Answer:(4.5, 6.5)
step-by-step explanation: this is the formular for finding a midpoint of a line.
- <u>midpoint=x1+y1?|/2 ,x2+y2/2</u>
Answer:
4cotα=tanα
4(1/tanα)=tanα
(4/tanα)=tanα
cross multiply
=> 4=tan²α
√4=√tan²α
±2=tanα
α=arc( tan) |2|
α=63.4° ( in first quadrant)
and
α=180+63.4=243.4 in the third quadrant
since we also found a negative answer( i.e –2) then α also lies in quadrants where it gives a negative value(i.e second and fourth quadrants)
α=180–63.4=116.6° in the second quadrant
α=360–63.4=296.6 in the fourth quadrant
therefore theta( in my case, alpha) lies in all four quadrants and is equal to:
α=63.4°,243.4°,116.6°and 296.6°
Given :
The diagonals of rhombus ABCD intersect at E.
∠CAD = 20°.
To Find :
The angle ∠CDA.
Solution :
We know, diagonals of a rhombus bisects each other perpendicularly.
So, ∠DEA = 90°.
In triangle ΔEAD :
∠EAD + ∠AED + ∠EDA = 180°
20° + 90° + ∠EDA = 180°
∠EDA = 70°
Now, we know diagonal of rhombus also bisect the angle between two sides .
So, ∠CDA = 2∠EDA
∠CDA = 2×70°
∠CDA =140°
Therefore, ∠CDA is 140°.
Answer:
The answer is 12, subtract 17 from 10 to get 7, then add 7 to 5