Diketahui tan A=3/4,tan B=5/12, A dan B sudut lancip. Nilai cos(A+B)=…

Halo Edwin, kaka bantu jawab ya... Jawaban: 33/65 Ingat kembali: ▶︎ tan x = depan x/samping x ▶︎ sin x = depan x /miring ▶︎ cos x = samping x/miring ▶︎ cos(A + B) = cosA.cosB - sinA.sinB ▶︎ teorema pythagoras c = √(a²+b²), dimana c: sisi miring dan a,b panjang sisi yang lain. Diketahui: *)tan A = 3/4 depan A = 3 samping A = 4 miring = √(3²+4²) -------> = √(9+16) -------> = √25 -------> = 5 Diperoleh: cos A = 4/5 sin A = 3/5 *)tan B=5/12 depan B = 5 samping B = 12 miring = √(5²+12²) -------> = √(25+144) -------> = √169 -------> = 13 Diperoleh: cos B = 12/13 sin B = 5/13 cos(A + B) = cosA . cosB - sinA . sinB cos(A + B) = (4/5)(12/13) - (3/5)(5/13) cos(A + B) = 48/65 - 15/65 cos(A + B) = 33/65 Jadi, nilai cos(A + B) = 33/65.