NCERT Solutions Class 7 Mathematics Triangles and its Properties Ex 6.3

Answer :
(i) 50^o + 𝓍 = 120^o [Exterior angle property of a Δ ]

⇒ 𝓍 = 120^o– 50^o= 70^o Now, 50^o+ 𝓍 + 𝓎 = 180^o [Angle sum property of a Δ ] ⇒ 50^o + 70^o + 𝓎 = 180^o

⇒ 120^o + 𝓎 = 180^o

⇒ 𝓎 = 180^o – 120^o = 60^o

(ii) 𝓎 = 80^o ……….(i) [Vertically opposite angle]

Now, 50^o + 𝓍 + 𝓎 = 180^o [Angle sum property of a Δ ]

⇒ 50^o + 80^o + 𝓎 = 180^o [From eq. (i)]

⇒ 130^o + 𝓎 = 180^o

⇒ 𝓎 = 180^o – 130^o = 50^o

(iii) 50^o + 60^o = 𝓍 [E𝓍terior angle property of a Δ ]

⇒ 𝓍 = 110^o

Now 50^o + 60^o + 𝓎 = 180^o [Angle sum property of a Δ ]

⇒ 110^o + 𝓎 = 180^o

⇒ 𝓎 = 180^o – 110^o

⇒ 𝓎 = 70^o

(iv) 𝓍 = 60^o ……….(i) [Vertically opposite angle]

Now, 30^o + 𝓍 + 𝓎 = 180^o [Angle sum property of a Δ ] ⇒ 50^o + 60^o + 𝓎 = 180^o [From eq. (i)]

⇒ 90^o + 𝓎 = 180^o

⇒ 𝓎 = 180^o – 90^o = 90^o

(v) 𝓎 = 90^o ……….(i) [Vertically opposite angle]

Now, 𝓎 + 𝓍 + 𝓍 = 180^o [Angle sum property of a Δ ]

⇒ 90^o + 2𝓍 = 180^o [From eq. (i)]

⇒ 2𝓍 = 180^o – 90^o

⇒ 2𝓍 = 90^o

⇒ 𝓍 = 90^o2 = 45^o

(vi) 𝓍 = 𝓎 ……….(i) [Vertically opposite angle]

Now, 𝓍 + 𝓍 + 𝓎 = 180^o [Angle sum property of a Δ ]

⇒ 2𝓍 + 𝓍 = 180^o [From eq. (i)]

⇒ 3𝓍 = 180^o

⇒ 𝓍 = 180^o3 = 60^o