In Fig. 3, AD = 4 cm, BD = 3 cm and CB = 12 cm,

Solution:

In Fig 3, if we have

AD = 4 cm,

BD = 3 cm, and

CB = 12 cm, then

cot θ = ?

We have the following diagram

If we apply Pythagoras theorem in triangle ADB, we get

hypotenus $=\sqrt{(\text { base })^{2}+(\text { perpendicular })^{2}}$

$\Rightarrow \quad A B=\sqrt{(B D)^{2}+(A D)^{2}}$

$\Rightarrow \quad A B=\sqrt{3^{2}+4^{2}}$

 

$\Rightarrow \quad A B=5$

Now,

$\cot \theta=\frac{\text { base }}{\text { perpendicular }}$

$=\frac{C B}{A B}$

$=\frac{12}{5}$

Hence the correct option is (d).