If $ \sin \theta =\dfrac{3}{5} $ , find the values of other trigonometric ratios.
Answer
645.6k+ views
Hint: Consider a right angled $ \Delta ABC $ , right angled at B. Use Pythagoras theorem then find the values of $ \cos \theta , \tan \theta $ and their inverse $ \csc \theta,\sec \theta $ and $ \cot \theta $
Complete step-by-step answer:
Let us consider a right angled triangle ABC. We know the Pythagoras theorem, also known as Pythagoras theorem; it is a fundamental relation in Euclidean geometry among the 3 sides of a right triangle.
It states that the area of the squares whose sides is the hypotenuse is equal to the sum of the areas of the squares on the other two sides.
$ A{{B}^{2}}+B{{C}^{2}}=A{{C}^{2}} $
We have been given, $ \sin \theta =\dfrac{3}{5} $ .
In the triangle ABC, $ \angle B=90{}^\circ $
And take $ \angle C=\theta $
Here, $ \sin \theta = $ opposite side/hypotenuse = $ \dfrac{AB}{BC} $
$ \cos \theta = $ Adjacent side/hypotenuse = $ \dfrac{BC}{AC} $
Given, $ \sin \theta =\dfrac{3}{5} $
AB=3 and AC=5
Using the Pythagoras theorem, $ A{{B}^{2}}+B{{C}^{2}}=A{{C}^{2}} $
$ \begin{align}
& {{5}^{2}}={{3}^{2}}+B{{C}^{2}} \\
& \Rightarrow B{{C}^{2}}={{5}^{2}}-{{3}^{2}} \\
& \Rightarrow BC=\sqrt{25-9}=\sqrt{16}=4 \\
& \therefore \cos \theta =\dfrac{4}{5} \\
\end{align} $
$ \tan \theta = $ Opposite side/adjacent side= $ \dfrac{AB}{BC}=\dfrac{3}{4} $
$ \csc\theta =\dfrac{1}{\sin \theta }=\dfrac{1}{\dfrac{3}{5}}=\dfrac{5}{3} $
$ \begin{align}
& \sec \theta =\dfrac{1}{\cos \theta }=\dfrac{1}{\dfrac{4}{5}}=\dfrac{5}{4} \\
& \cot \theta =\dfrac{1}{\tan \theta }=\dfrac{1}{\dfrac{3}{4}}=\dfrac{4}{3} \\
& \therefore \sin \theta =\dfrac{3}{5},\cos \theta =\dfrac{4}{5},\tan \theta =\dfrac{3}{4} \\
& \csc\theta =\dfrac{5}{3},\sec \theta =\dfrac{5}{4},\cot \theta =\dfrac{4}{3} \\
\end{align} $
Note: There are three types of special right triangle, 30-60-90 triangle, 45-45-90 triangle and Pythagoras triple triangles. This problem was based on trigonometric ratios. So we just need to remember all the formulas of trigonometric ratio in the form of perpendicular, base and hypotenuse.
Complete step-by-step answer:
Let us consider a right angled triangle ABC. We know the Pythagoras theorem, also known as Pythagoras theorem; it is a fundamental relation in Euclidean geometry among the 3 sides of a right triangle.
It states that the area of the squares whose sides is the hypotenuse is equal to the sum of the areas of the squares on the other two sides.
$ A{{B}^{2}}+B{{C}^{2}}=A{{C}^{2}} $
We have been given, $ \sin \theta =\dfrac{3}{5} $ .
In the triangle ABC, $ \angle B=90{}^\circ $
And take $ \angle C=\theta $
Here, $ \sin \theta = $ opposite side/hypotenuse = $ \dfrac{AB}{BC} $
$ \cos \theta = $ Adjacent side/hypotenuse = $ \dfrac{BC}{AC} $
Given, $ \sin \theta =\dfrac{3}{5} $
AB=3 and AC=5
Using the Pythagoras theorem, $ A{{B}^{2}}+B{{C}^{2}}=A{{C}^{2}} $
$ \begin{align}
& {{5}^{2}}={{3}^{2}}+B{{C}^{2}} \\
& \Rightarrow B{{C}^{2}}={{5}^{2}}-{{3}^{2}} \\
& \Rightarrow BC=\sqrt{25-9}=\sqrt{16}=4 \\
& \therefore \cos \theta =\dfrac{4}{5} \\
\end{align} $
$ \tan \theta = $ Opposite side/adjacent side= $ \dfrac{AB}{BC}=\dfrac{3}{4} $
$ \csc\theta =\dfrac{1}{\sin \theta }=\dfrac{1}{\dfrac{3}{5}}=\dfrac{5}{3} $
$ \begin{align}
& \sec \theta =\dfrac{1}{\cos \theta }=\dfrac{1}{\dfrac{4}{5}}=\dfrac{5}{4} \\
& \cot \theta =\dfrac{1}{\tan \theta }=\dfrac{1}{\dfrac{3}{4}}=\dfrac{4}{3} \\
& \therefore \sin \theta =\dfrac{3}{5},\cos \theta =\dfrac{4}{5},\tan \theta =\dfrac{3}{4} \\
& \csc\theta =\dfrac{5}{3},\sec \theta =\dfrac{5}{4},\cot \theta =\dfrac{4}{3} \\
\end{align} $
Note: There are three types of special right triangle, 30-60-90 triangle, 45-45-90 triangle and Pythagoras triple triangles. This problem was based on trigonometric ratios. So we just need to remember all the formulas of trigonometric ratio in the form of perpendicular, base and hypotenuse.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 10 English: Engaging Questions & Answers for Success

Trending doubts
Explain the Treaty of Vienna of 1815 class 10 social science CBSE

Why is it 530 pm in india when it is 1200 afternoon class 10 social science CBSE

What is the full form of POSCO class 10 social science CBSE

Define Potential, Developed, Stock and Reserved resources

The speaker of the Lok Sabha is elected by the APresident class 10 social science CBSE

Complete the sentence with the most appropriate word class 10 english CBSE

