Question

In the fig. AD=4cm BD = 3cm and BC=12cm, find cot theta.

Answer 1

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Answer 2

Answer:

$cot{\theta}=\frac{12}{5}$

Step-by-step explanation:

In ΔABD, we have

$(AB)^{2}=(BD)^{2}+(AD)^{2}$

⇒$(AB)^{2}=(4)^{2}+(3)^{2}$

⇒$(AB)^{2}=16+9$

⇒$(AB)^{2}=25$

⇒$AB=\sqrt{25}$

⇒$AB=5$.

Now, from ΔABC, we have

⇒$(AC)^{2}=(AB)^{2}+(BC)^{2}$

⇒$(AC)^{2}=(5)^{2}+(12)^{2}$

⇒$(AC)^{2}=169$

⇒$AC=13$

Now, from ΔABC, $cot{\theta}=\frac{Base}{Perpendicular}$

$cot{\theta}=\frac{BC}{AB}$

$cot{\theta}=\frac{12}{5}$