Question

In a triangle abc angle a + angle b is 110° , angle c + angle a is 135° . find angle value of abc

Answer 1

a+b=110°

c+a=135°

we know that

a+b+c=180°

110°+c=180°

c=70°

given, c+a=135°

70°+a=135°   [c=70°]

a=135°-70°

a=65°

so,b+a=110°

b+65°=110° [a=65°]

b=110°-65°

b=45°

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Answer 2

Given :

=> <A + <B = 110°

=> <C + <A = 135°

Solution :

=> <A + <B = 110° -----(1) { Given }

=> <A + <B = 110° -----(1) { Given }=> <A + <B + <C = 180° { Angle sum property of ∆le }

=> <A + <B = 110° -----(1) { Given }=> <A + <B + <C = 180° { Angle sum property of ∆le }=> 110° + <C = 180° { From (1) }

=> <A + <B = 110° -----(1) { Given }=> <A + <B + <C = 180° { Angle sum property of ∆le }=> 110° + <C = 180° { From (1) }=> <C = 180° - 110°

=> <A + <B = 110° -----(1) { Given }=> <A + <B + <C = 180° { Angle sum property of ∆le }=> 110° + <C = 180° { From (1) }=> <C = 180° - 110°=> <C = 70°

=> <C + <A = 135° { Given }

=> 70° + <A = 135°

=> <A = 135° - 70°

=> <A = 65°

=> <A + <B = 110° { Given }

=> 65° + <B = 110°

=> <B = 110° - 65°

=> <B = 45°

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