Answer
Verified
425.4k+ views
Hint: The sine of the point is the proportion of the length of the side inverse the point partitioned by the length of the hypotenuse. The cosine of the point is the proportion of the length of the side near the point partitioned by the length of the hypotenuse. The digression of the point is the proportion of the length of the side inverse the point isolated by the length of side adjoining the point.
Complete step by step answer:
You consider that,
$\dfrac{{10\pi }}{3} = 2\pi + \pi + \dfrac{\pi }{3}$, so that
$ \Rightarrow $sin$(\dfrac{{10\pi }}{3})$$ = $ sin$(2\pi + \pi + \dfrac{\pi }{3}) = $ sin$(\pi + \dfrac{\pi }{3})$
$ \Rightarrow $sin$\pi $cos$(\dfrac{\pi }{3})$$ + $cossin$(\dfrac{\pi }{3})$
$ \Rightarrow - $sin$\dfrac{\pi }{3}$$ = - \dfrac{{\sqrt 3 }}{2}$
$ \Rightarrow $Cos$(\dfrac{{10\pi }}{3}) = $cos$(2\pi + \pi + \dfrac{\pi }{3}) = $cos$(\pi + \dfrac{\pi }{3})$
$ \Rightarrow $cos$\pi $cos$(\dfrac{\pi }{3})$$ - $sin$\pi $sin$(\dfrac{\pi }{3})$$ = $$ - $cos$(\dfrac{\pi }{3})$
$ \Rightarrow - \dfrac{1}{2}$
$ \Rightarrow $Tan$(\dfrac{{10\pi }}{3}) = \dfrac{{\sin (\dfrac{{10\pi }}{3})}}{{\cos (\dfrac{{10\pi }}{3})}} = \dfrac{{ - \dfrac{{\sqrt 3 }}{2}}}{{ - \dfrac{1}{2}}} = \sqrt 3 $
Note: In arithmetic, geometrical capacities called round capacities, point capacities or goniometric capacities are genuine capacities which relate the point of a correct point triangle to proportions of two side lengths.
Complete step by step answer:
You consider that,
$\dfrac{{10\pi }}{3} = 2\pi + \pi + \dfrac{\pi }{3}$, so that
$ \Rightarrow $sin$(\dfrac{{10\pi }}{3})$$ = $ sin$(2\pi + \pi + \dfrac{\pi }{3}) = $ sin$(\pi + \dfrac{\pi }{3})$
$ \Rightarrow $sin$\pi $cos$(\dfrac{\pi }{3})$$ + $cossin$(\dfrac{\pi }{3})$
$ \Rightarrow - $sin$\dfrac{\pi }{3}$$ = - \dfrac{{\sqrt 3 }}{2}$
$ \Rightarrow $Cos$(\dfrac{{10\pi }}{3}) = $cos$(2\pi + \pi + \dfrac{\pi }{3}) = $cos$(\pi + \dfrac{\pi }{3})$
$ \Rightarrow $cos$\pi $cos$(\dfrac{\pi }{3})$$ - $sin$\pi $sin$(\dfrac{\pi }{3})$$ = $$ - $cos$(\dfrac{\pi }{3})$
$ \Rightarrow - \dfrac{1}{2}$
$ \Rightarrow $Tan$(\dfrac{{10\pi }}{3}) = \dfrac{{\sin (\dfrac{{10\pi }}{3})}}{{\cos (\dfrac{{10\pi }}{3})}} = \dfrac{{ - \dfrac{{\sqrt 3 }}{2}}}{{ - \dfrac{1}{2}}} = \sqrt 3 $
Note: In arithmetic, geometrical capacities called round capacities, point capacities or goniometric capacities are genuine capacities which relate the point of a correct point triangle to proportions of two side lengths.
Recently Updated Pages
Identify the feminine gender noun from the given sentence class 10 english CBSE
Your club organized a blood donation camp in your city class 10 english CBSE
Choose the correct meaning of the idiomphrase from class 10 english CBSE
Identify the neuter gender noun from the given sentence class 10 english CBSE
Choose the word which best expresses the meaning of class 10 english CBSE
Choose the word which is closest to the opposite in class 10 english CBSE
Trending doubts
A rainbow has circular shape because A The earth is class 11 physics CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Which are the Top 10 Largest Countries of the World?
Change the following sentences into negative and interrogative class 10 english CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Write a letter to the principal requesting him to grant class 10 english CBSE