Question

Find $\tan {53^0}38'$.
$(a){\text{ 1}}{\text{.23}}$
$(b){\text{ 0}}{\text{.45}}$
$(c){\text{ 0}}{\text{.56}}$
$(d){\text{ 1}}{\text{.35}}$

Answer 1

Hint: In the above given question, we are asked to find the value of
${\text{tan5}}{{\text{3}}^0}38'$. Here, the value 53 is in degrees whereas the value 38 is given in minutes and we must know that, one degree is equal to sixty minutes and one minute is equal to sixty seconds respectively.

Complete step-by-step answer:
We know that the degree is divided into sixty parts called minutes.

That is, $60' \to {1^0}$

So, $1' \to {\left( {\dfrac{1}{{60}}} \right)^0}$

$\therefore 38' \to {\left( {\dfrac{{38}}{{60}}} \right)^0} = {0.63^0}$

By putting the value of 38’ in the given term that is, \[\tan {53^0}38\prime \].

$ \Rightarrow \tan {53^0}38' = \tan {53.63^0}$

$ \Rightarrow \tan {53.63^0} = 1.35$

Hence, the solution is the option$(d){\text{ 1}}{\text{.35}}$.

Note: When we face such a type of question, the key point is to have a good understanding of the concepts like degree, minutes, seconds, etc. We have to simply convert the given minutes value in the degrees by using the definition of minutes. Then, with the simple manipulations we will obtain the required solution.