Question

Side BC of ΔABC is extended to E. DBC is another Δ such that DB intersects AC at P. Find the angles marked x and y and arrange the sides of ΔPCD in ascending order.

Answer 1

Given,

Side BC of ΔABC is extended to E.

DBC is another Δ such that DB intersects AC at P.

In ΔABC, ∠ABC + ∠BAC + ∠BCA = 180°   ...(Sum of an interior angle of the triangle)

55° + 55° + 40° + ∠BCA = 180°

∠BCA = 30°

∠ACE = ∠CAB + ∠ABC

x + x = 40° + 55° + 55°

2x = 150°

x = 75°

In ΔABP, ∠BAP + ∠ABP + ∠APB = 180°  ...(Sum of an interior angle of the triangle)

40° + 55° + ∠APB = 180°

∠APB = 85°

∠APB and ∠CPD are vertically opposite angle.

So, ∠CPD = 85°

In ΔDPC, ∠CPD + ∠PDC + ∠DCP = 180°

85° + y + 75° = 180°

y = 20°

In ΔPCD, ∠CPD = 40°, ∠PDC = 20°, ∠DCP = 75°

We know that, the side opposite to the largest angle is the longest in triangles.

Arranging the sides in ascending order, PC < PD < DC.

Hence, x = 75°, y = 20° and arrangement of side in ΔPCD is PC < PD < DC.