Pls help me not fail
1 answer:
Answer:
D
Step-by-step explanation:
Using the Cosine rule to find AC
AC² = BC² + AB² - (2 × BC × AB × cosB )
= 18² + 12² - ( 2 × 18 × 12 × cos75° )
= 324 + 144 - 432cos75°
= 468 - 111.8
= 356.2 ( take the square root of both sides )
AC = ≈ 18.9
-----------------------------------------
Using the Sine rule to find ∠ A
=
( cross- multiply )
18.9 sinA = 18 sin75° ( divide both sides by 18.9 )
sinA = , then
∠ A = (
) ≈ 66.9°
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Answer:
Equation of other line passing through point (6 , - 11) and parallel to given line is y = - 23 x + 127 .
Step-by-step explanation:
Given as :
The equation of one line
y = - 23 x + 12
∵ The equation of line in slope-intercept form is
y = m x + c
where m is the slope of line and c is y-intercept
Now, Compare given line with standard line equation
So, The slope of given line = m = - 23
Now, Again
Other line is passing through point (6 , - 11) and is parallel to given line
so, both the lines are parallel
For parallel line condition , Slope of both lines are equal
Let The slope of other line = M
So, M = m = - 23
Now, Equation of other line passing through point (6 , - 11) and slope - 23 in slope-point form
y - = M × (x -
)
i.e y - ( - 11) = (- 23) × (x - 6)
Or, y + 11 = - 23 × x + 138
Or, y = - 23 x + 138- 11
i.e y = - 23 x + 127
So, The equation of other line y = - 23 x + 127
Hence, Equation of other line passing through point (6 , - 11) and parallel to given line is y = - 23 x + 127 . Answer