Elements of Mathematics Solutions for Class 11th ( Part - 1 ) - Unbox Goodies

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Solutions Start From Here ( Part – 1 )

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Solutions Start From Here ( Part – 1 )

Question. In which quadrants do the following angles lie ?
(i)750° (ii) – 870°

Solution. (i) 750° = 2 x 360° + 30°

Now, 360° = 1 complete revolution.

Thus the revolving line, starting from OX, makes two complete
revolutions in the positive direction and further traces out an angle of 30° in the same direction.

Hence the revolving line lies in the first quadrant.

(ii) — 870° = – 2 x 360° – 150°

Thus the revolving line, starting from OX, makes two complete
revolutions in the negative direction and moves further through angle of 150° in the same direction. Hence, the revolving line lies in the third quadrant.

Question. Express in radians the following angles:

(i) 45° (ii) 530° (iii) 40° 20°
(iv) 104° 36′ (v) – 37° 30′ (vi) 15° 15’ 15”

Solution. (i) ${{45}^{\circ }}=45\times \frac{\pi }{{180}}$ radians $=\frac{\pi }{4}$ radians.

$\left[ {\because {{1}^{\circ }}=\frac{\pi }{{180}}\text{ }\!\!~\!\!\text{ radians }\!\!~\!\!\text{ }} \right]$

(ii)

${{530}^{\circ }}=530\times \frac{\pi }{{180}}$ radians $=\frac{{53\pi }}{{18}}$ radians.

(iii) ${{20}^{\text{ }\!\!'\!\!\text{ }}}$ $={{\left( {\frac{{20}}{{60}}} \right)}^{\circ }}={{\left( {\frac{1}{3}} \right)}^{\circ }}$ $\left[ {\because {{{60}}^{\text{ }\!\!'\!\!\text{ }}}={{1}^{\circ }}} \right]$ ${{40}^{\circ }}{{20}^{\text{ }\!\!'\!\!\text{ }}}={{\left( {40\frac{1}{3}} \right)}^{\circ }}$ $={{\left( {\frac{{121}}{3}} \right)}^{\circ }}$ $=\frac{{121}}{3}\times \frac{\pi }{{180}}\text{ }\!\!~\!\!\text{ radians}$ $\text{=}\frac{{121\pi }}{{540}}\text{ }\!\!~\!\!\text{ radians}$

(iv)

${{36}^{\text{ }\!\!'\!\!\text{ }}}={{\left( {\frac{{36}}{{60}}} \right)}^{\circ }}={{\left( {\frac{3}{5}} \right)}^{\circ }}$

$\left[ {\because {{{60}}^{\text{ }\!\!'\!\!\text{ }}}={{1}^{\circ }}} \right]$

${{104}^{\circ }}{{36}^{\text{ }\!\!'\!\!\text{ }}}={{\left( {104\frac{3}{5}} \right)}^{\circ }}$ $={{\left( {\frac{{523}}{5}} \right)}^{\circ }}$

$=\frac{{523}}{5}\times \frac{\pi }{{180}}\text{ }\!\!~\!\!\text{ radians}$

$\left[ {\because {{1}^{\circ }}=\frac{\pi }{{180}}\text{ }\!\!~\!\!\text{ radians }\!\!~\!\!\text{ }} \right]$

$=\frac{{523}}{{900}}\pi \text{ }\!\!~\!\!\text{ radians}\text{. }\!\!~\!\!\text{ }$

(v)

${{30}^{\text{ }\!\!'\!\!\text{ }}}={{\left( {\frac{{30}}{{60}}} \right)}^{\circ }}={{\left( {\frac{1}{2}} \right)}^{\circ }}$

$\left[ {\because {{{60}}^{\text{ }\!\!'\!\!\text{ }}}={{1}^{\circ }}} \right]$ $\begin{array}{*{20}{c}} {} \\ {-{{{37}}^{\circ }}{{{30}}^{\text{ }\!\!'\!\!\text{ }}}=\text{ }\!\!~\!\!\text{ }-{{{\left( {37\frac{1}{2}} \right)}}^{0}}=\text{ }\!\!~\!\!\text{ }-{{{\left( {\frac{{75}}{2}} \right)}}^{\circ }}} \end{array}$

