Question

in the following figure find the value of X, Y and Z​

Answer 1

Answer:

z=50

y=80

x=120

Step-by-step explanation:

apply angle sum propert in triangle BFC

THEN z=180-90-40=50

apply in triangle ADC

then angle FDC = 100

then Y=180-100=80

apply in triangle BDF

then angle BFD = 60

then X = 180-60 = 120

Answer 2

Answer:

x = 120°

y = 80°

z = 50°

Step-by-step explanation:

Given : ∠DAC = 30°

∠EBC = 40°

∠BEC = 90°

To Find : ∠BCE, ∠ADB, ∠AFB

Now, in ∆ BEC,

∠BEC + ∠EBC + ∠BCE = 180°

(Angle Sum Property)

Putting known values, we get

→ 90° + 40° + ∠BCE = 180°

→ 130° + ∠BCE = 180°

→ ∠BCE = 180° + 130°

→ ∠BCE = 50°

→ z = 50°

Now, in ∆ ADC,

∠DAC + ∠ADC + ACD = 180°

(Angle Sum Property)

Putting known values, we get

→ 30° + ∠ADC + 50° = 180°

(∠ADC = ∠BCE = 50°)

→ 80° + ∠ADC = 180°

→ ∠ADC = 180° - 80°

→ ∠ADC = 100°

As we know that,

∠ADC + ADB = 180°

(Linear Pair)

Putting known value, we get

→ 100° + ∠ADB = 180°

→ ∠ADB = 180° - 100°

→ ∠ADB = 80°

→ y = 80°

Now, in ∆ BDF,

∠BFD + ∠BDF + ∠DBF = 180°

(Angle Sum Property)

Putting known values, we get

→ ∠BFD + 80° + 40° = 180°

(∠BDF = ∠ADB = 80°)

→ ∠BFD + 120° = 180°

→ ∠BFD = 180° - 120°

→ ∠BFD = 60°

As we know that,

∠BFD + AFB = 180°

(Linear Pair)

Putting known value, we get

→ 60° + ∠AFB = 180°

→ ∠AFB = 180° - 60°

→ ∠AFB = 120°

→ x = 120°