Question

Find the radian measure corresponding to the following degree measures:

(i) ${300}^{\circ }$

(ii) ${35}^{\circ }$

(iii) $-{56}^{\circ }$

(iv) ${135}^{\circ }$

(v) $-{300}^{\circ }$

(vi) $7^{\circ }{30}^{\prime }$

(vii) ${125}^{\circ }{30}^{\prime }$

(viii) $-{47}^{\circ }{30}^{\prime }$

Answer 1

(i) ${300}^{\circ }$
We have,
${180}^{\circ }={\pi }^c\therefore 1^{\circ }={\left(\frac{\pi }{180}\right)}^c$
Now,
${300}^{\circ }=300\times \frac{\pi }{180}=\frac{5\pi }{3}$
(ii) ${35}^{\circ }$
We have,
${180}^{\circ }={\pi }^c\therefore 1^{\circ }={\left(\frac{\pi }{180}\right)}^c$
Now,
${35}^{\circ }=35\times \frac{\pi }{180}=\frac{7\pi }{36}$
(iii) $-{56}^{\circ }$
We have,
${180}^{\circ }={\pi }^c\therefore 1^{\circ }={\left(\frac{\pi }{180}\right)}^c$

Now,

$-{56}^{\circ }=-56\times \frac{\pi }{180}=\frac{-14\pi }{45}$

(iv) ${135}^{\circ }$

We have,

${180}^{\circ }={\pi }^c{1}^{\circ }={\left(\frac{\pi }{180}\right)}^c$

Now,

${135}^{\circ }=135\times \frac{\pi }{180}=\frac{3\pi }{4}$

(v) $-{300}^{\circ }$

We have,

${180}^{\circ }={\pi }^c{1}^{\circ }={\left(\frac{\pi }{180}\right)}^c$

Now,

$-{300}^{\circ }=-300\times \frac{\pi }{180}=\frac{-5\pi }{3}$

(vi) $7^{\circ }$ 30'

We have,

${180}^{\circ }={\pi }^c{1}^{\circ }={\left(\frac{\pi }{180}\right)}^c{7}^{\circ }{30}^{\prime }={(7\times \frac{\pi }{180})}^c\times {\left(\frac{30}{60}\right)}^{\circ }={\left(7\frac{1}{2}\right)}^{\circ }\times {\left(\frac{\pi }{180}\right)}^c={(\frac{15}{2}\times \frac{\pi }{180})}^c=\frac{\pi }{24}$

(vii) ${125}^{\circ }$ 30'

We have

${180}^{\circ }={\pi }^c{1}^{\circ }={\left(\frac{\pi }{180}\right)}^c{125}^{\circ }{30}^{\prime }={125}^{\circ }{\left(\frac{30}{60}\right)}^{\circ }={\left(125\frac{1}{2}\right)}^{\circ }={(\frac{251}{2}\times \frac{\pi }{180})}^c=\frac{251\pi }{360}$

(viii) $-{47}^{\circ }{30}^{\prime }$

We have,

${180}^{\circ }={\pi }^c{1}^{\circ }={\left(\frac{\pi }{180}\right)}^c-{47}^{\circ }{30}^{\prime }=-{47}^{\circ }{\left(\frac{30}{60}\right)}^{\circ }={(-47\frac{1}{2})}^{\circ }={\left(\frac{-95}{2}\right)}^{\circ }={(\frac{-95}{2}\times \frac{\pi }{180})}^{\circ }=\frac{-19\pi }{72}$