$=\frac{{-75}}{2}\times \frac{\pi }{{180}}\text{ }\!\!~\!\!\text{ radians}$ $\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ = }\!\!~\!\!\text{ -}\frac{{5\pi }}{{24}}\text{ }\!\!~\!\!\text{ radians}\text{. }\!\!~\!\!\text{ }$

(vi)

${{15}^{{\text{ }\!\!'\!\!\text{ }\!\!'\!\!\text{ }}}}={{\left( {\frac{{15}}{{60}}} \right)}^{\text{ }\!\!'\!\!\text{ }}}=\text{ }\!\!~\!\!\text{ }{{\left( {\frac{1}{4}} \right)}^{\text{ }\!\!'\!\!\text{ }}}$

$\left[ {\because {{{60}}^{{\text{ }\!\!'\!\!\text{ }\!\!'\!\!\text{ }}}}={1}'} \right]$

${{15}^{\text{ }\!\!'\!\!\text{ }}}{{15}^{{\text{ }\!\!'\!\!\text{ }\!\!'\!\!\text{ }}}}=\text{ }\!\!~\!\!\text{ }{{\left( {15\frac{1}{4}} \right)}^{\text{ }\!\!'\!\!\text{ }}}~$

$\begin{array}{*{20}{r}} {} & ~ \\ {} & {={{{\left( {\frac{{61}}{4}} \right)}}^{\text{ }\!\!'\!\!\text{ }}}=\text{ }\!\!~\!\!\text{ }{{{\left( {\frac{{61}}{4}\times \frac{1}{{60}}} \right)}}^{0}}} \\ {} & ~ \end{array}$

$\left[ {\because {{{60}}^{\text{ }\!\!'\!\!\text{ }}}={{1}^{\circ }}} \right]$

$={{\left( {\frac{{61}}{{240}}} \right)}^{\circ }}$

${{15}^{\circ }}{{15}^{\text{ }\!\!'\!\!\text{ }}}{{15}^{{\text{ }\!\!'\!\!\text{ }\!\!'\!\!\text{ }}}}={{\left( {15\frac{{61}}{{240}}} \right)}^{\circ }}$

$={{\left( {\frac{{3661}}{{240}}} \right)}^{\circ }}$

$=\frac{{3661}}{{240}}\times \frac{\pi }{{180}}\text{ }\!\!~\!\!\text{ radians}$

$\left[ {\because {{1}^{\circ }}=\frac{\pi }{{180}}\text{ }\!\!~\!\!\text{ radians }\!\!~\!\!\text{ }} \right]$

$=\frac{{3661\pi }}{{43200}}\text{ }\!\!~\!\!\text{ radians}$

Question. Find the degree measures corresponding to the following radian measures :

(i) ${{\left( {\frac{\pi }{8}} \right)}^{\text{e}}}$

(ii) ${{\left( {\frac{{7\pi }}{{12}}} \right)}^{e}}$

(iii) $\text{ }\!\!~\!\!\text{ }{{(-2)}^{\text{e}}}$

(iv) ${{\left( {\frac{3}{4}} \right)}^{e}}$

(v) ${{(6)}^{\text{e}}}$

Solution. Since $\pi$ radians $={{180}^{\circ }}$ , therefore

(i) $\begin{array}{*{20}{r}} {\left( {\frac{\pi }{8}} \right)\text{ }\!\!~\!\!\text{ radians }\!\!~\!\!\text{ }} & {} \\ {} & {} \end{array}$ $~={{\left( {\frac{\pi }{8}\times \frac{{180}}{\pi }} \right)}^{\circ }}={{\left( {\frac{{180}}{8}} \right)}^{\circ }}$

$={{\left( {\frac{{45}}{2}} \right)}^{\circ }}={{\left( {22\frac{1}{2}} \right)}^{\circ }}$ $~={{22}^{\circ }}{{30}^{\text{ }\!\!'\!\!\text{ }}}$

(ii) $\left( {\frac{{7\pi }}{{12}}} \right)$ radians $={{\left( {\frac{{7\pi }}{{12}}\times \frac{{180}}{\pi }} \right)}^{\circ }}={{105}^{\circ }}$ .

