2.diketahui sin A=3/5,dan Cos B= – 5/13 dimana 0 derajat < A < 90 derajat dan 90 derajat < B <

Pertanyaan:

2.diketahui sin A=3/5,dan Cos B= – 5/13 dimana 0 derajat < A < 90 derajat dan 90 derajat < B < 180 derajat .nilai dari sin ( A-B ) adalah

a. – 63/65b.- 33/65c. 14/65d. 33/65e. 63/65​

Diketahui sin ( A ) = 3 / 5 sine open paren cap A close paren equals 3 / 5 s i n ( A ) = 3 / 5 dengan 0 ∘ < A < 90 ∘ 0 raised to the composed with power is less than cap A is less than 90 raised to the composed with power 0 ∘ < A < 9 0 ∘ (kuadran I). Nilai cos ( A ) cosine open paren cap A close paren c o s ( A ) dihitung menggunakan identitas trigonometri sin 2 ( A ) + cos 2 ( A ) = 1 sine squared open paren cap A close paren plus cosine squared open paren cap A close paren equals 1 s i n 2 ( A ) + c o s 2 ( A ) = 1 : cos ( A ) = 1 − sin 2 ( A ) = 1 − ( 3 / 5 ) 2 = 1 − 9 / 25 = 16 / 25 = 4 / 5 cosine open paren cap A close paren equals the square root of 1 minus sine squared open paren cap A close paren end-root equals the square root of 1 minus open paren 3 / 5 close paren squared end-root equals the square root of 1 minus 9 / 25 end-root equals the square root of 16 / 25 end-root equals 4 / 5 c o s ( A ) = 1 − s i n 2 ( A ) √ = 1 − ( 3 / 5 ) 2 √ = 1 − 9 / 2 5 √ = 1 6 / 2 5 √ = 4 / 5 Karena A cap A A di kuadran I, cos ( A ) cosine open paren cap A close paren c o s ( A ) bernilai positif.