Question

Match the following angles in sexagesmial form to their degrees equivalent measurements(upto 2 decimal places)

• A. ${9.36}^{\circ }$
• B. ${60.01}^{\circ }$
• C. ${20.85}^{\circ }$
• D. ${30}^{\circ }$

Answer 1

Given angles in minutes and seconds and we have to convert it into it's equivalent degree measure.
We know: ${60}^{\prime }=1^{\circ }\&{60}^{\prime \prime }=1^{\prime }$
$\Rightarrow 1^{\prime }=\frac{1^{\circ }}{60}\cdots \left(i\right)$
$\Rightarrow 1^{\prime \prime }=\frac{1^{\prime }}{60}=\frac{1^{\circ }}{3600}\cdots \left(ii\right)$

Now, using this relation, let's find the equivalent angles.
$a.{1800}^{\prime }$
$\Rightarrow {1800}^{\prime }=\frac{1^{\circ }}{60}\times 1800={30}^{\circ }$

$b.{3600}^{\prime }{50}^{\prime \prime }$
$\Rightarrow {3600}^{\prime }{50}^{\prime \prime }={3600}^{\prime }+{50}^{\prime \prime }$
Using $\left(i\right)\&\left(ii\right)$
${3600}^{\prime }=3600\times \frac{1^{\circ }}{60}={60}^{\circ }$
Also, ${50}^{\prime \prime }=50\times \frac{1^{\circ }}{3600}=0.01388..{.}^{\circ }\approx {0.0139}^{\circ }$
$\Rightarrow {3600}^{\prime }{50}^{\prime \prime }={60}^{\circ }+{0.0139}^{\circ }={60.0139}^{\circ }\approx {60.01}^{\circ }$

$c.{560}^{\prime }{90}^{\prime \prime }$
$\Rightarrow {560}^{\prime }{90}^{\prime \prime }={560}^{\prime }+{90}^{\prime \prime }$
Using $\left(i\right)\&\left(ii\right)$
${560}^{\prime }=560\times \frac{1^{\circ }}{60}={9.333}^{\circ }$
Also, ${90}^{\prime \prime }=90\times \frac{1^{\circ }}{3600}={0.025}^{\circ }$
$\Rightarrow {560}^{\prime }{90}^{\prime \prime }={9.333}^{\circ }+{0.025}^{\circ }={9.358}^{\circ }\approx {9.36}^{\circ }$

$d.{1250}^{\prime }{60}^{\prime \prime }$
$\Rightarrow {1250}^{\prime }{60}^{\prime \prime }={1250}^{\prime }+{60}^{\prime \prime }$
Using $\left(i\right)\&\left(ii\right)$
${1250}^{\prime }=1250\times \frac{1^{\circ }}{60}={20.833}^{\circ }$
Also, ${60}^{\prime \prime }=60\times \frac{1^{\circ }}{3600}={0.017}^{\circ }$
$\Rightarrow {1250}^{\prime }{60}^{\prime \prime }={20.833}^{\circ }+{0.017}^{\circ }\approx {20.85}^{\circ }$