(iii) $\left( {-2} \right)\text{ }\!\!~\!\!\text{ radians }\!\!~\!\!\text{ }$ $=$ $-{{\left( {\frac{{2\times 180\times 7}}{{22}}} \right)}^{\circ }}$

$~=\text{ }\!\!~\!\!\text{ }-{{\left( {\frac{{1260}}{{11}}} \right)}^{\circ }}=\text{ }\!\!~\!\!\text{ }-{{\left( {114\frac{6}{{11}}} \right)}^{\circ }}$

$=-\left[ {{{{114}}^{\circ }}{{{\left( {\frac{{6\times 60}}{{11}}} \right)}}^{\text{ }\!\!'\!\!\text{ }}}} \right]=-\left[ {{{{114}}^{\circ }}{{{\left( {32\frac{8}{{11}}} \right)}}^{\text{ }\!\!'\!\!\text{ }}}} \right]$

$=-{{114}^{\circ }}{{32}^{\text{ }\!\!'\!\!\text{ }}}{{\left( {\frac{{8\times 60}}{{11}}} \right)}^{{\text{ }\!\!'\!\!\text{ }\!\!'\!\!\text{ }}}}$ $=-{{114}^{\circ }}{{32}^{\text{ }\!\!'\!\!\text{ }}}{{44}^{{\text{ }\!\!'\!\!\text{ }\!\!'\!\!\text{ }}}}$ (approx.)

(iv)

$\left( {\frac{3}{4}} \right)\text{ }\!\!~\!\!\text{ radians }\!\!~\!\!\text{ }$ $~={{\left( {\frac{3}{4}\times \frac{{180}}{\pi }} \right)}^{\circ }}={{\left( {\frac{{135}}{\pi }} \right)}^{\circ }}$

$={{\left( {\frac{{135\times 7}}{{22}}} \right)}^{0}}={{\left( {\frac{{945}}{{22}}} \right)}^{\circ }}$

$={{\left( {42\frac{{21}}{{22}}} \right)}^{\circ }}={{42}^{\circ }}+{{\left( {\frac{{21\times 60}}{{22}}} \right)}^{\text{ }\!\!'\!\!\text{ }}}$

$={{42}^{\circ }}+{{57}^{\text{ }\!\!'\!\!\text{ }}}+{{\left( {\frac{{3\times 60}}{{11}}} \right)}^{{\text{ }\!\!'\!\!\text{ }\!\!'\!\!\text{ }}}}={{42}^{\circ }}{{57}^{\text{ }\!\!'\!\!\text{ }}}{{17}^{{\text{ }\!\!'\!\!\text{ }\!\!'\!\!\text{ }}}}\left( {\text{ }\!\!~\!\!\text{ approx }\!\!~\!\!\text{ }} \right)$

(v)

$6\text{ }\!\!~\!\!\text{ radians }\!\!~\!\!\text{ }={{\left( {6\times \frac{{180}}{\pi }} \right)}^{\circ }}$

$={{\left( {\frac{{1080\times 7}}{{22}}} \right)}^{\circ }}={{\left( {343\frac{7}{{11}}} \right)}^{\circ }}$

$={{343}^{\circ }}{{\left( {\frac{{7\times 60}}{{11}}} \right)}^{\text{ }\!\!'\!\!\text{ }}}\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }$

$\text{ }\!\!~\!\!\text{ }\left[ {\because {{1}^{\circ }}={{{60}}^{\text{ }\!\!'\!\!\text{ }}}} \right]$

$={{343}^{\circ }}{{\left( {38\frac{2}{{11}}} \right)}^{\text{ }\!\!'\!\!\text{ }}}={{343}^{\circ }}{{38}^{\text{ }\!\!'\!\!\text{ }}}{{\left( {\frac{2}{{11}}} \right)}^{\text{ }\!\!'\!\!\text{ }}}$

$={{343}^{\circ }}{{38}^{\text{ }\!\!'\!\!\text{ }}}{{\left( {\frac{{2\times 60}}{{11}}} \right)}^{{\text{ }\!\!'\!\!\text{ }\!\!'\!\!\text{ }}}}$ $\left[ {\because {1}'={{{60}}^{{\text{ }\!\!'\!\!\text{ }\!\!'\!\!\text{ }}}}} \right]$

$={{343}^{\circ }}{{38}^{\text{ }\!\!'\!\!\text{ }}}{{11}^{{\text{ }\!\!'\!\!\text{ }\!\!'\!\!\text{ }}}}\text{ }\!\!~\!\!\text{ (approx}\text{.) }\!\!~\!\!\text{ }$

Question. (i) If $l=25\text{ }\!\!~\!\!\text{ cm},r=\frac{1}{2}\text{ }\!\!~\!\!\text{ m}$ , then find the value of $\theta$ .

