In the Given Figure, D Is the Mid-Point of BC, Then the Value of `"Cot Y °"/"Cot X °"` Is - Mathematics

In the given figure, D is the mid-point of BC, then the value of `"cot 𝑦 °"/"cot 𝑥 °"` is

(a) 2

(b) `frac{1}{2}`

(c) `frac{1}{3}`

(d) `frac{1}{4}`

(b) `frac{1}{2}` 

Explanation:

= `\frac{cot y^{0}}{cot x^{0}} = \frac{\frac{1}{tan y^{0}}}{\frac{1}{tan x^{0}}`

= `\frac{1}{tan y^{0}}\times\frac{tan x^{0}}{1}`

= `\frac{tan x^{0}}{tan y^{0}}`

Now,

(1) `tan x^{0}`

= `tan x^{0}` = `frac{CD}{AC}` = `frac{a}{AC}`

(2) `tan x^{0}`

= `tan y^{0}` = `frac{CB}{AC}` 

= `frac{a + a}{AC}`

= `frac{2a}{AC}`

then,

= `\frac{cot y^{0}}{cot x^{0}}=\frac{tan x^{0}}{tan y^{0}}`

= `\frac{\frac{a}{AC}}{\frac{2a}{AC}}`

= `"a"/"AC"\times"AC"/"2a"`

= `"a"/"2a"`

= `1/2`

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