In $\triangle A B C, A C=8.4 \mathrm{~cm}$ and $B C=14 \mathrm{~cm}$. $\mathrm{P}$ is a point on $\mathrm{AB}$ such that $\mathrm{CP}=11.2 \mathrm{~cm}$ and $\angle A C P=\angle B$. What is the length (in $\mathrm{cm}$) of $\mathrm{BP}$ ?

3.78

It is given that ∠ACP = ∠B

In Triangle ACP and ABC ,

∠ACP = ∠B

∠A is common in both

So, △ACP ∼ △ABC

⇒ \(\frac{AC}{AB}\) = \(\frac{PC}{BC}\) = \(\frac{AP}{AC}\)

\(\frac{AC}{AB}\) = \(\frac{PC}{BC}\)

\(\frac{8.4}{AB}\) = \(\frac{11.2}{14}\)

AB = \(\frac{8.4 × 14 }{11.2}\)

AB = 10.5

Now, \(\frac{PC}{BC}\) = \(\frac{AP}{AC}\)

  \(\frac{11.2}{14}\) = \(\frac{AP}{ 8.4}\)

AP = 6.72

Now, BP = AB - AP = 10.5 - 6.72

= 3.78