Given,  b  is the mean proportion between  a  and  c.

$$\begin{gathered} \Rightarrow \frac{\mathrm{a}}{\mathrm{b}} = \frac{\mathrm{b}}{\mathrm{c}} = \mathrm{k} ( \mathrm{Say} ) \\ \Rightarrow \mathrm{a} = \mathrm{b} \mathrm{k}, \mathrm{b} = \mathrm{c} \mathrm{k} \\ \Rightarrow \mathrm{a} = ( \mathrm{c} \mathrm{k} ) \mathrm{k} = \mathrm{c} {\mathrm{k}}^2, \mathrm{b} = \mathrm{c} \mathrm{k} \\ \mathrm{L}.\mathrm{H}.\mathrm{S}. = \frac{{\mathrm{a}}^4 + {\mathrm{a}}^2 {\mathrm{b}}^2 + {\mathrm{b}}^4}{{\mathrm{b}}^4 + {\mathrm{b}}^2 {\mathrm{c}}^2 + {\mathrm{c}}^4} \\ = \frac{( \mathrm{c} {\mathrm{k}}^2 {)}^4 +( \mathrm{c} {\mathrm{k}}^2 {)}^2 ( \mathrm{c} \mathrm{k} {)}^2 + ( \mathrm{c} \mathrm{k} {)}^4}{( \mathrm{c} \mathrm{k} {)}^4 + ( \mathrm{c} \mathrm{k} {)}^2 {\mathrm{c}}^2 + {\mathrm{c}}^4} \end{gathered}$$

           $$\begin{gathered} = \frac{ {\mathrm{c}}^4 {\mathrm{k}}^8 + ( {\mathrm{c}}^2 {\mathrm{k}}^4 ) ( {\mathrm{c}}^2 {\mathrm{k}}^2 ) + {\mathrm{c}}^4 {\mathrm{k}}^4}{ {\mathrm{c}}^4 {\mathrm{k}}^4 + ( {\mathrm{c}}^2 {\mathrm{k}}^2 ) {\mathrm{c}}^2 + {\mathrm{c}}^4} \\ = \frac{ {\mathrm{c}}^4 {\mathrm{k}}^8 + {\mathrm{c}}^4 {\mathrm{k}}^6 + {\mathrm{c}}^4 {\mathrm{k}}^4}{ {\mathrm{c}}^4 {\mathrm{k}}^4 + {\mathrm{c}}^4 {\mathrm{k}}^2 + {\mathrm{c}}^4} \\ = \frac{{\mathrm{c}}^4 {\mathrm{k}}^4 \left( {\mathrm{k}}^4 + {\mathrm{k}}^2 + 1 \right)}{{\mathrm{c}}^4 \left( {\mathrm{k}}^4 + {\mathrm{k}}^2 + 1 \right)} \\ = {\mathrm{k}}^4 \end{gathered}$$

$$\begin{gathered} \mathrm{R}.\mathrm{H}.\mathrm{S}. = \frac{{\mathrm{a}}^2}{{\mathrm{c}}^2} \\ = \frac{( \mathrm{c} {\mathrm{k}}^2 {)}^2}{{\mathrm{c}}^2} \\ = \frac{ {\mathrm{c}}^2 {\mathrm{k}}^4}{{\mathrm{c}}^2} \\ = {\mathrm{k}}^4 \end{gathered}$$

Hence,   L.H.S.  =  R.H.S.

Given equation is   4 x² - 5 x - 3 = 0.

Comparing with  ax² + b x + c = 0,    we get

a = 4,    b = - 5   and     c = - 3

$$\begin{gathered} \therefore \mathrm{x} = \frac{- \mathrm{b} \pm \sqrt{ {\mathrm{b}}^2 - 4 \mathrm{a} \mathrm{c}}}{2 \mathrm{a}} \\ = \frac{- ( - 5 ) \pm \sqrt{ ( - 5 {)}^2 - 4 ( 4 ) ( - 3 )}}{2 \times 4} \\ = \frac{5 \pm \sqrt{ 25 + 48}}{8} \\ = \frac{5 \pm \sqrt{ 73}}{8} \\ = \frac{5 \pm 8.54}{8} \end{gathered}$$

       $$\begin{gathered} = \frac{5 + 8.54}{8} \mathrm{or} \frac{5 - 8.54}{8} \\ = \frac{13.54}{8} \mathrm{or} \frac{- 3.54}{8} \end{gathered}$$

       = 1.6925      or     - 0.4425

       = 1.69          or      - 0.44

Join  OA  and  OC.

Since the perpendicular from the centre of the circle to a chord bisects the chord.

Therefore,  N  and  M  are the mid-points of  AB  and  CD  respectively.

