Question

Two line segments ac and bd intersect each other at the point p such that pa=6cm pb=3cm pc=2.5cm pd=5cm angle apb=50 angle cdp=30 then pba is equal to

Answer 1

Answer:

∠apb=∠cpd=50°→→vertically opposite angles

∠cdp=30°,

Also, ∠c p d + ∠ c d p+∠d c p=180°→→Angle sum property of triangle

50° + 30° + ∠d cp = 180°

∠ d c p= 100° ----(1)

In Δa p b and Δ d p c

ap=6 cm, p d= 5 cm, p b=3 cm, p c= 2.5 cm

∠ a p b= ∠ c p d= 50°→→Vertically opposite angles

Also, $\frac{ap}{pd}=\frac{b p}{p c}\\\\ \frac{6}{5}=\frac{3}{2.5}$

Δa p b ~ Δ d p c →→[S A S]

∠p b a=∠p c d→→→[Corresponding angle of two similar triangles]

∠ p b a= 100°------[using (1)]