Question

How do you sketch the angle whose terminal side in standard position passes through \[\left( {6,8} \right)\] and how do you find sin and cos?

Answer 1

Hint: To sketch the angle whose terminal side in standard position passes through \[\left( {6,8} \right)\], we need to find \[r = \sqrt {{a^2} + {b^2}} \] of the terminal such that to find the angles of sin and cos we need to substitute the value of r with the given points in their respective formulas.

Formula used:
\[r = \sqrt {{a^2} + {b^2}} \]
\[\sin \left( \theta \right) = \dfrac{b}{r}\] and \[\cos \theta = \dfrac{a}{r}\]

Complete step by step solution:
To sketch the angle in standard position, one side of the angle is the positive x axis (the right side of the horizontal axis). That is the initial side.
The terminal side has one end at the origin (the point \[\left( {0,0} \right)\], also the intersection of the two axis) and goes through the point \[\left( {6,8} \right)\]. Hence, to locate the point \[\left( {6,8} \right)\], starting at the origin and count 6 to the right and up 8. That will get you to the point \[\left( {6,8} \right)\]. Put a dot there.
Now draw a line from the origin through the point \[\left( {6,8} \right)\]. The graph is represented as:

Give the angle a name, let us use angle as \[\theta \].
In this, question we have \[\left( {a,b} \right) = \left( {6,8} \right)\], If the point \[\left( {a,b} \right)\] lies on the terminal side of an angle in standard position, then let:
\[r = \sqrt {{a^2} + {b^2}} \]
Now, substitute the given points of a and b as:
\[ \Rightarrow r = \sqrt {{{\left( 6 \right)}^2} + {{\left( 8 \right)}^2}} \]
Squaring the terms, we get:
\[ \Rightarrow r = \sqrt {36 + 64} \]
\[ \Rightarrow r = \sqrt {100} \]
We know that, \[\sqrt {100} = 10\], hence we get:
\[ \Rightarrow r = 10\]
Now, we need to find sin and cos using the formula with respect to the points we have:
\[\sin \left( \theta \right) = \dfrac{b}{r}\] and \[\cos \theta = \dfrac{a}{r}\]
We know that, Sine of angle is equal to the ratio of opposite side and hypotenuse whereas cosine of an angle is equal to ratio of adjacent side and hypotenuse i.e.,

\[\sin \theta = \dfrac{{opposite}}{{hypotenuse}}\] and \[\cos \theta = \dfrac{{adjacent}}{{hypotenuse}}\]
As, we have the points given as \[a,b:\left( {6,8} \right)\] and \[r = 10\], hence substitute these values in their respective formulas as:
To find, sine of \[\theta \] as:
\[\sin \left( \theta \right) = \dfrac{b}{r}\]
We, have b=8 and r=10, hence substituting these we get:
\[ \Rightarrow \sin \left( \theta \right) = \dfrac{8}{{10}}\]
Evaluating the terms, we get the value of \[\sin \theta \] as:
\[ \Rightarrow \sin \theta = \dfrac{4}{5}\].
Now, we need to find cos of \[\theta \] as:
\[\cos \theta = \dfrac{a}{r}\]
We, have a=6 and r=10, hence substituting these we get:
\[ \Rightarrow \cos \theta = \dfrac{6}{{10}}\]
Evaluating the terms, we get the value of \[\cos \theta \] as:
\[ \Rightarrow \cos \theta = \dfrac{3}{5}\].
Therefore, the value of sin and cos is:
\[\sin \theta = \dfrac{4}{5}\] and \[\cos \theta = \dfrac{3}{5}\].
If we are asked to find the angle \[\theta \] of sin and cos, then we have:
\[\sin \theta = \dfrac{4}{5}\] i.e., \[\left( {\dfrac{4}{5} = 0.8} \right)\]
\[ \Rightarrow \theta = {\sin ^{ - 1}}\left( {0.8} \right) = 53.13^\circ \]
And, for \[\cos \theta = \dfrac{3}{5}\] i.e., \[ \left( {\dfrac{3}{5} = 0.6} \right)\]
\[ \Rightarrow \theta = {\cos ^{ - 1}}\left( {0.6} \right) = 53.13^\circ \]
Hence, angles are the same for both sin and cos function.

Note: We must note that:
If the point \[\left( {a,b} \right)\] lies on the terminal side of an angle in standard position, then let:
\[r = \sqrt {{a^2} + {b^2}} \]; and the angle has sine \[\dfrac{b}{r}\] and it has cosine \[\dfrac{a}{r}\].