Answer:
Step-by-step explanation:
Given: quadrilateral ABCD inscribed in a circle
To Prove:
1. ∠A and ∠C are supplementary.
2. ∠B and ∠D are supplementary.
Construction : Join AC and BD.
Proof: As, angle in same segment of circle are equal.Considering AB, BC, CD and DA as Segments, which are inside the circle,
∠1=∠2-----(1)
∠3=∠4-----(2)
∠5=∠6-------(3)
∠7=∠8------(4)
Also, sum of angles of quadrilateral is 360°.
⇒∠A+∠B+∠C+∠D=360°
→→∠1+∠2+∠3+∠4+∠5+∠6+∠7+∠8=360°→→→using 1,2,3,and 4
→→→2∠1+2∠4+2∠6+2∠8=360°
→→→→2( ∠1 +∠6) +2(∠4+∠8)=360°⇒Dividing both sides by 2,
→→→∠B + ∠D=180°as, ∠1 +∠6=∠B , ∠4+∠8=∠B------(A)
As, ∠A+∠B+∠C+∠D=360°
∠A+∠C+180°=360°
∠A+∠C=360°-180°------Using A
∠A+∠C=180°
Hence proved.
credit: someone else
Answer:
- EF = 4.1
- DE = 9.1
- m∠F = 66°
Step-by-step explanation:
The hypotenuse and one acute angle are given. The relevant relations are ...
Sin = Opposite/Hypotenuse
Cos = Adjacent/Hypotenuse
__
For the given triangle, these tell us ...
sin(24°) = EF/DF = EF/10
EF = 10·sin(24°) ≈ 4.1
and ...
cos(24°) = DE/DF = DE/10
DE = 10·cos(24°) ≈ 9.1
The remaining acute angle is the complement of the given one:
F = 90° -D = 90° -24°
∠F = 66°
Answer:
0.4374
Step-by-step explanation:
B the answer is B - hope this helps you out
