Mathematics
Answer
From figure,
∠ABD = ∠BAD = 36° (As angles opposite to equal sides are equal.)
∠BDA = 180° - (36° + 36°) = 180° - 72° = 108°.
From figure,
∠BDA + ∠ADC = 180°
108° + ∠ADC = 180°
∠ADC = 72°.
In △ADC,
As AD = AC,
∴ ∠ADC = ∠ACD = 72° (As angles opposite to equal sides are equal.)
∠ADC + ∠ACD + ∠DAC = 180°
72° + 72° + ∠DAC = 180°
∠DAC = 180° - 144° = 36°.
From figure,
∠BAD + ∠DAC + x = 180°
36° + 36° + x = 180°
72° + x = 180°
x = 108°.
Hence, x = 108°.
Related Questions
In the adjoining figure, TR = TS, ∠1 = 2∠2 and ∠4 = 2∠3. Prove that RB = SA.
In △ABC, D is a point on BC such that AD is the bisector of ∠BAC. CE is drawn parallel to DA to meet BD produced at E. Prove that △CAE is isosceles.
In the figure (1) given below, find the value of x.
In the adjoining figure, ABC is a right angled triangle at B. ADEC and BCFG are squares. Prove that AF = BE.
