segitiga klm memiliki koordinat k 5

Pembahasan

Diketahui :

Ditanya : cos open parentheses straight L close parentheses space dan space tan open parentheses straight M close parentheses ?

Jawab:

  • Menentukan panjang K L comma space L M comma space M K space.

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar K L close vertical bar end cell equals cell open vertical bar L minus K close vertical bar end cell row blank equals cell square root of open parentheses 3 plus 5 close parentheses squared plus left parenthesis negative 2 plus 2 right parenthesis squared end root end cell row blank equals cell square root of 8 squared end root end cell row blank equals 8 row cell open vertical bar L M close vertical bar end cell equals cell open vertical bar M minus L close vertical bar end cell row blank equals cell square root of open parentheses negative 5 minus 3 close parentheses squared plus left parenthesis 4 plus 2 right parenthesis squared end root end cell row blank equals cell square root of left parenthesis negative 8 right parenthesis squared plus left parenthesis 6 right parenthesis squared end root end cell row blank equals cell square root of 64 plus 36 end root end cell row blank equals cell square root of 100 end cell row blank equals 10 row cell open vertical bar M K close vertical bar end cell equals cell open vertical bar K minus M close vertical bar end cell row blank equals cell square root of open parentheses negative 5 plus 5 close parentheses squared plus left parenthesis negative 2 minus 4 right parenthesis squared end root end cell row blank equals cell square root of left parenthesis negative 6 right parenthesis squared end root end cell row blank equals 6 end table

  • Menentukan c o s open parentheses L close parentheses space dan space t a n open parentheses M close parentheses

table attributes columnalign right center left columnspacing 0px end attributes row cell cos open parentheses L close parentheses end cell equals cell samping over miring end cell row blank equals cell fraction numerator K L over denominator L M end fraction end cell row blank equals cell 8 over 10 end cell row blank equals cell 4 over 5 end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row cell tan open parentheses M close parentheses end cell equals cell depan over samping end cell row blank equals cell fraction numerator K L over denominator M K end fraction end cell row blank equals cell 8 over 6 end cell row blank equals cell 4 over 3 end cell end table

Oleh karena itu, jawaban yang benar adalah A. 

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