Question

Question 5 please fast reply

Answer 1

Answer:

(i) x=42°

(ii) x=40°

Step-by-step explanation:

(i)IN ∆BAD

angle BAD = 48°

[angle opp. to equal sides are equal]

angle BAD+ angle BDA+ angle DBA =180°

48°+48°+angle BDA=180° [ASP]

angle BDA = 180°-96°

= 84°

84°+ angle ADC=180° [Straight line]

angle ADC = 180°-84°

= 96°

IN ∆CAD

96°+x+x= 180°

[angle opp. to equal sides are equal]

2x=180°-96°

2x=84°

x=84°/2=42°

x=42°

(ii)IN ∆ACD

angle ACD = 180°-130°

= 50°

therefore angle DAC = 50°

[angle opp. to equal sides are equal]

50°+50°+ angle ADC = 180° [ASP]

angle ADC = 180°-100°

= 80°

80°+ angle ADB=180 [straight line]

angle ADB = 180°-80°

= 100°

IN ∆BAD

x+x+100°= 180°

[angle opp. to equal sides are equal]

2x= 180°-100°

= 80°

x= 80°/2= 40°