# In Fig. 5.48, AD = 4 cm, BD = 3 cm and CB = 12 cm, find the cot θ.

Question:

In Fig. $5.48, A D=4 \mathrm{~cm}, B D=3 \mathrm{~cm}$ and $C B=12 \mathrm{~cm}$, find the $\cot \theta$.

(a) $\frac{12}{5}$

(b) $\frac{5}{12}$

(C) $\frac{13}{12}$

(d) $\frac{12}{13}$

Solution:

We have the following given data in the figure, $A D=4 \mathrm{~cm}, B D=3 \mathrm{~cm}, C B=12 \mathrm{~cm}$

Now we will use Pythagoras theorem in $\triangle A B D$,

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

$=5 \mathrm{~cm}$

Therefore,

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

$=\frac{12}{5}$

So the answer is $(a)$