Answer:
The angles are <u>155°</u> and <u>25°</u>.
Step-by-step explanation:
Given:
Two supplementary angles are in the ratio of 31:5.
Now, to find the angles.
The sum of two supplementary angles = 180°
Let the ratio of the angles be 
.
So, according to question:
<em>Dividing both sides by 36 we get:</em>
So, 
And, 
Therefore, the angles are 155° and 25°.
ΔAOB is a right angled triangle. Therefore the Pythagorean Theorem applies in this situation.
θ is the angle from a standard position of the line OA
The length of the y component is √(1-0)2 +(-3-(-3))2] =√(12+ 02) = 1 A(-3,1) to B(-3,0) which is opposite
Then the length of the x-component is √[(-3-0)2 +(0-0)2] = √(9+0)= 3 B(-3,0) to O(0,0) which is adjacent
The length of vector OA is √[(-3-0)2 + (1-0)2] = √(9+1) = √(10) A(-3,1) to O(0,0) which is the hypotenuse of the triangle
θ = 180 - α
sinθ = sin(180-α) = opposite/hypotenuse = 1/√10
cosθ = adjacent/hypotenuse = -3/√10
tanθ = opposite/adjacent = 1/-3 = -1/3
α= arcsin(1/√10) ≈ 18
θ =180 -18 ≈162
Note: Consider we need to find the vertices of the triangle A'B'C'
Given:
Triangle ABC is rotated 90 degrees clockwise about the origin to create triangle A'B'C'.
Triangle A,B,C with vertices at A(-3, 6), B(2, 9), and C(1, 1).
To find:
The vertices of the triangle A'B'C'.
Solution:
If triangle ABC is rotated 90 degrees clockwise about the origin to create triangle A'B'C', then
Using this rule, we get
Therefore, the vertices of A'B'C' are A'(6,3), B'(9,-2) and C'(1,-1).
All I know is that m=9 and p=1
Answer:
Because we have a point and slope, we can use at the beginning the point-slope form: y-y1=m(x-x1)
Step-by-step explanation:
m=3 , x1=1 , y=-3
y-(-3)=3(x-1)
y+3=3x-3 subtract 3 from both sides
y=3x-6 the answer in the slope-intercept form y=mx+b
