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(SNMPTN 2012) Lingkaran ( x − 3 ) 2 + ( y − 4 ) 2 = 25 memotong sumbu-X di titik A dan B. jika P adalah titik pusat lingkaran tersebut, maka cos ∠ A PB = ….

(SNMPTN 2012)

Lingkaran  memotong sumbu-X di titik A dan B. jika P adalah titik pusat lingkaran tersebut, maka  ….

  1. 7 over 25

  2. 8 over 25

  3. 12 over 25

  4. 16 over 25

  5. 18 over 25

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I. Roy

Master Teacher

Mahasiswa/Alumni Universitas Negeri Surabaya

Jawaban terverifikasi

Pembahasan

begin mathsize 12px style left parenthesis x minus 3 right parenthesis squared plus left parenthesis y minus 4 right parenthesis squared equals 25 space P u s a t space equals space left parenthesis 3 comma space 4 right parenthesis space d a n space j a r i minus j a r i space equals space 5 space space  M e n c a r i space t i t i k space p o t o n g space l i n g k a r a n space d a n space s u m b u minus X comma m a k a space y space equals space 0 comma space s e h i n g g a left parenthesis x minus 3 right parenthesis squared plus left parenthesis 0 minus 4 right parenthesis squared equals 25 space left parenthesis x minus 3 right parenthesis squared plus 16 equals 25 space left parenthesis x minus 3 right parenthesis squared equals 9 space x minus 3 equals plus-or-minus 3 space x equals 0 space a t a u space x equals 6 space S e h i n g g a space t i t i k space p o t o n g space l i n g k a r a n space t e r h a d a p space s u m b u minus X a d a l a h space d i space t i t i k space A left parenthesis 0 comma space 0 right parenthesis space d a n space B left parenthesis 6 comma space 0 right parenthesis. space  P e r h a t i k a n space g a m b a r space d i space b a w a h space i n i. end style

begin mathsize 12px style P a n j a n g space A P space equals space B P space equals space j a r i minus j a r i space l i n g k a r a n space equals space 5 space P a n j a n g space A B space equals space j a r a k space t i t i k space left parenthesis 0 comma space 0 right parenthesis space k e space t i t i k space left parenthesis 6 comma space 0 right parenthesis space space equals square root of left parenthesis 6 minus 0 right parenthesis squared plus left parenthesis 0 minus 0 right parenthesis squared end root space space equals square root of 36 plus 0 end root space space equals square root of 36 space space equals 6 space space S e h i n g g a space b e s a r space angle A P B space b i s a space d i t e n t u k a n space d e n g a n colon cos angle A P B equals fraction numerator A P squared plus P B squared minus A B squared over denominator 2. A P. P B end fraction space equals fraction numerator 5 squared plus 5 squared minus 6 squared over denominator 2.5.5 end fraction equals fraction numerator 25 plus 25 minus 36 over denominator 50 end fraction space space equals 14 over 50 space space equals 7 over 25 J a d i comma space space cos angle A P B equals 7 over 25 end style

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(SBMPTN 2016) Titik (0 , b ) adalah titik potong garis singgung persekutuan luar lingkaran x 2 + y 2 = 16 dan ( x − 8 ) 2 + ( y − 8 ) 2 = 16 dengan sumbu-Y. Nilai b adalah ….

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