Question

in the figure given AN = AC, /_ BAC =52°, /_ ACK = 84° ans BCK is a straight line. Prove that NB = NC​

Answer 1

Answer:

Given:

AN=AC

hence ∆ANC is an isoceles triangle

/_BAC=52°

/_ACK=84°

BCK is a straight line

To prove:

NB = NC

Proof:

In ∆ANC

/_ANC=/_ACN (AN=AC)

52°+/_ANC+/_ANC=180° (Sum of a triangle is 180°)

2/_ANC=180°-52°

/_ANC=128°/2

=64°

therefore,/_ACN=64°

Now,

/_BCK=180°(BCK is a straight line)

/_ACK+/_ACN+/_NCB=180°

84°+64°+/_NCB=180°

/_NCB=180°-148°

=32°

In ∆NBC,

/_CNB+/_ANC=180° (Linear pair)

/_CNB=180°-64°

=116°

Now,

/_CNB+/_NBC+/_NCB=180° (Sum of a triangle is 180°)

116°+32°+/_NBC=180°

/_NBC=180°-148°

=32°

Therefore,/_NBC=/_NCB

i.e. NB=NC

Hence proved.