Question

The circumference of a circle is divided into 6 parts such that The lengths are in AP if the smallest and largest arcs are in the ratio 1:5, then find the angle substended by the smallest arc.

Answer 1

$\sf\red{\underline{\underline{Answer:}}}$

$\sf{The \ angle \ substended \ by \ smallest \ arc}$

$\sf{is \ 20^\circ.}$

$\sf{\longmapsto{The \ circumference \ of \ a \ circle \ is}}$

$\sf{divided \ into \ 6 \ parts \ such \ that \ the \ lengths}$

$\sf{are \ in \ AP.}$

$\sf{\longmapsto{The \ smallest \ arc \ and \ the \ largest \ arc}}$

$\sf{are \ in \ the \ ratio \ of \ 1:5}$

$\sf\pink{To \ find:}$

$\sf{The \ angle \ substended \ by \ the \ smallest \ arc.}$

$\sf\green{\underline{\underline{Solution:}}}$

$\sf{Let \ the \ lengths \ of \ 6 \ arcs \ be}$

$\sf{a-5d, \ a-3d, \ a-d, \ a+d, \ a+3d \ and \ a+5d}$

$\sf{We, \ know \ sum \ of \ 6 \ parts=Circumference}$

$\sf{\mapsto{(a-5d)+(a-3d)+(a-d)+(a+d)+(a+3d)+(a+5d)=2\pi \times \ r}}$

$\sf{\mapsto{\therefore{6a=2\pi\times \ r}}}$

$\sf{\mapsto{\therefore{a=\dfrac{\pi\times \ r}{3}...(1)}}}$

$\sf{According \ to \ the \ given \ condition. }$

$\sf{\dfrac{a-5d}{a+5d}=\dfrac{1}{5}}$

$\sf{\therefore{5(a-5d)=a+5d}}$

$\sf{But \ a=\dfrac{\pi\times \ r}{3}}$

$\sf{\therefore{\dfrac{5\pi\times \ r}{3}-25d=\dfrac{pi\times \ r}{3}+5d}}$

$\sf{\therefore{30d=\dfrac{5\pi\times \ r}{3}-\dfrac{pi\times \ r}{3}}}$

$\sf{\therefore{30d=\dfrac{4\pi\times \ r}{3}}}$

$\sf{\therefore{d=\dfrac{2\pi\times \ r}{45}}}$

$\sf{The \ smallest \ arc=a-5d}$

$\sf{=\dfrac{\pi\times \ r}{3}-5(\dfrac{2\pi\times \ r}{45})}$

$\sf{=\dfrac{\pi\times \ r}{3}-\dfrac{2\pi\times \ r}{9}}$

$\sf{=\dfrac{3\pi\times \ r-2\pi\times \ r}{9}}$

$\sf{=\dfrac{\pi\times \ r}{9}}$

$\sf{We \ know,}$

$\sf{Length \ of \ arc=\theta^{c}\times \ r}$

$\sf{=\dfrac{\pi}{9}\times \ r}$

$\sf{On \ comparing \ we \ get,}$

$\sf{\theta^{c}=\dfrac{\pi}{9}}$

$\sf{\theta^{c}=(\theta\times\dfrac{180}{\pi})^\circ}$

$\sf{=\dfrac{\pi}{9}\times\dfrac{180}{\pi}}$

$\sf{=\dfrac{180}{9}}$

$\sf{=20^\circ}$

$\sf\purple{\tt{\therefore{The \ angle \ substended \ by \ smallest \ arc}}}$

$\sf\purple{\tt{is \ 20^\circ.}}$