Find the measure of each exterior angle of a regular polygon of
(i) 9 sides (ii) 15 sides
Sum of the interior angles of a polygon having $n$ sides = $(n - 2) \times 180^\circ$.
(i) For $n = 9$:
Sum of interior angles = $(9 - 2) \times 180^\circ = 7 \times 180^\circ = 1260^\circ$.
Each interior angle = $\dfrac{1260^\circ}{9} = 140^\circ$.
Each exterior angle = $180^\circ - 140^\circ = 40^\circ$.
(Alternate short method
Each exterior angle = $\dfrac{360^\circ}{9} = 40^\circ$.
Answer (i): $\boxed{40^\circ}$
(ii) For $n = 15$:
Sum of interior angles = $(15 - 2) \times 180^\circ = 13 \times 180^\circ = 2340^\circ$.
Each interior angle = $\dfrac{2340^\circ}{15} = 156^\circ$.
Each exterior angle = $180^\circ - 156^\circ = 24^\circ$.
(Alternate short method
Each exterior angle = $\dfrac{360^\circ}{15} = 24^\circ$.
Answer (ii): $\boxed{24^\circ}$
नियमित बहुभुज के प्रत्येक बाह्य कोण का मान ज्ञात कीजिए —
(क) 9 भुजाएँ (ख) 15 भुजाएँ
$n$ भुजाओं वाले बहुभुज के आंतरिक कोणों का योग = $(n - 2) \times 180^\circ$।
(क) जब $n = 9$:
आंतरिक कोणों का योग = $(9 - 2) \times 180^\circ = 7 \times 180^\circ = 1260^\circ$
प्रत्येक आंतरिक कोण = $\dfrac{1260^\circ}{9} = 140^\circ$
प्रत्येक बाह्य कोण = $180^\circ - 140^\circ = 40^\circ$
(वैकल्पिक छोटा तरीका
प्रत्येक बाह्य कोण = $\dfrac{360^\circ}{9} = 40^\circ$
उत्तर (क): $\boxed{40^\circ}$
(ख) जब $n = 15$:
आंतरिक कोणों का योग = $(15 - 2) \times 180^\circ = 13 \times 180^\circ = 2340^\circ$
प्रत्येक आंतरिक कोण = $\dfrac{2340^\circ}{15} = 156^\circ$
प्रत्येक बाह्य कोण = $180^\circ - 156^\circ = 24^\circ$
(वैकल्पिक छोटा तरीका
प्रत्येक बाह्य कोण = $\dfrac{360^\circ}{15} = 24^\circ$
उत्तर (ख): $\boxed{24^\circ}$