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Question: In triangle \[ABC\], right-angled at \[B\], \[AB = 24{\text{ cm}}\] and \[BC = 7{\text{ cm}}\]. Dete...

In triangle ABCABC, right-angled at BB, AB=24 cmAB = 24{\text{ cm}} and BC=7 cmBC = 7{\text{ cm}}. Determine
(i) sinA\sin A, cosA\cos A
(ii) sinC\sin C, cosC\cos C

Explanation

Solution

Here, we need to find the value of the sine and cosine of angles AA and CC. First, we will find the length of the hypotenuse using the Pythagoras’s theorem. Then, using the lengths of the sides of the right angled triangle, we can find the required trigonometric ratios using the formulae sinθ=PerpendicularHypotenuse\sin \theta = \dfrac{{{\text{Perpendicular}}}}{{{\text{Hypotenuse}}}} and cosθ=BaseHypotenuse\cos \theta = \dfrac{{{\text{Base}}}}{{{\text{Hypotenuse}}}}.
Formula Used: We will use the following formula to solve the question:
1.The Pythagoras’s theorem states that the square of the hypotenuse of a right angled triangle is equal to the sum of squares of the other two sides.
2.The sine of an angle θ\theta of a right angled triangle is given by sinθ=PerpendicularHypotenuse\sin \theta = \dfrac{{{\text{Perpendicular}}}}{{{\text{Hypotenuse}}}}.
3.The cosine of an angle θ\theta of a right angled triangle is given by cosθ=BaseHypotenuse\cos \theta = \dfrac{{{\text{Base}}}}{{{\text{Hypotenuse}}}}.

Complete step-by-step answer:
(i)
First, we will draw the diagram using the given information.

Here, B=90\angle B = 90^\circ , AB=24 cmAB = 24{\text{ cm}} and BC=7 cmBC = 7{\text{ cm}}.
We know that the side opposite to the right angle of a right angled triangle is the hypotenuse.
Therefore, ACAC is the hypotenuse.
Now, we will use the Pythagoras’s theorem to get the length of the hypotenuse.
The Pythagoras’s theorem states that the square of the hypotenuse of a right angled triangle is equal to the sum of squares of the other two sides, that is Hypotenusee2=Base2+Perpendicularr2{\text{Hypotenuse}}{{\text{e}}^2} = {\text{Bas}}{{\text{e}}^2} + {\text{Perpendicular}}{{\text{r}}^2}.
Therefore, in triangle ABCABC, we get
AC2=AB2+BC2A{C^2} = A{B^2} + B{C^2}
Substituting AB=24AB = 24 cm and BC=7BC = 7 cm in the equation, we get
AC2=242+72\Rightarrow A{C^2} = {24^2} + {7^2}
Applying the exponents on the bases, we get
AC2=576+49\Rightarrow A{C^2} = 576 + 49
Adding 576 and 49, we get
AC2=625\Rightarrow A{C^2} = 625
Taking the square root of both sides, we get

\Rightarrow AC = 25{\text{ cm}} \\\\$$ Therefore, the length of the hypotenuse $$AC$$ is 25 cm. Now, we will use the sides of the triangle to find the required trigonometric ratios. The sine of an angle $$\theta $$ of a right angled triangle is given by $$\sin \theta = \dfrac{{{\text{Perpendicular}}}}{{{\text{Hypotenuse}}}}$$. Therefore, we get the sine of $$A$$ as $$ \Rightarrow \sin A = \dfrac{{BC}}{{AC}}$$ Substituting $$BC = 7$$ cm and $$AC = 25$$ cm in the equation, we get $$ \Rightarrow \sin A = \dfrac{7}{{25}}$$ The cosine of an angle $$\theta $$ of a right angled triangle is given by $$\cos \theta = \dfrac{{{\text{Base}}}}{{{\text{Hypotenuse}}}}$$. Therefore, we get the cosine of $$A$$ as $$ \Rightarrow \cos A = \dfrac{{AB}}{{AC}}$$ Substituting $$AB = 24$$ cm and $$AC = 25$$ cm in the equation, we get $$ \Rightarrow \cos A = \dfrac{{24}}{{25}}$$ Therefore, we get the values of $$\sin A$$ and $$\cos A$$ as $$\dfrac{7}{{25}}$$ and $$\dfrac{{24}}{{25}}$$ respectively. (ii) We will use the sides of the triangle to find the required trigonometric ratios. In triangle $$ABC$$, we get the sine of $$C$$ as $$ \Rightarrow \sin C = \dfrac{{AB}}{{AC}}$$ Substituting $$AB = 24$$ cm and $$AC = 25$$ cm in the equation, we get $$ \Rightarrow \sin C = \dfrac{{24}}{{25}}$$ Also, in triangle $$ABC$$, we get the cosine of $$C$$ as $$ \Rightarrow \cos C = \dfrac{{BC}}{{AC}}$$ Substituting $$BC = 7$$ cm and $$AC = 25$$ cm in the equation, we get $$ \Rightarrow \cos C = \dfrac{7}{{25}}$$ $$\therefore $$ We get the values of $$\sin C$$ and $$\cos C$$ as $$\dfrac{{24}}{{25}}$$ and $$\dfrac{7}{{25}}$$ respectively. **Note:** We can also use the trigonometric ratios of complementary angles to find the sine and cosine of $$C$$. Using the angle sum property of a triangle, we get $$\angle A + \angle B + \angle C = 180^\circ $$ Therefore, we get $$\Rightarrow \angle A + 90^\circ + \angle C = 180^\circ \\\ \Rightarrow \angle A + \angle C = 90^\circ \\\ \Rightarrow \angle C = 90^\circ - \angle A \\\\$$ Therefore, the angles $$A$$ and $$C$$ are complementary angles. We know that the sine and cosine of complementary angles can be written as $$\sin \left( {90^\circ - \theta } \right) = \cos \theta $$ and $$\cos \left( {90^\circ - \theta } \right) = \sin \theta $$. Rewriting the sine of angle $$C$$, we get $$\Rightarrow \sin C = \sin \left( {90^\circ - A} \right) \\\ \Rightarrow \sin C = \cos A \\\ \therefore \sin C = \dfrac{{24}}{{25}} \\\\$$ Similarly, rewriting the cosine of angle $$C$$, we get $$\Rightarrow \cos C = \cos \left( {90^\circ - A} \right) \\\ \Rightarrow \cos C = \sin A \\\ \Rightarrow \cos C = \dfrac{7}{{25}} \\\\$$ $$\therefore $$ We get the values of $$\sin C$$ and $$\cos C$$ as $$\dfrac{{24}}{{25}}$$ and $$\dfrac{7}{{25}}$$ respectively.