Question

If the difference between two acute angles of a right angled triangle is (2pi/5)c, find these two angles in radians.
if (2x/3)g, (3x/2)degree, (pix/75)c are the angles of a triangle, find them in degrees.

Answer 1

$\large\underline{\sf{Solution-1}}$

☆ Let assume that the two acute angle in radians measure be x and y and x > y.

We know,

☆ Sum of angles of a triangle is supplementary.

$\rm :\longmapsto\:x + y + \dfrac{\pi}{2} = \pi$

$\rm :\longmapsto\:x + y = \pi - \dfrac{\pi}{2}$

$\bf\implies \:x + y =\dfrac{\pi}{2} - - - - (1)$

Also,

Given that,

$\rm :\longmapsto\:x - y = \dfrac{\pi}{5} - - - - (2)$

☆ On adding equation (1) and equation (2), we get

$\rm :\longmapsto\:2x = \dfrac{\pi}{2} + \dfrac{\pi}{5}$

$\rm :\longmapsto\:2x = \dfrac{5\pi + 2\pi}{10}$

$\rm :\longmapsto\:2x = \dfrac{7\pi}{10}$

$\rm :\longmapsto\:x = \dfrac{7\pi}{20} - - - - (3)$

☆ On substituting the value of x in equation (1) we get

$\rm :\longmapsto\:\dfrac{7\pi}{20} + y = \dfrac{\pi}{2}$

$\rm :\longmapsto\: y = \dfrac{\pi}{2} - \dfrac{7\pi}{20}$

$\rm :\longmapsto\: y = \dfrac{10\pi - 7\pi}{20}$

$\rm :\longmapsto\: y = \dfrac{3\pi}{20}$

Hence,

$\begin{gathered}\begin{gathered}\bf\: Acute \: angles \: are \: -\begin{cases} &\sf{x = \dfrac{7\pi}{20} } \\ \\ &\sf{y = \dfrac{3\pi}{20} } \end{cases}\end{gathered}\end{gathered}$

$\red{\large\underline{\bf{Solution-2}}}$

$\begin{gathered}\begin{gathered}\bf\: Let \: the \: angles \: be-\begin{cases} &\sf{x = {\bigg(\dfrac{2x}{3} \bigg) }^{g} } \\ \\ &\sf{y = {\bigg(\dfrac{3x}{2} \bigg) }^{ \degree} } \\ \\ &\sf{z = \dfrac{\pi \: x}{75} } \end{cases}\end{gathered}\end{gathered}$

We know,

$\rm :\longmapsto\: {1}^{g} = \dfrac{9}{10}^{\degree}$

$\rm :\longmapsto\: {\bigg(\dfrac{2x}{3} \bigg) }^{g} = {\bigg(\dfrac{2x}{3} \times \dfrac{9}{10} \bigg) }^{\degree}$

$\red{\rm :\longmapsto\: {\bigg(\dfrac{2x}{3} \bigg) }^{g} = {\bigg(\dfrac{3x}{5} \bigg) }^{\degree}}$

Also,

we know that

$\rm :\longmapsto\: {1}^{c} = {\bigg(\dfrac{180}{\pi} \bigg) }^{\degree}$

$\rm :\longmapsto\:\dfrac{\pi \: x}{75} = {\bigg(\dfrac{\pi \: x}{75} \times \dfrac{180}{\pi} \bigg) }^{\degree}$

$\red{\rm :\longmapsto\:\dfrac{\pi \: x}{75} = {\bigg(\dfrac{12\: x}{5} \bigg) }^{\degree}}$

$\begin{gathered}\begin{gathered}\bf\:So \: the \: angles \: be-\begin{cases} &\sf{x = {\bigg(\dfrac{2x}{3} \bigg) }^{g} = {\bigg(\dfrac{3\: x}{5} \bigg) }^{\degree}} \\ \\ &\sf{y = {\bigg(\dfrac{3x}{2} \bigg) }^{ \degree} } \\ \\ &\sf{z = \dfrac{\pi \: x}{75} ={\bigg(\dfrac{12\: x}{5} \bigg) }^{\degree} } \end{cases}\end{gathered}\end{gathered}$

We know that,

☆ Sum of interior angles of a triangle is 180°

$\rm :\longmapsto\:x + y + z = 180\degree$

$\rm :\longmapsto\:{\bigg(\dfrac{3\: x}{5} \bigg) }^{\degree} + {\bigg(\dfrac{3\: x}{2} \bigg) }^{\degree} + {\bigg(\dfrac{12\: x}{5} \bigg) }^{\degree} = 180\degree$

$\rm :\longmapsto\:{\bigg(\dfrac{3\: x + 12x}{5} \bigg) }^{\degree} + {\bigg(\dfrac{3\: x}{2} \bigg) }^{\degree} = 180\degree$

$\rm :\longmapsto\:{\bigg(\dfrac{15\: x}{5} \bigg) }^{\degree} + {\bigg(\dfrac{3\: x}{2} \bigg) }^{\degree} = 180\degree$

$\rm :\longmapsto\:{\bigg(3x \bigg) }^{\degree} + {\bigg(\dfrac{3\: x}{2} \bigg) }^{\degree} = 180\degree$

$\rm :\longmapsto\:{\bigg(\dfrac{6x + 3\: x}{2} \bigg) }^{\degree} = 180\degree$

$\rm :\longmapsto\:{\bigg(\dfrac{9x}{2} \bigg) }^{\degree} = 180\degree$

$\bf\implies \:x = 40\degree$

Hence,

$\begin{gathered}\begin{gathered}\bf\:So \: the \: angles \: are-\begin{cases} &\sf{x = {\bigg(\dfrac{3\: x}{5} \bigg) }^{\degree}} = 24\degree \\ \\ &\sf{y = {\bigg(\dfrac{3x}{2} \bigg) }^{ \degree} = 60\degree} \\ \\ &\sf{z = ={\bigg(\dfrac{12\: x}{5} \bigg) }^{\degree} = 96\degree} \end{cases}\end{gathered}\end{gathered}$