(ii) Find the radius of the circle in which a central angle of ${{60}^{\circ }}$ intercepts an arc of $37.4\text{ }\!\!~\!\!\text{ cm}$ (Use $\left. {\pi =22/7} \right)$ .
[NCERT]

(iii) Find the length of an arc of a circle of radius $10\text{ }\!\!~\!\!\text{ cm}$ which subtends an angle of ${{45}^{\circ }}$ at the centre.

Solution. (i) Here $l=25\text{ }\!\!~\!\!\text{ cm}$ and $r=\frac{1}{2}\text{ }\!\!~\!\!\text{ m=50 }\!\!~\!\!\text{ cm}$ $\theta =\frac{l}{r}=\frac{{25}}{{50}}=\frac{1}{2}\text{ }\!\!~\!\!\text{ radians}\text{. }\!\!~\!\!\text{ }$

(ii) Let $r$ be the radius of the circle

Here $l=37.4\text{ }\!\!~\!\!\text{ cm}.,\text{ }\!\!~\!\!\text{ }~~~~~\theta ={{60}^{\circ }}=60\times \frac{\pi }{{180}}$ radian $=\frac{\pi }{3}$ radian
Now,

$r=\frac{l}{\theta }=\frac{{37\cdot 4}}{{\frac{\pi }{3}}}$ $=\frac{{374\times 3\times 7}}{{220}}=35\cdot 7\text{ }\!\!~\!\!\text{ cm}$

(iii) Here $\theta ={{45}^{\circ }},\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }l=?,\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }r=10\text{ }\!\!~\!\!\text{ cm}$
Now,

$~\theta ~=\frac{l}{r}\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\Rightarrow \text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }{{45}^{\circ }}=\frac{l}{{10}}$ $\begin{array}{*{20}{r}} {} & {} \\ l & {~={{{450}}^{\circ }}\times \frac{\pi }{{180}}=\frac{{5\pi }}{2}\text{ }\!\!~\!\!\text{ cm}\text{.}} \end{array}$

Question. The difference between two acute angles of a rt. angled triangle is $\frac{\pi }{9}$ radians. Find the angles in degrees.

Solution. The sum of two acute angles of the $\text{rt}$ . angled triangle $={{90}^{\circ }}$ . Let one acute angle be ${{x}^{\circ }}$

Then the other angle is ${{(90-x)}^{\circ }}$

Now, difference of these two angles $=\frac{\pi }{9}$ radians

$={{\left( {\frac{\pi }{9}\times \frac{{180}}{\pi }} \right)}^{\circ }}={{20}^{\circ }}$ $\therefore \text{ }\!\!~\!\!\text{ }{{(90-x)}^{\circ }}-{{x}^{\circ }}={{20}^{\circ }}$ ${{90}^{\circ }}-2x={{20}^{\circ }}\Rightarrow 2x={{70}^{\circ }}\Rightarrow x={{35}^{\circ }}$ $\therefore \text{ }\!\!~\!\!\text{ }$ Other angle $={{90}^{\circ }}-{{35}^{\circ }}={{55}^{\circ }}$ .

Hence the two angles are ${{35}^{\circ }}$ and ${{55}^{\circ }}$ .

Question. (i) Express both in degrees and radians, the angles of a triangle, whose angles to each other are in the ratio $1:2:3$ .

(ii) The angles of a triangle are in A.P., the greatest of them being ${{80}^{\circ }}$ ; find all the three angles in radians.

(iii) The angles of a triangle are in A.P. If one of them is ${{36}^{\circ }}$ , find all angles in radians.

