Question

find the range of x such that given quadrilateral can exist.Explain with steps​

Answer 1

✪ᴀɴsᴡᴇʀ✪

• $\boxed{\bf{ 2.5^o < x < 15^o}}$

• A more accurate range of x is $\boxed{\bf{2.5^o < x ≤ 14.11^o}}$

⍟ᴇxᴘʟᴀɪɴᴀᴛɪᴏɴ⍟

Approximate Range

Since Angle is always positive,

$\:\:\:\:\: 4x-10 > 0$

$\to 4x > 10$

$\to x > 2.5^o...(1)$

Let

• AC = CE = a
• BC = CD = b

Using cosine formula,

$\:\:\:\:\: \cos(50)-\cos(4x-10)$

$= \frac{a^2 + b^2 - 15^2}{2ab} -\frac{a^2 + b^2 - 14^2}{2ab}$

$= \frac{(a^2 + b^2 - 15^2) -(a^2 + b^2 - 14^2)}{2ab}$

$=\frac{- 15^2 + 14^2}{2ab}$

$= \frac{- 29}{2ab} < 0$

$\to \cos(50) < \cos(4x-10)$

$\to 50 > 4x-10$

$\to 4x-10 < 50$

$\to 4x < 60$

$\to x < 15^o...(2)$

From (1) and (2), we get

$\:\:\:\:\: 2.5^o < x < 15^o$

A More Accurate Range

From the above analysis

$\cos(50) = \frac{a^2 + b^2 - 15^2}{2ab}.. (3)$

$\cos(4x-10) = \frac{a^2 + b^2 - 14^2}{2ab}.. (4)$

Substracting (3) from (4), we get

$\to \cos(4x-10)-\cos(50) = \frac{29}{2ab}$

$\to 2ab = \frac{29}{\cos(4x-10)-\cos(50)}$

$\to a^2b^2 = \frac{29^2}{4[\cos(4x-10)-\cos(50)]^2}$

Substituting value of ab in (3),we get

$\to \cos(50) = \frac{a^2 + b^2 - 15^2}{2ab}$

$\to a^2 + b^2 = 15^2 + 2abcos(50)$

$\to a^2 + b^2 = 15^2 + \frac{29cos(50)}{\cos(4x-10)-\cos(50)}$

$\to a^2 + b^2 = \frac{225\cos(4x-10)-196cos(50)}{\cos(4x-10)-\cos(50)}$

Quadratic equation whose roots are $a^2, b^2$ is of the form

$\:\:\:\:\: t^2 - (a^2+b^2)t + a^2 b^2 = 0$

Since $a, b \in \R, Determinant ≥ 0$

$\to (a^2+b^2)^2 - 4a^2 b^2 ≥ 0$

$\to \left[\frac{225\cos(4x-10)-196cos(50)}{\cos(4x-10)-\cos(50)}\right]^2$

$\:\:\:\:\: - \frac{4\times 29^2}{4[\cos(4x-10)-\cos(50)]^2} ≥ 0$

$\to \left[225\cos(4x-10)-196cos(50)\right]^2$

$\:\:\:\:\: - 29^2 ≥ 0$

$\to \left[225\cos(4x-10)-196cos(50) - 29\right]$

$\left[225\cos(4x-10)-196cos(50) + 29\right] ≥ 0$

$\to \cos(4x-10)≥ \frac{196cos(50) + 29}{225}$

or $cos(4x-10)≤\frac{196cos(50) - 29}{225}$

$\to \cos(4x-10) ≥ 0.6889$

or $cos(4x-10) ≤0.4310$

We already know that

$\to \cos(4x-10)>cos(50)$

$\to \cos(4x-10)>0.643$

Combining the two results we get

$\to \cos(4x-10) ≥ 0.6889$

$\to 0 < 4x-10 ≤ cos^{-1}(0.6889)$

$\to 0 < 4x-10 ≤ 46.458$

$\to 10 < 4x ≤ 56.458$

$\to \boxed{\bf{2.5 < x ≤ 14.11}}$