Question

solve it! fast ...........​

Answer 1

Given,

$\longrightarrow\sf{f(x)=ax^2+bx+c\quad\quad\dots(1)}$

for $\sf{a,\ b,\ c\in\mathbb{Z},}$ such that,

$\longrightarrow\sf{f\left(\cos\left(\dfrac{\pi}{7}\right)\right)=\sin\left(\dfrac{\pi}{7}\right)\sin\left(\dfrac{3\pi}{7}\right)+\sin\left(\dfrac{3\pi}{7}\right)\sin\left(\dfrac{5\pi}{7}\right)+\sin\left(\dfrac{5\pi}{7}\right)\sin\left(\dfrac{\pi}{7}\right)}$

By product - to - sum rule, we have,

$\sf{\sin A\sin B=\dfrac{1}{2}\big[\cos(A-B)-\cos(A+B)\big]}$

Then,

$\begin{aligned}\longrightarrow\sf{f\left(\cos\left(\dfrac{\pi}{7}\right)\right)}&=\sf{\dfrac{1}{2}\left[\cos\left(\dfrac{\pi}{7}-\dfrac{3\pi}{7}\right)-\cos\left(\dfrac{\pi}{7}+\dfrac{3\pi}{7}\right)\right]}\\\\&+\sf{\dfrac{1}{2}\left[\cos\left(\dfrac{3\pi}{7}-\dfrac{5\pi}{7}\right)-\cos\left(\dfrac{3\pi}{7}+\dfrac{5\pi}{7}\right)\right]}\\\\&+\sf{\dfrac{1}{2}\left[\cos\left(\dfrac{5\pi}{7}-\dfrac{\pi}{7}\right)-\cos\left(\dfrac{5\pi}{7}+\dfrac{\pi}{7}\right)\right]}\\\\&\end{aligned}$

Since $\sf{\cos(-x)=\cos x,}$

$\begin{aligned}\longrightarrow\sf{f\left(\cos\left(\dfrac{\pi}{7}\right)\right)}&=\sf{\dfrac{1}{2}\left[\cos\left(\dfrac{2\pi}{7}\right)-\cos\left(\dfrac{4\pi}{7}\right)\right]}\\\\&+\sf{\dfrac{1}{2}\left[\cos\left(\dfrac{2\pi}{7}\right)-\cos\left(\dfrac{8\pi}{7}\right)\right]}\\\\&+\sf{\dfrac{1}{2}\left[\cos\left(\dfrac{4\pi}{7}\right)-\cos\left(\dfrac{6\pi}{7}\right)\right]}\\\\&\end{aligned}$

$\begin{aligned}\longrightarrow\sf{f\left(\cos\left(\dfrac{\pi}{7}\right)\right)}&=\sf{\dfrac{1}{2}\left[\cos\left(\dfrac{2\pi}{7}\right)-\cos\left(\dfrac{4\pi}{7}\right)\right]}\\\\&+\sf{\dfrac{1}{2}\left[\cos\left(\dfrac{2\pi}{7}\right)-\cos\left(\pi+\dfrac{\pi}{7}\right)\right]}\\\\&+\sf{\dfrac{1}{2}\left[\cos\left(\dfrac{4\pi}{7}\right)-\cos\left(\pi-\dfrac{\pi}{7}\right)\right]}\\\\&\end{aligned}$

Since  $\sf{\cos(\pi-x)=\cos(\pi+x)=-\cos x,}$

$\begin{aligned}\longrightarrow\sf{f\left(\cos\left(\dfrac{\pi}{7}\right)\right)}&=\sf{\dfrac{1}{2}\left[\cos\left(\dfrac{2\pi}{7}\right)-\cos\left(\dfrac{4\pi}{7}\right)\right]}\\\\&+\sf{\dfrac{1}{2}\left[\cos\left(\dfrac{2\pi}{7}\right)+\cos\left(\dfrac{\pi}{7}\right)\right]}\\\\&+\sf{\dfrac{1}{2}\left[\cos\left(\dfrac{4\pi}{7}\right)+\cos\left(\dfrac{\pi}{7}\right)\right]}\\\\&\end{aligned}$

$\longrightarrow\sf{f\left(\cos\left(\dfrac{\pi}{7}\right)\right)=\cos\left(\dfrac{2\pi}{7}\right)+\cos\left(\dfrac{\pi}{7}\right)}$

Since $\sf{\cos(2x)=2\cos^2x-1,}$

$\longrightarrow\sf{f\left(\cos\left(\dfrac{\pi}{7}\right)\right)=2\cos^2\left(\dfrac{\pi}{7}\right)+\cos\left(\dfrac{\pi}{7}\right)-1\quad\quad\dots(2)}$

Hence, on comparing (2) with (1), we get,

$\longrightarrow\sf{f(x)=2x^2+x-1}$

Therefore,

$\longrightarrow\sf{f(2)=2(2)^2+2-1}$

$\longrightarrow\sf{\underline{\underline{f(2)=9}}}$

Hence 9 is the answer.

Answer 2

Answer:

Given :

Area of rhombus = 120 cm²

Length of diagonal = 8 cm

To find :

Length of another diagonal

According to the question,

$\sf{ : \implies Area \: of \: rhombus = \dfrac{1}{2} \times d _{1} \times d_{2} }$

$\sf : \implies{ {120 \: cm}^{2} = \dfrac{1}{2} \times 8 \: cm \times x}$

$\sf : \implies{ {120 \: cm}^{2} \times 2 = 8 \: cm \times x }$

$\sf : \implies{ {240 \: cm}^{2} = 8 \: cm \times x}$

$\sf : \implies{ \dfrac{240}{8} \: cm = x}$

${ \underline{ \boxed{ \sf \pink{ : \implies{ \bm3 \bm0 \: c m =x}}}}}$

${ \therefore{ \underline{\sf{So \:,the \: length \: of \: other \: diagonal \: is \: 3 0 \: cm}}}}$