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Sagot :
Sure, let me guide you step-by-step to find the equation of the line which cuts off an intercept of 8 from the [tex]\( y \)[/tex]-axis and makes an angle of [tex]\( 60^\circ \)[/tex] with the [tex]\( y \)[/tex]-axis.
1. Identify the [tex]\( y \)[/tex]-intercept ([tex]\( c \)[/tex]):
We're given that the [tex]\( y \)[/tex]-intercept is [tex]\( 8 \)[/tex]. This means the line intersects the [tex]\( y \)[/tex]-axis at point [tex]\((0, 8)\)[/tex].
2. Determine the angle with respect to the [tex]\( x \)[/tex]-axis:
Given the angle is [tex]\( 60^\circ \)[/tex] with the [tex]\( y \)[/tex]-axis, we need to find its complementary angle with the [tex]\( x \)[/tex]-axis. Since the sum of the angles between the [tex]\( x \)[/tex]-axis and the [tex]\( y \)[/tex]-axis is [tex]\( 90^\circ \)[/tex], the angle between the line and the [tex]\( x \)[/tex]-axis is:
[tex]\[ 90^\circ - 60^\circ = 30^\circ \][/tex]
3. Calculate the slope ([tex]\( m \)[/tex]) of the line:
The slope of the line is defined as the tangent of the angle it makes with the [tex]\( x \)[/tex]-axis. Therefore, the slope [tex]\( m \)[/tex] is:
[tex]\[ m = \tan(30^\circ) \][/tex]
4. Find the value of [tex]\( \tan(30^\circ) \)[/tex]:
We know from trigonometry that:
[tex]\[ \tan(30^\circ) = \frac{1}{\sqrt{3}} \approx 0.577 \][/tex]
Therefore, the slope [tex]\( m \)[/tex] is approximately:
[tex]\[ m \approx 0.577 \][/tex]
5. Construct the equation in slope-intercept form:
The slope-intercept form of a linear equation is:
[tex]\[ y = mx + c \][/tex]
Here, [tex]\( m \)[/tex] is the slope and [tex]\( c \)[/tex] is the [tex]\( y \)[/tex]-intercept.
6. Substitute the values of [tex]\( m \)[/tex] and [tex]\( c \)[/tex]:
Given [tex]\( m = 0.577 \)[/tex] and [tex]\( c = 8 \)[/tex], the equation of the line becomes:
[tex]\[ y = 0.577x + 8 \][/tex]
7. Provide the final equation:
Thus, the equation of the line which cuts off an intercept of [tex]\( 8 \)[/tex] from the [tex]\( y \)[/tex]-axis and makes an angle of [tex]\( 60^\circ \)[/tex] with the [tex]\( y \)[/tex]-axis is:
[tex]\[ y = 0.58x + 8 \][/tex]
So, the equation of the line is [tex]\( y = 0.58x + 8 \)[/tex].
1. Identify the [tex]\( y \)[/tex]-intercept ([tex]\( c \)[/tex]):
We're given that the [tex]\( y \)[/tex]-intercept is [tex]\( 8 \)[/tex]. This means the line intersects the [tex]\( y \)[/tex]-axis at point [tex]\((0, 8)\)[/tex].
2. Determine the angle with respect to the [tex]\( x \)[/tex]-axis:
Given the angle is [tex]\( 60^\circ \)[/tex] with the [tex]\( y \)[/tex]-axis, we need to find its complementary angle with the [tex]\( x \)[/tex]-axis. Since the sum of the angles between the [tex]\( x \)[/tex]-axis and the [tex]\( y \)[/tex]-axis is [tex]\( 90^\circ \)[/tex], the angle between the line and the [tex]\( x \)[/tex]-axis is:
[tex]\[ 90^\circ - 60^\circ = 30^\circ \][/tex]
3. Calculate the slope ([tex]\( m \)[/tex]) of the line:
The slope of the line is defined as the tangent of the angle it makes with the [tex]\( x \)[/tex]-axis. Therefore, the slope [tex]\( m \)[/tex] is:
[tex]\[ m = \tan(30^\circ) \][/tex]
4. Find the value of [tex]\( \tan(30^\circ) \)[/tex]:
We know from trigonometry that:
[tex]\[ \tan(30^\circ) = \frac{1}{\sqrt{3}} \approx 0.577 \][/tex]
Therefore, the slope [tex]\( m \)[/tex] is approximately:
[tex]\[ m \approx 0.577 \][/tex]
5. Construct the equation in slope-intercept form:
The slope-intercept form of a linear equation is:
[tex]\[ y = mx + c \][/tex]
Here, [tex]\( m \)[/tex] is the slope and [tex]\( c \)[/tex] is the [tex]\( y \)[/tex]-intercept.
6. Substitute the values of [tex]\( m \)[/tex] and [tex]\( c \)[/tex]:
Given [tex]\( m = 0.577 \)[/tex] and [tex]\( c = 8 \)[/tex], the equation of the line becomes:
[tex]\[ y = 0.577x + 8 \][/tex]
7. Provide the final equation:
Thus, the equation of the line which cuts off an intercept of [tex]\( 8 \)[/tex] from the [tex]\( y \)[/tex]-axis and makes an angle of [tex]\( 60^\circ \)[/tex] with the [tex]\( y \)[/tex]-axis is:
[tex]\[ y = 0.58x + 8 \][/tex]
So, the equation of the line is [tex]\( y = 0.58x + 8 \)[/tex].
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