Question

A ray $OB$ rotates ${48.58}^{\circ }$ in counterclockwise direction with respect to another ray $OA$ and then ${157.32}^{\circ }$ in the clockwise direction. The final angle between the rays $OA\&OB$ is(in degrees minutes and seconds)

• A. $-{108}^{\circ }{44}^{\prime }{24}^{\prime \prime }$
• B. $-{205}^{\circ }{54}^{\prime }$
• C. ${108}^{\circ }{44}^{\prime }{24}^{\prime \prime }$
• D. ${205}^{\circ }{54}^{\prime }$

Answer 1

The correct option is A $-{108}^{\circ }{44}^{\prime }{24}^{\prime \prime }$

Given: A ray $OB$ making an angle of ${48.58}^{\circ }$ w.r.t $OA$ in counterclockwise directtion which can be drawn as:

Now, $OB$ rotates ${157.32}^{\circ }$ in clockwise direction as shown:

Now, final angle between the two rays $OA\&OB$
$=\angle BO{B}^{\prime }+\angle AO{B}^{\prime }={48.58}^{\circ }+(-{157.32}^{\circ })$
$=-{108.74}^{\circ }$
Now, we have to find the answeer in terms of degrees, minutes and seconds
Now, ${108.74}^{\circ }={108}^{\circ }+{0.74}^{\circ }$
${0.74}^{\circ }=0.74\times {60}^{\prime }={44.4}^{\prime }={44}^{\prime }+{0.4}^{\prime }$
Also, $1^{\prime }={60}^{\prime \prime }\Rightarrow {0.4}^{\prime }=0.4\times {60}^{\prime \prime }={24}^{\prime \prime }$
$\Rightarrow {108.74}^{\circ }={108}^{\circ }{44}^{\prime }{24}^{\prime \prime }$
Now, including the sign, we get the angle made by $OB$ w.r.t $OA=-{108}^{\circ }{44}^{\prime }{24}^{\prime \prime }$