Question

the angles of a triangle are in AP and the greatest angle is double the least. find all the angles in degrees and radians​

Answer 1

$\huge\bf\red{\underline{\underline{Given}}}\::$

• $\sf\gray{The\: angles\: of\: a\: triangle\: are \:in\: AP}$

• $\sf\gray{The\: greatest \:angle\: is\: double\: the\: least}$

$\huge\bf\red{\underline{\underline{To\:Find}}}\::$

• $\sf\gray{All\: the\: angles \:in \:degrees\: \& \:radians}$

$\huge\bf\red{\underline{\underline{Solution}}}\::$

$\star\:\bf\pink{Let\:} \begin{cases}\sf\green{The \: least \: angle\: =\:x^{\circ}}\\ \sf\purple{The \: greatest \:angle\: be\:(2x)^{\circ}}\end{cases}$

$\to\:\:\sf\blue{x^{\circ},\:y^{\circ},\:(2x)^{\circ}}$

$\to\:\:\sf\purple{y\:-\:x\:=\:2x\:-\:y}$

$\to\:\:\sf\blue{y\:+\:y\:=\:2x\:+\:x}$

$\to\:\:\sf\purple{2y\:=\:3x}$

$\to\:\:{\underline{\boxed{\sf{\blue{y\:=\:\bigg( \dfrac{3x}{2} \bigg)^{\circ} }}}}}$

$\star\:\sf\underline\red{Sum\:of\:angles\:of\:a\: triangle\:=\:180^{\circ}}$

$\mapsto\:\:\sf\orange{x\:+\:\dfrac{3x}{2}\:+\:2x\:=\:180}$

$\mapsto\:\:\sf\blue{2x\:+\:3x\:4x\:=\:2\:\times\:180}$

$\mapsto\:\:\sf\orange{9x\:=\:360^{\circ}}$

$\mapsto\:\:\sf\blue{x\:=\:\dfrac{\cancel{360}}{\cancel{9}}}$

$\mapsto\:\:{\underline{\boxed{\sf{\orange{x\:=\:40^{\circ} }}}}}$

$\mapsto\:\:\sf\blue{40\:\times\:\dfrac{\pi}{180}}$

$\mapsto\:\:\sf\orange{\dfrac{2\pi}{9}\:(radian)}$

$\star\:\sf\underline\red{2nd\:angle\:y\:=\:\dfrac{3x}{2}}$

$\hookrightarrow\:\:\sf\pink{\dfrac{3\:\times\:\cancel{40}}{\cancel{2}}}$

$\hookrightarrow\:\:\sf\green{3\:\times\:20}$

$\hookrightarrow\:\:\sf\pink{60^{\circ}}$

$\hookrightarrow\:\:\sf\green{60\:\times\:\dfrac{\pi}{180}}$

$\hookrightarrow\:\:\sf\pink{\dfrac{\pi}{3}\:(radian)}$

$\star\:\sf\underline\red{Largest\:angle\:=\:2x}$

$\Rightarrow\:\:\sf\purple{2\:\times\:40\:=\:80^{\circ}}$

$\Rightarrow\:\:\sf\green{80\:\times\:\dfrac{\pi}{180}}$

$\Rightarrow\:\:\sf\purple{\dfrac{4\pi}{9}\:(radian)}$

Answer 2

$\pink{\large{\underline{\underline{ \rm{Given: }}}}}$

◉ The angles are in AP.

◉ The greatest angle is double the least.

$\pink{\large{\underline{\underline{ \rm{To \: Find: }}}}}$

◉ All the angles in degrees and radians.

$\pink{\large{\underline{\underline{ \rm{Let's \: Find \: Out: }}}}}$

As, the required angles are in AP.

So let the required angles be ( a - d )°, a°, ( a + d )°.

We know, the sum of angles of a triangle is 180°.

So on adding all the angles, we have:

➯ $\sf(a - d) + a +(a + d) = 180 \degree$

➯ $\sf a - d + a+ a + d = 180 \degree$

➯ $\sf3a = 180 \degree$

➯ $\sf{a = \dfrac{180 \degree}{3}}$

➯ $\sf{a = 60 \degree}$ .......①

It is given that the greatest angle is double the least that means the greatest angle is twice the least.

We have:

➩ $\sf{a + d = 2(a - d)}$

➩ $\sf{a + d = 2a - 2d}$

➩ $\sf{d + 2d = 2a - a}$

➩ $\sf3d = a$

We know a = 60° from ①

Substitute the value of a, we have:

➩ $\sf3d = 60 \degree$

➩ $\sf{d = \dfrac{60 \degree}{3} }$

➩ $\sf{d = 20 \degree}$

Put this value in angles, we get:-

✷ $\sf{(a - d) \degree = 60 \degree - 20 \degree}$

${ \sf{(a -d ) \degree = 40 \degree}}$

✷ $\sf{a = 60 \degree}$

✷ ${ \sf{(a + d) \degree = 60 \degree + 20 \degree}}$

${ \sf{(a + d) \degree = 80 \degree}}$

Hence, the angles are 40°, 60°, 80° ( in degrees ).

Angles in radians :-

❍ $\sf{(40 \times { \dfrac{\pi}{180}) }^{c} }$

$\sf= ({ \dfrac{2\pi}{9}) }^{c} \: or \: \dfrac{2\pi}{9} rad$

❍ $\sf{(60 \times { \dfrac{\pi}{180})}^{c}}$

$\sf = ({\dfrac{\pi}{3})}^{c} \: or \: \dfrac{\pi}{3} rad$

❍ $\sf{80 \times { \dfrac{\pi}{180} }^{c} }$

$\sf = ({ \dfrac{4\pi}{9})}^{c} \: or \: \dfrac{4\pi}{9} rad$

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