UG Element Math Solutions Part 2

Elements of Mathematics Solutions for Class 11th ( Part – 2 ) – Unbox Goodies

Class 11thElements of Math Solutions

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Solutions Start From Here ( Part – 2 )

Question. In which quadrants do the following angles lie ?

  1. 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.

  1. — 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] \begin{array}{*{20}{r}} {} & {~\text{ }\!\!~\!\!\text{ }} \\ {{{{40}}^{\circ }}{{{20}}^{\text{ }\!\!'\!\!\text{ }}}} & {~={{{\left( {40\frac{1}{3}} \right)}}^{\circ }}={{{\left( {\frac{{121}}{3}} \right)}}^{\circ }}} \\ {} & \text{ }\!\!~\!\!\text{ } \\ {} & ~ \\ {} & {} \end{array} =\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{ }

(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] \begin{array}{*{20}{r}} {} & ~ \\ {{{{15}}^{\text{ }\!\!'\!\!\text{ }}}{{{15}}^{{\text{ }\!\!'\!\!\text{ }\!\!'\!\!\text{ }}}}} & {=\text{ }\!\!~\!\!\text{ }{{{\left( {15\frac{1}{4}} \right)}}^{\text{ }\!\!'\!\!\text{ }}}={{{\left( {\frac{{61}}{4}} \right)}}^{\text{ }\!\!'\!\!\text{ }}}} \\ {} & ~ \end{array} =\text{ }\!\!~\!\!\text{ }{{\left( {\frac{{61}}{4}\times \frac{1}{{60}}} \right)}^{0}} \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 }} \begin{array}{*{20}{r}} {} & ~ \\ {} & {~=\frac{{3661}}{{240}}\times \frac{\pi }{{180}}\text{ }\!\!~\!\!\text{ radians }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }} \\ {} & {} \end{array}

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

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

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{ }} & {~={{{\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 }}} \\ {} & {} \end{array} ~={{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 }} \begin{array}{*{20}{r}} {} & ~ \\ {} & {~=\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]} \end{array} =-{{114}^{\circ }}{{32}^{\text{ }\!\!'\!\!\text{ }}}{{\left( {\frac{{8\times 60}}{{11}}} \right)}^{{\text{ }\!\!'\!\!\text{ }\!\!'\!\!\text{ }}}} =-{{114}^{\circ }}{{32}^{\text{ }\!\!'\!\!\text{ }}}{{44}^{{\text{ }\!\!'\!\!\text{ }\!\!'\!\!\text{ }}}}

(iv)

\left( {\frac{3}{4}} \right)\text{ }\!\!~\!\!\text{ radians}={{\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 }}\text{ }\!\!~\!\!\text{ } ={{\left( {42\frac{{21}}{{22}}} \right)}^{\circ }}={{42}^{\circ }}+{{\left( {\frac{{21\times 60}}{{22}}} \right)}^{\text{ }\!\!'\!\!\text{ }}} \begin{array}{*{20}{r}} {} & ~ \\ {} & ~ \\ {} & {~={{{42}}^{\circ }}+{{{57}}^{\text{ }\!\!'\!\!\text{ }}}+{{{\left( {\frac{{3\times 60}}{{11}}} \right)}}^{{\text{ }\!\!'\!\!\text{ }\!\!'\!\!\text{ }}}}} \end{array} ={{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{ } \begin{array}{*{20}{r}} {} & ~ \\ {} & {\text{ }\!\!~\!\!\text{ }\left[ {\because {{1}^{\circ }}={{{60}}^{\text{ }\!\!'\!\!\text{ }}}} \right]} \\ {} & ~ \\ {} & ~ \\ {} & ~ \end{array} ={{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 cm (Use \left. {\pi =22/7} \right).

[NCERT]

(iii) Find the length of an arc of a circle of radius 10 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] \begin{array}{*{20}{r}} {\therefore \text{ }\!\!~\!\!\text{ }x+2x+3x} & {~={{{180}}^{\circ }}} \\ {\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 }}

  1. 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 }}. \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,

  1. 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]

  1. e.,
    {{(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

\begin{array}{*{20}{r}} {} & {\left[ {{{{180}}^{\circ }}-{{{\left( {\frac{{360}}{{5x}}} \right)}}^{\circ }}} \right]-\left[ {{{{180}}^{\circ }}-{{{\left( {\frac{{360}}{{4x}}} \right)}}^{\circ }}} \right]={{9}^{\circ }}} \\ {} & {~\Rightarrow {{{\left( {\frac{{360}}{{4x}}} \right)}}^{\circ }}-{{{\left( {\frac{{360}}{{5x}}} \right)}}^{\circ }}={{9}^{\circ }}} \end{array}

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