Question

Please help me.
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Answer 1

Step-by-step explanation:

In ∆ BAD,

angle BAD = 32° (GIVEN)

BD=AD (GIVEN)

∆BAD IS AN ISOSCELES TRIANGLE

ANGLE BAD = ANGLE DBA = 32° (ANGLE OPPOSITE TO CONGRUENT SIDES ARE CONGRUENT)

x = 32°

In ∆ BAD,

Angle B + Angle A + Angle D = 180°

32° + 32° + Angle D = 180°

64° + Angle D = 180

Angle D = 180 - 64

Angle D = 106°

Angle ADB = 106°

ANGLE ADB + ANGLE CDB = 180° (ANGLES IN A LINEAR PAIR)

106 + ANGLE CDB = 180

ANGLE CDB = 180 - 106

ANGLE CDB = 74°

y = 74°

IN ∆ CDB,BC = BD ( GIVEN )

ANGLE CDB = ANGLE BCD = 74° (ANGLE OPPOSITE TO CONGRUENT SIDES ARE CONGRUENT)

In ∆ CDB,

ANGLE C + ANGLE D + ANGLE B = 180°

74 + 74 + ANGLE B = 180°

148 + ANGLE B = 180°

ANGLE B = 180 - 148

ANGLE B = 32°

z = 32°