C-16-Q094mediumsingle_mcq

The half-angle at a vertex of the isosceles triangle formed by joining the centre of a regular $n$ -gon to two adjacent vertices is —

a $\dfrac{180°}{n}$
b $90° - \dfrac{180°}{n}$
c $\dfrac{360°}{n}$
d $90° - \dfrac{90°}{n}$

ব্যাখ্যা

The central angle of each isosceles triangle is $\dfrac{360°}{n}$ , so the two equal base angles sum to $180°-\dfrac{360°}{n}$ . Each base angle (the half-angle at the polygon's vertex) is therefore $90°-\dfrac{180°}{n}$ .

এই অধ্যায়ের আরও প্রশ্ন

Chapter 16: Mensuration — সব প্রশ্ন দেখুন (210টি) →

The half-angle at a vertex of the isosceles triangle formed by joining the centre of a regular $n$ -gon to two adjacent vertices is —

\dfrac{180°}{n}

90° - \dfrac{180°}{n}

\dfrac{360°}{n}

90° - \dfrac{90°}{n}

The half-angle at a vertex of the isosceles triangle formed by joining the centre of a regular nnn-gon to two adjacent vertices is —

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

The half-angle at a vertex of the isosceles triangle formed by joining the centre of a regular nnn-gon to two adjacent vertices is —

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

The half-angle at a vertex of the isosceles triangle formed by joining the centre of a regular $n$ -gon to two adjacent vertices is —

The half-angle at a vertex of the isosceles triangle formed by joining the centre of a regular $n$ -gon to two adjacent vertices is —