4 square units.
The easiest way to find this is by looking at the triangle as if NL was the base. In this case PM would be the height. Since a triangle is the same shape even when we move it or rotate it, this is a common strategy to finding areas when the direction is tough for us.
Now that we have this we can measure the lengths of the base and height and use them in the triangle area formula.
Base = 4
Height = 2
Area of a triangle = 1/2bh
Area of a triangle = 1/2(4)(2)
Area of a triangle = 2(2)
Area of a triangle = 4
Answer:
(1, 3 )
Step-by-step explanation:
Given the 2 equations
y = 3x → (1)
x + y = 4 → (2)
Substitute y = 3x into (2)
x + 3x = 4, that is
4x = 4 ( divide both sides by 4 )
x = 1
Substitute x = 1 into (1) for corresponding value of y
y = 3 × 1 = 3
Solution is (1, 3 )
Answer:
(1-cos2A) /(1+cos2A) =tan²A
Proof:
We know that,
cos(A+B) =cosA.cosB-sinA.sinB
=>cos2A=cos(A+A)
=>cos2A=cosA.cosA - sinA.sinA
=>cos2A=cos²A-sin²A
=>cos2A=(cos²A-sin²A)/(cos²A+sin²A
Since {cos²A+sin²A=1}
Divide the numerator & the denominator by (cos²A) to get,
cos2A = {(cos²A-sin²A) ÷cos²A} / {(cos²A+sin²A) ÷cos²A}
cos2A ={(1-tan²A)/(1+tan²A)}
Then,
1-cos2A = 1-[{(1–tan²A)/(1+tan²A)}]
1-cos2A =(1+tan²A-1+tan²A)/(1+tan²A)
1-cos2A=(2tan²A)/(1+tan²A)
And now.......
1+cos2A=1+[{(1-tan²A)/(1+tan²A)}]
1+cos2A={1+tan²A+1-tan²A}/{1+tan²A}
1+cos2A=2/(1+tan²A)
So now,
(1-cos2A)/(1+cos2A)= {2tan²A/(1+tan²A)}÷{2/(1+tan²A)}
={(2tan²A)(1+tan²A)}÷{2(1+tan²A)}
=tan²A
Step-by-step explanation:
make me as brain liest
Answer:
Step-by-step explanation:
Converse of alternate interior angles theorem:
This theorem states that if two lines are intersected by a transversal and the alternate interior angles are congruent, lines will be parallel.
Therefore, p║n.
Perpendicular transversal theorem:
By this theorem,
In a plane, if a line is perpendicular to one of two parallel lines, then it will be perpendicular to the other line.
Therefore, l ⊥ p.
Options selected in the blank spaces are correct.
Sorry don't follow what I just said I misread the question.
