Question

coty is In the given figure, if D is mid-point of BC, then value of cot y/cot x is

(a) 2
(b) 1/4
(c) 1/3
(d)1/2


please answer the question asap and please explain the answer​

Answer 1

Answer:

2 is the answer i think so

Answer 2

Concept:-

Here, It is given that a right ΔACB with median AD. And because of median we got two right angled triangle's   i.e., ΔACD and ΔACB.

Given:-

• AD is the median of ΔACB,  i.e., CD = DB

• x = ∠DAC and y = ∠BAC.

To Find:-

• $\sf{\dfrac{\cot\ y}{\cot\ x}} =\ ?}$

Solution:-

Here, We first find the value of cot y.

$\sf{\longrightarrow \cot\ y =\ \dfrac{base}{perpendicular} }$

$\sf{\longrightarrow \cot\ y =\ \dfrac{AC}{BC} }$                           [ ∵ Taking AC as base ]

$\sf{\longrightarrow \cot\ y =\ \dfrac{AC}{CD + DB} }$  

$\sf{\longrightarrow \cot\ y =\ \dfrac{AC}{2CD}}$                ----- (i)           [ ∵ CD = DB ]

Now, Value of cot x.

$\sf{\longrightarrow \cot\ x =\ \dfrac{base}{perpendicular} }$

$\sf{\longrightarrow \cot\ x =\ \dfrac{AC}{CD} }$               ----- (ii)          [ ∵ Taking AC as base ]

From equation (i) and (ii) we have,

$\sf{\Longrightarrow \dfrac{\cot\ y}{\cot\ x}} =\ \dfrac{\dfrac{AC}{2CD}}{\dfrac{AC}{CD}} }$

$\sf{\Longrightarrow \dfrac{\cot\ y}{\cot\ x}} =\ \dfrac{AC}{2CD} \times \dfrac{CD}{AC}}$

$\sf{\Longrightarrow \dfrac{\cot\ y}{\cot\ x}} =\ \dfrac{1}{2} \times \dfrac{1}{1}}$

$\sf{\Longrightarrow \dfrac{\cot\ y}{\cot\ x}} =\ \dfrac{1}{2}$

$\sf{Hence,\ We\ got\ the\ answer\ \dfrac{\cot\ y}{\cot\ x}} =\ \dfrac{1}{2}.$

Option (d) is correct.