Solution. (i) Let the angles of the triangle be ${{x}^{\circ }},2{{x}^{\circ }}$ and $3{{x}^{\circ }}$
[Being in the ratio $1:2:3$ ] $\therefore \text{ }\!\!~\!\!\text{ }x+2x+3x={{180}^{\circ }}$ $\begin{array}{*{20}{r}} {} & ~ \\ {\Rightarrow \text{ }\!\!~\!\!\text{ }6x} & {~={{{180}}^{\circ }}\Rightarrow \text{ }\!\!~\!\!\text{ }x={{{30}}^{\circ }}} \end{array}$ $\therefore$ Angles in degree are ${{30}^{\circ }},{{60}^{\circ }},{{90}^{\circ }}$

e., $\text{ }\!\!~\!\!\text{ }30\times \frac{\pi }{{180}},\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }60\times \frac{\pi }{{180}},\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }90\times \frac{\pi }{{180}}$ radians.

i.e., $\frac{\pi }{6},\frac{\pi }{3},\frac{\pi }{2}$ radians.

(ii) Let the angles of triangle be ${{(a-d)}^{\circ }},{{a}^{\circ }},{{(a+d)}^{\circ }}$ $\therefore \left( {a-d} \right)+a+a+d={{180}^{\circ }}\text{ }\!\!~\!\!\text{ }$ [Sum of angles of a triangle $={{180}^{\circ }}$ ] $3a={{180}^{\circ }}\Rightarrow a={{60}^{\circ }}$

Also, greatest of all three angles $=a+d={{80}^{\circ }}$ $\therefore \text{ }\!\!~\!\!\text{ }$ Third angle $={{180}^{\circ }}-\left( {{{{80}}^{\circ }}+{{{60}}^{\circ }}} \right)={{40}^{\circ }}.\text{ }\!\!~\!\!\text{ }\left[ \because \right.$ One angle $=a={{60}^{\circ }}]$ $\therefore$ Angles are ${{40}^{\circ }},{{60}^{\circ }},{{80}^{\circ }}$ i.e.,

$40\times \frac{\pi }{{180}},60\times \frac{\pi }{{180}},80\times \frac{\pi }{{180}}$ radians,

e., $\frac{{2\pi }}{9},\frac{\pi }{3}$ and $\frac{{4\pi }}{9}$ radians.

(iii) Let the angles be ${{(a-d)}^{\circ }},\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }{{a}^{\circ }},\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }{{(a+d)}^{\circ }}$ $\therefore \left( {a-d} \right)+a+\left( {a+d} \right)\text{ }\!\!~\!\!\text{ }={{180}^{\circ }}$ $\text{ }\!\!~\!\!\text{ }\Rightarrow \text{ }\!\!~\!\!\text{ }3a={{180}^{\circ }}\text{ }\!\!~\!\!\text{ }\Rightarrow \text{ }\!\!~\!\!\text{ }a={{60}^{\circ }}$

Also one of the angle is ${{36}^{\circ }}$
[Given] $\therefore$ The third angle is ${{180}^{\circ }}-\left( {{{{60}}^{\circ }}+{{{36}}^{\circ }}} \right)={{84}^{\circ }}$

Hence the three angles are ${{36}^{\circ }},{{60}^{\circ }}$ and ${{84}^{\circ }}$

i.e., $\frac{\pi }{5},\frac{\pi }{3}$ and $\frac{{7\pi }}{{15}}$ radians.

[ $\because {{1}^{\circ }}=\frac{\pi }{{180}}$ radians $]$

Question. The angles of a quadrilateral are in A.P. and the greatest is double the least. Find the least angle in radians.

Solution. Let the angles of the quadrilateral be

${{(a-3d)}^{\circ }},{{(a-d)}^{\circ }},{{(a+d)}^{\circ }},{{(a+3d)}^{\circ }}$

Now the sum of the angles of the quadrilateral $={{360}^{\circ }}$ $\therefore \text{ }\!\!~\!\!\text{ }\left( {a-3d} \right)+\left( {a-d} \right)+\left( {a+d} \right)+\left( {a+3d} \right)={{360}^{\circ }}$ $4a={{360}^{\circ }}\Rightarrow a={{90}^{\circ }}$ $\therefore$ The angles are ${{(90-3d)}^{\circ }},{{(90-d)}^{\circ }},{{(90+d)}^{\circ }},{{(90+3d)}^{\circ }}$

Greatest angle of the quadrilateral $={{(90+3d)}^{\circ }}$ .

