The difference between two acute angles of a right-angled triangle is 9 find the angles in degree

π radians = 180° \(\frac { \pi }{ 9 }\)  radian =  \(\frac { 180 }{ \pi } \times \frac { \pi }{ 9 }\)  = 20° Let the two acute angles of the right-angled triangle be x and y (x > y). Then x + y = 90° …(i) and x – y = 20° …(ii) Solving (i) and (ii) we get 2x = 110° ⇒ x = 55° 55° + y = 90°  ⇒ y = 35° Hence, two angles are 55° and 35° respectively.

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The difference between the two acute angles of a right angled triangle is 2 π/5 radians. Express the angles in degrees.

Solution

Let ${\theta }_{1}and{\theta }_2$ be two acute angles of a right angled triangle.

$\therefore$ different of acute angles

${\theta }_1-{\theta }_2=\frac{2\pi }{5}\text{radians}$

$\because$ in a right angled triangle,

${\theta }_1+{\theta }_2=\frac{\pi }{2}$

${\theta }_1-{\theta }_2=\frac{2\pi }{5}$ . . .(i)

${\theta }_1+{\theta }_2=\frac{\pi }{2}$ . . . (ii)

On solving

$2{\theta }_1=\frac{2\pi }{5}+\frac{\pi }{2}$

${\theta }_1=\frac{9\pi }{20}$

From equation (ii)

${\theta }_2=\frac{\pi }{20}$

So angles in degrees

${\theta }_1=\frac{9\pi }{20}\times \frac{180}{\pi }={81}^{\circ }$

and ${\theta }_2=\frac{\pi }{20}\times \frac{180}{\pi }=9^{\circ }$

Mathematics

RD Sharma

Standard XI