Question

In an isosceles triangle ABC ,AB‌=BC,angle B=20 . M and N are on AB and BC respectively such that angle MCA =60, angleNAC =50.find angle MNC

Answer 1

Answer:

∡MNC = $100^{o}$

Step-by-step explanation:

Given;

(i) ABC is an Isosceles Triangle.

(ii) ∡B = $20^{o}$  

(iii) AB = BC, which means AC is the base of the triangle and AB and BC are the legs; as shown in the figure at the attachment.

Also, ∡A = ∡C (apex angles are equal)

And, ∡A + ∡B + ∡C = $180^{o}$ (Sum of angles in  Triangle are equal)

    => ∡A + ∡C = $180^{o}$ - $20^{o}$

    => ∡A + ∡C = $160^{o}$

    => ∡A = ∡C = $80^{o}$

Also given;

(i) ∡MCA = $60^{o}$, ∡NAC = $50^{o}$

   Now,

   ∡BAC = ∡BAN + ∡NAC

   $80^{o}$ = ∡BAN + $50^{o}$

  => ∡BAN = 30

   ∡BCA = ∡BCM + ∡MCA

   $80^{o}$ = ∡BCM + $60^{o}$

  => ∡BCM = $20^{o}$

Also;

MN // to AC

So,

∡BMN = ∡BAC & ∡BNM = ∡BCA (Corresponding Angles)

∡BMN = $80^{o}$, ∡BNM = $80^{o}$

Now;

Consider Triangle ∡ANC,

∡NAC + ∡ACN + ∡CNA = $180^{o}$ (Sum of angles in a Triangle = $180^{o}$)

=> $50^{o}$ + $80^{o}$ + ∡CNA = $180^{o}$

=> ∡CNA = $50^{o}$

Also,

∡BNM + ∡MNA + ∡NAC = $180^{o}$ (BC is a straight line)

=> $80^{o}$ + ∡MNA + $50^{o}$ = $180^{o}$

=> ∡MNA = $50^{o}$

And,

∡MNC = ∡MNA + ∡ANC

   = $50^{o}$ + $50^{o}$

   = $100^{o}$

Answer 2

Step-by-step explanation:

Answer:

∡MNC = 100^{o}100o

Step-by-step explanation:

Given;

(i) ABC is an Isosceles Triangle.

(ii) ∡B = 20^{o}20o  

(iii) AB = BC, which means AC is the base of the triangle and AB and BC are the legs; as shown in the figure at the attachment.

Also, ∡A = ∡C (apex angles are equal)

And, ∡A + ∡B + ∡C = 180^{o}180o (Sum of angles in  Triangle are equal)

    => ∡A + ∡C = 180^{o}180o - 20^{o}20o

    => ∡A + ∡C = 160^{o}160o

    => ∡A = ∡C = 80^{o}80o

Also given;

(i) ∡MCA = 60^{o}60o , ∡NAC = 50^{o}50o

   Now,

   ∡BAC = ∡BAN + ∡NAC

   80^{o}80o = ∡BAN + 50^{o}50o

  => ∡BAN = 30

   ∡BCA = ∡BCM + ∡MCA

   80^{o}80o = ∡BCM + 60^{o}60o

  => ∡BCM = 20^{o}20o

Also;

MN // to AC

So,

∡BMN = ∡BAC & ∡BNM = ∡BCA (Corresponding Angles)

∡BMN = 80^{o}80o , ∡BNM = 80^{o}80o

Now;

Consider Triangle ∡ANC,

∡NAC + ∡ACN + ∡CNA = 180^{o}180o (Sum of angles in a Triangle = 180^{o}180o )

=> 50^{o}50o + 80^{o}80o + ∡CNA = 180^{o}180o

=> ∡CNA = 50^{o}50o

Also,

∡BNM + ∡MNA + ∡NAC = 180^{o}180o (BC is a straight line)

=> 80^{o}80o + ∡MNA + 50^{o}50o = 180^{o}180o

=> ∡MNA = 50^{o}50o

And,

∡MNC = ∡MNA + ∡ANC

   = 50^{o}50o + 50^{o}50o

   = 100^{o}100o