° = 121°

consider ΔPBE,

PB and PE are radii therefore isosceles triangles

In isosceles triangles the angles opposite equal sides are equal

if the angle is x°, then there are 2 angles with values x each

x° +x° + 121° = 180°

2x° = 59°

x = 29.5°

therefore m∠3 = 29.5°

next connect PC and consider ΔPCB,

according to given information, C∠PB = 72°

as PC and PB are radii ΔPCB is an isosceles triangle, then the angles (y°) opposite radii are equal

2y° + 72° = 180°

y° = 54°

then consider ΔPCD, again an isosceles triangle

as a straight line has an angle of 180°

B∠PC + C∠PD = 180°

72° + C∠PD = 180°

C∠PD = 108°

as ΔPCD is an isosceles triangle, P∠CD and P∠DC are equal (take this angle as z°)

2z° + 108° = 180°

z° = 36°

B∠CD is a sum of y and z angles

therefore m∠8 = y + z

m∠8 = 54° + 36° = 90°

there is also a tangent line that cuts point D of the circle. this tangent cuts D and is perpendicular to the PD radius

this indicates that lines PD and HD are perpendicular to each other

hence, forms a right angle where it cuts each other.

therefore H∠DP / m∠4 = 90°

C∠DH /m∠5 is the difference between P∠DH and P∠DC

m∠5 = 90° - 36° = 54°

m∠1 and m∠6 together make a straight line therefore the sum of these 2 angles is 180°

by looking at the options of answers given the 2 angle values given that add up to 180° are 116.5° and 63.5°.

m∠1 is the acute angle therefore m∠1 = 63.5°

m∠6 is the obtuse angle therefore m∠6 = 116.5°

consider ΔDFB

D∠FB is 180° - m∠1 = 180° - °63.5 = 116.5°

D∠FB + F∠DB(w°) + D∠BF (m∠3)= 180°

116.5 + w° + 29.5° = 180°

w° = 34°

m∠2 = m∠4 + w°

= 90° + 34°

= 124°

finally all the answers are

m∠1= 63.5°

m∠2 = 124°

m∠3 = 29.5°

m∠4 = 90°

m∠5 = 54°

m∠6 = 116.5°

m∠7 = 121°

m∠8 = 90°

the sum of the internal angles of a triangle is 180°.

when the angles of arcs have been given(e.g: BA) that means from the centre of the circle to point B has been rotated by 53° to point A.

Next from point A to point E, the angle rotated is 68°

the addition of these 2 angles give the angle rotated from point B to point E,

that is angle no.7

m∠7 = 53° + 68° = 121°

consider ΔPBE,

PB and PE are radii therefore isosceles triangles

In isosceles triangles the angles opposite equal sides are equal

if the angle is x°, then there are 2 angles with values x each

x° +x° + 121° = 180°

2x° = 59°

x = 29.5°

therefore m∠3 = 29.5°

next connect PC and consider ΔPCB,

according to given information, C∠PB = 72°

as PC and PB are radii ΔPCB is an isosceles triangle, then the angles (y°) opposite radii are equal

2y° + 72° = 180°

y° = 54°

then consider ΔPCD, again an isosceles triangle

as a straight line has an angle of 180°

B∠PC + C∠PD = 180°

72° + C∠PD = 180°

C∠PD = 108°

as ΔPCD is an isosceles triangle, P∠CD and P∠DC are equal (take this angle as z°)

2z° + 108° = 180°

z° = 36°

B∠CD is a sum of y and z angles

therefore m∠8 = y + z

m∠8 = 54° + 36° = 90°

there is also a tangent line that cuts point D of the circle. this tangent cuts D and is perpendicular to the PD radius

this indicates that lines PD and HD are perpendicular to each other

hence, forms a right angle where it cuts each other.

therefore H∠DP / m∠4 = 90°

C∠DH /m∠5 is the difference between P∠DH and P∠DC

m∠5 = 90° - 36° = 54°

m∠1 and m∠6 together make a straight line therefore the sum of these 2 angles is 180°

by looking at the options of answers given the 2 angle values given that add up to 180° are 116.5° and 63.5°.

m∠1 is the acute angle therefore m∠1 = 63.5°

m∠6 is the obtuse angle therefore m∠6 = 116.5°

consider ΔDFB

D∠FB is 180° - m∠1 = 180° - °63.5 = 116.5°

D∠FB + F∠DB(w°) + D∠BF (m∠3)= 180°

116.5 + w° + 29.5° = 180°

w° = 34°

m∠2 = m∠4 + w°

= 90° + 34°

= 124°

finally all the answers are

m∠1= 63.5°

m∠2 = 124°

m∠3 = 29.5°

m∠4 = 90°

m∠5 = 54°

m∠6 = 116.5°

m∠7 = 121°

m∠8 = 90°