Question

In $\Delta ABC$, right-angled at $B$, $AB = 24{\text{ cm, }}BC = 7{\text{ cm}}$. Determine:
(i) $\sin A,{\text{ }}\cos A{\text{ }}$
(ii) $\sin C,{\text{ }}\cos C{\text{ }}$

Answer 1

Hint: First calculate the third side of the triangle by applying Pythagoras Theorem for right angled triangle, ${c^2} = {a^2} + {b^2}$, where $c$ is the length of the hypotenuse and $a$ and $b$ are the lengths of other two sides. Then apply trigonometric formulas $\sin \theta = \dfrac{{{\text{Perpendicular}}}}{{{\text{Hypotenuse}}}}$ and $\cos \theta = \dfrac{{{\text{Base}}}}{{{\text{Hypotenuse}}}}$ to calculate the required values.

Complete step-by-step answer:
According to the question, in a right angled triangle $\Delta ABC$, right angled at $B$, we have been given lengths of two sides.
$ \Rightarrow AB = 24{\text{ cm, }}BC = 7{\text{ cm}}$

We can apply Pythagoras Theorem to calculate the length of the third side. According to this theorem, in a right angled triangle, ${c^2} = {a^2} + {b^2}$, where $c$ is the length of the hypotenuse and $a$ and $b$ are the lengths of other two sides.
Applying this theorem, we’ll get:
$ \Rightarrow A{C^2} = A{B^2} + B{C^2}$
Putting values of $AB$ and $BC$, we have:
$
   \Rightarrow A{C^2} = {\left( {24} \right)^2} + {\left( 7 \right)^2} \\
   \Rightarrow A{C^2} = 576 + 49 = 625 \\
   \Rightarrow AC = 25{\text{ }}.....{\text{(1)}}
 $
Also $AC$ is the perpendicular for both the angles $A$ and $C$ as it is evident from the figure.
In the first case, we have to determine the values of $\sin A$ and $\cos A$.
In the above diagram, for angle $A$, base is $AB$ and perpendicular is $BC$. Further, we know that $\sin \theta = \dfrac{{{\text{Perpendicular}}}}{{{\text{Hypotenuse}}}}$ and $\cos \theta = \dfrac{{{\text{Base}}}}{{{\text{Hypotenuse}}}}$. Using these formulas, we’ll get:
$
   \Rightarrow \sin A = \dfrac{{BC}}{{AC}} \\
   \Rightarrow \cos A = \dfrac{{AB}}{{AC}}
 $
Putting $AB = 24{\text{ cm, }}BC = 7{\text{ cm}}$ and $AC = 25{\text{ cm}}$, we’ll get:
$
   \Rightarrow \sin A = \dfrac{7}{{25}} \\
   \Rightarrow \cos A = \dfrac{{24}}{{25}}
 $
In the second case, we have to determine the values of $\sin C$ and $\cos C$.
Again in the above diagram, for angle $C$, base is $BC$ and perpendicular is $AB$. Again applying the formulas $\sin \theta = \dfrac{{{\text{Perpendicular}}}}{{{\text{Hypotenuse}}}}$ and $\cos \theta = \dfrac{{{\text{Base}}}}{{{\text{Hypotenuse}}}}$, we’ll get:
\[
   \Rightarrow \sin C = \dfrac{{AB}}{{AC}} \\
   \Rightarrow \cos C = \dfrac{{BC}}{{AC}}
 \]
Putting $AB = 24{\text{ cm, }}BC = 7{\text{ cm}}$ and $AC = 25{\text{ cm}}$, we’ll get:
$
   \Rightarrow \sin C = \dfrac{{24}}{{25}} \\
   \Rightarrow \cos C = \dfrac{7}{{25}}
 $

So the required values are:
$
   \Rightarrow \sin A = \dfrac{7}{{25}} \\
   \Rightarrow \cos A = \dfrac{{24}}{{25}} \\
   \Rightarrow \sin C = \dfrac{{24}}{{25}} \\
   \Rightarrow \cos C = \dfrac{7}{{25}}
 $

Note: The formulas of first three trigonometric ratios are:
\[ \Rightarrow \sin \theta = \dfrac{{{\text{Perpendicular}}}}{{{\text{Hypotenuse}}}},{\text{ }}\cos \theta = \dfrac{{{\text{Base}}}}{{{\text{Hypotenuse}}}}{\text{ and }}\tan \theta = \dfrac{{{\text{Perpendicular}}}}{{{\text{Base}}}}\].
The other three trigonometric ratios are reciprocal of these three as shown below:
$ \Rightarrow \operatorname{cosec} = \dfrac{1}{{\sin \theta }},{\text{ }}\sec \theta = \dfrac{1}{{\cos \theta }}{\text{ and }}\cot \theta = \dfrac{1}{{\tan \theta }}$.
Thus remembering the first three formulas and taking their reciprocal, we can determine all the six trigonometric ratios.