Consequently,

$$\begin{gathered} \mathrm{AN} = \mathrm{NB} = \frac{1}{2} \mathrm{AB} = \frac{1}{2} \times 24 = 12 \mathrm{cm} \mathrm{and} \\ \mathrm{CM} = \mathrm{MD} = \frac{1}{2} \mathrm{CD} = \frac{1}{2} \times 10 = 5 \mathrm{cm} \end{gathered}$$

In right-angled triangles  ANO  and  CMO,  we have

      ( OA )² = ( ON )² + ( AN )²         and      ( OC )² = ( OM )² + ( CM )² 

$\Rightarrow$  ( 13 )² = ( ON )² + ( 12 )²          and      ( 13 )² = ( OM )² + ( 5 )² 

$\Rightarrow$  ( ON )² = ( 13 )² - ( 12 )²           and       ( OM )²  =  ( 13 )² - ( 5 )² 

$\Rightarrow$  ( ON )² = 169 - 144                    and       ( OM )²  =  169 - 25

$\Rightarrow$  ( ON )²  =  25                            and        ( OM )²  = 144

$\Rightarrow$  ON = 5                                      and         OM = 12

Now,   NM = ON + OM

               = 5 + 12

               = 17 cm.

Hence, the distance between the two chords is  17 cm.  

sin² $28°$+  sin² $62°$  +  tan² $38°$-  cot² $52°$ + $\frac{1}{4}$ sec² $30°$.

= sin² $28°$+  sin² ( 90 - 28 )$°$ +  tan² 3$8°$  -  cot² ( 90 - 38 )$°$ + $\frac{1}{4}$ sec² $30°$.

$$\begin{gathered} = \left( {\sin }^2 28° + {\cos }^2 28° \right) + {\tan }^2 38° - {\tan }^2 38° + \frac{1}{4} \times {\left( \frac{2}{\sqrt{3}} \right)}^2 \\ = 1 + 0 + \frac{1}{4} \times \frac{4}{3} \\ = 1 + \frac{1}{3} \\ = \frac{4}{3} \end{gathered}$$

$$\begin{gathered} \mathrm{Given}: \mathrm{A} = \left[ \begin{array}{cc}1& 3\\ 3& 4\end{array} \right], \mathrm{B} = \left[ \begin{array}{cc}- 2 & 1\\ - 3 & 2\end{array} \right] \mathrm{and} {\mathrm{A}}^2 - 5 {\mathrm{B}}^2 = 5 \mathrm{C} \\ \mathrm{Now}, {\mathrm{A}}^2 = \mathrm{A} \times \mathrm{A} = \left[ \begin{array}{cc}1& 3\\ 3& 4\end{array} \right] \times \left[ \begin{array}{cc}1& 3\\ 3& 4\end{array} \right] \\ = \left[ \begin{array}{cc}1 \mathrm{x} 1 + 3 \mathrm{x} 3& 1 \mathrm{x} 3 + 3 \mathrm{x} 4\\ 3 \mathrm{x} 1 + 4 \mathrm{x} 3& 3 \mathrm{x} 3 + 4 \mathrm{x} 4\end{array} \right] \\ = \left[ \begin{array}{cc}1 + 9& 3 + 12\\ 3 + 12& 9 + 16\end{array} \right] \end{gathered}$$

                        = $\left[\begin{array}{cc} 10& 15\\ 15& 25\end{array} \right]$

$$\begin{gathered} \mathrm{And}, {\mathrm{B}}^2 = \mathrm{B} \times \mathrm{B} = \left[ \begin{array}{cc}- 2 & 1\\ - 3 & 2\end{array} \right] \times \left[ \begin{array}{cc}- 2 & 1\\ - 3 & 2\end{array} \right] \\ = \left[ \begin{array}{cc}- 2 \mathrm{x} ( - 2 ) + 1 \mathrm{x} ( - 3 )& - 2 \mathrm{x} 1 + 1 \mathrm{x} 2\\ - 3 \mathrm{x} ( - 2 ) + 2 \mathrm{x} ( - 3 )& - 3 \mathrm{x} 1 + 2 \mathrm{x} 2\end{array} \right] \\ = \left[ \begin{array}{cc}4 - 3& - 2 + 2 \\ 6 - 6& - 3 + 4 \end{array} \right] \\ = \left[ \begin{array}{cc}1& 0\\ 0& 1\end{array} \right] \end{gathered}$$

$$\begin{gathered} \mathrm{Now}, {\mathrm{A}}^2 - 5 {\mathrm{B}}^2 = \left[ \begin{array}{cc}10& 15\\ 15& 25\end{array} \right] - 5 \left[ \begin{array}{cc}1& 0\\ 0& 1\end{array} \right] \\ = \left[ \begin{array}{cc}10& 15\\ 15& 25\end{array} \right] - \left[ \begin{array}{cc}5& 0\\ 0& 5\end{array} \right] \\ = \left[ \begin{array}{cc}5& 15\\ 15& 20\end{array} \right] \\ = 5 \left[ \begin{array}{cc}1& 3\\ 3& 4\end{array} \right] \\ = 5 \mathrm{C} \\ \mathrm{Hence}, \mathrm{C} = \left[ \begin{array}{cc}1& 3\\ 3& 4\end{array} \right] \end{gathered}$$