Question

In a \[\Delta ABC\] right angled at B, \[AB = 24cm\], \[BC = 7cm\]. Determine
A) \[\sin A,\cos A\]
B) \[\sin C,\cos C\]

Answer 1

Hint:
In this question we are given a triangle \[\Delta ABC\] which is right angled at B and we are provided with the length of the sides that is \[AB = 24cm\] and \[BC = 7cm\]. We are to determine \[\sin A,\cos A\] and \[\sin C,\cos C\] for doing this we have to first find the third side of the triangle that is by Pythagoras theorem and later on finding the \[\sin A,\cos A\] and \[\sin C,\cos C\] components of the triangle let us see the detailed solution of this question.

Complete step by step solution:
Now we are given a triangle \[\Delta ABC\] right angled at B, such that \[AB = 24cm\] and \[BC = 7cm\] as depicted in the figure downwards.

Now applying Pythagoras theorem to find \[AC\]that is
\[{\left( {hypotenuse} \right)^2}\] = \[{\left( {Base} \right)^2} + {\left( {perpendicular} \right)^2}\]
Where base is the base of the triangle and perpendicular is the side opposite to the angle in the consideration and the hypotenuse is the longest side of the triangle
Now in the given triangle in the above question we are provided with the\[AB = 24cm\]and \[BC = 7cm\]so applying the Pythagoras theorem we get-
\[\;{\left( {AC} \right)^2} = {\left( {BC} \right)^2} + {\left( {AB} \right)^2}\]
\[\;{\left( {AC} \right)^2} = {\left( 7 \right)^2} + {\left( {24} \right)^2}\]
\[
  \;{\left( {AC} \right)^2} = {\left( {625} \right)^2} \\
  \left( {AC} \right) = \left( {25} \right) \\
 \]
Now we get AC as \[25cm\]
Now for finding \[\sin A,\cos A\]
As The \[\sin \theta \] component of any angle of the triangle is \[\dfrac{{{\text{Opposite}}}}{{{\text{Hypotenuse}}}}\] where opposite is the side opposite to the angle in consideration that is A in this case similarly \[\cos \theta \] component of any angle of the triangle is \[\dfrac{{Base}}{{{\text{Hypotenuse}}}}\]Also similar formulas are to be used when angle C is in consideration
So now \[\sin A\]=\[\dfrac{{{\text{Opposite}}}}{{{\text{Hypotenuse}}}}\]
That is
\[\sin A\]= \[
  \dfrac{{BC}}{{AC}} \\
  \dfrac{7}{{25}} \\
 \]
Now for \[\cos A\]that is -
\[\cos A\]\[ = \dfrac{{Base}}{{{\text{Hypotenuse}}}}\]
\[\cos A\]=\[ = \dfrac{{BC}}{{AC}}\]
\[\cos A\]\[ = \dfrac{{24}}{{25}}\]
Now for finding \[\sin C,\cos C\]

Now angle C is in consideration that is for \[\sin C\]-
\[
  \sin C = \dfrac{{{\text{Opposite}}}}{{{\text{Hypotenuse}}}} \\
  \sin C = \dfrac{{AB}}{
  AC \\
   = \dfrac{{24}}{{25}} \\
 } \\
 \]
Now for \[\cos C\]it becomes –
\[
  \cos C = \dfrac{{Base}}{{{\text{Hypotenuse}}}} \\
  \cos C = \dfrac{{BC}}{
  AC \\
   = \dfrac{7}{{25}} \\
 } \\
 \]
Hence we get our required answers.

Note:
While solving such kind of questions the concentration should be paid to the fact that which side becomes perpendicular and base while considering different angles to get the desired answer correctly. Here we can also the trigonometry identity of \[{\sin ^2}\theta + {\cos ^2}\theta = 1\]to find the \[\cos C\] and \[\cos A\] after finding the \[\sin C\] and \[\sin A\] which would save the time.