Question

57. त्रिभुज ABC में, ZBAC = 50° तथा ZABC
तथा LACB के विभाजक बिंदु P पर मिलते हैं।
ZBPC का मान (डिग्री में) कितना होगा?​

Answer 1

Answer :ZBPC=115°

Step by step explanation :

For finding ZBPC, we have ZBAC = 50° and P is the bisector of ZABC and ZACB, so __

In triangle ABC.__

ZBAC +ZABC +ZACB = 180° ( Sum property of triangle)

ZABC + ZACB = 180° - ZBAC

ZABC + ZACB = 180° - 50°

ZABC + ZACB = 130°

1/2( ZABC + ZACB) = 1/2(130°) { P is the bisector of ZABC + ZACB}

ZPBC + ZPCB = 65°

In triangle PBC __

(ZPBC + ZPCB)+ ZBPC = 180° ( Sum property of triangle)

65° + ZBPC = 180°

ZBPC = 180° - 65°

ZBPC = 115° (answer)