and $\text{ }\!\!~\!\!\text{ least }\!\!~\!\!\text{ angle }\!\!~\!\!\text{ =(90-3}d{{)}^{\circ }}$

Now, $\text{ }\!\!~\!\!\text{ }$ least angle $=\frac{1}{2}\times$ greatest angle
[Given]

${{(90-3d)}^{\circ }}=\frac{1}{2}{{(90+3d)}^{\circ }}$ $\Rightarrow {{180}^{\circ }}-6d={{90}^{\circ }}+3d$ $\Rightarrow \text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }9d={{90}^{\circ }}\Rightarrow d=10.$ $\therefore$ Least angle $={{(90-3d)}^{\circ }}={{(90-30)}^{\circ }}$ $={{60}^{\circ }}=60\times \frac{\pi }{{180}}=\frac{\pi }{3}$ radians.

Question. (i) Find the magnitude, in radians and degrees, of the interior angle of a regular pentagon.

(ii) Express in radians as well as in degrees the angle of a regular polygon of 40 sides.

Solution. (i) Number of sides in a regular pentagon $=5$ $\therefore$ Number of exterior angles in a regular pentagon $=5$

Sum of exterior angles of a regular pentagon $={{360}^{\circ }}$ $\therefore$ Each exterior angle $=\frac{{{{{360}}^{\circ }}}}{5}={{72}^{\circ }}\text{ }\!\!~\!\!\text{ }$ [See footnote on next page] $\therefore$ Each interior angle $={{180}^{\circ }}-{{72}^{\circ }}={{108}^{\circ }}$ $=\left( {108\times \frac{\pi }{{180}}} \right)\text{ }\!\!~\!\!\text{ radians }\!\!~\!\!\text{ =}\left( {\frac{{3\pi }}{5}} \right)\text{ }\!\!~\!\!\text{ radians }\!\!~\!\!\text{ }$

(ii) The sum of 40 exterior angles of a regular polygon $={{360}^{\circ }}$ $\therefore \text{ }\!\!~\!\!\text{ }$ Each exterior angle $=\frac{{{{{360}}^{\circ }}}}{{40}}={{9}^{\circ }}$ $\therefore \text{ }\!\!~\!\!\text{ }$ Each interior angle $={{180}^{\circ }}-{{9}^{\circ }}={{171}^{\circ }}$ $=\left( {117\times \frac{\pi }{{180}}} \right)\text{ }\!\!~\!\!\text{ radians }\!\!~\!\!\text{ =}\left( {\frac{{19\pi }}{{20}}} \right)\text{ }\!\!~\!\!\text{ radians}\text{. }\!\!~\!\!\text{ }$

Question. The number of sides of two regular polygons are $5:4$ and the difference between their angles is ${{9}^{\circ }}$ . Find the number of sides in the polygons.

Solution. Let the number of sides of two regular polygons be $5x$ and $4x$ .

$\therefore \text{ }\!\!~\!\!\text{ }$ Each exterior angle of first polygon $={{\left( {\frac{{360}}{{5x}}} \right)}^{\circ }}$

and each exterior angle of second polygon $={{\left( {\frac{{360}}{{4x}}} \right)}^{\circ }}$ $\therefore \text{ }\!\!~\!\!\text{ }$ Each interior angle of first polygon $={{180}^{\circ }}-{{\left( {\frac{{360}}{{5x}}} \right)}^{\circ }}$

and $\text{ }\!\!~\!\!\text{ }$ each interior angle of second polygon $={{180}^{\circ }}-{{\left( {\frac{{360}}{{4x}}} \right)}^{\circ }}$

According to the given condition, we have

$\left[ {{{{180}}^{\circ }}-{{{\left( {\frac{{360}}{{5x}}} \right)}}^{\circ }}} \right]-\left[ {{{{180}}^{\circ }}-{{{\left( {\frac{{360}}{{4x}}} \right)}}^{\circ }}} \right]={{9}^{\circ }}$

$\begin{array}{*{20}{r}} {} & {} \\ {} & {~\Rightarrow {{{\left( {\frac{{360}}{{4x}}} \right)}}^{\circ }}-{{{\left( {\frac{{360}}{{5x}}} \right)}}^{\circ }}={{9}^{\circ }}} \end{array}$