if A,B,C,Dare angles of a cyclic quadrilateral .prove that cosA+cosB+cosC+cosD=0​

Question

if A,B,C,Dare angles of a cyclic quadrilateral .prove that cosA+cosB+cosC+cosD=0​

in progress 0
Natalia 3 weeks 2021-09-04T18:24:43+00:00 1 Answer 0 views 0

Answers ( )

    0
    2021-09-04T18:25:47+00:00

    Answer:

    Here is your answer mate

    Step-by-step explanation:

    If A,B, C,D are cylic quadrilateral then prove that cosA+cosB+cosC+cosD=0?

    Assume that the given cyclic quadrilateral ABCD is convex. Then we know from one of its many properties that, the opposite angles of ABCD are supplementary, that is

    A + C = π = 180°……………….………………….(1)

    B + D = π = 180°……………….………………….(2)

    Transposing C from left to right in eq. (1),

    A = π – C

    Taking cosines on both sides,

    cos A = cos (π – C) = cos π . cos C + sin π . sin C = -1 .cos C + 0.sin C = – cos C

    Or, cos A + cos C = 0………………………..……(3)

    Similarly it can be shown from eq.(2) that

    cos B + cos D = 0……………………..……………(4)

    Adding eq.(3) and eq.(4), we get

    cos A + cos B + cos C + cos D = 0 (Proved)

    Please follow and mark as brainliest

Leave an answer

Browse
Browse

18:9+8+9*3-7:3-1*13 = ? ( )