Antwoord:
Zie de uitleg hieronder
Uitleg:
# 6sinA + 8cosA = 10 #
Beide kanten verdelen #10#
# 3 / 5sinA + 4 / 5cosA = 1 #
Laat # Cosalpha = 3/5 # en # Sinalpha = 4/5 #
# Cosalpha = cosalpha / sinalpha = (3/5) / (05/04) = 3/4 #
daarom
# SinAcosalpha + sinalphacosA = sin (A + a) = 1 #
Zo, # A + a = pi / 2 #, #mod 2pi #
# A = pi / 2-alpha #
# Tana = tan (pi / 2-a) = cotalpha = 3/4 #
# Tana = 3/4 #
# QED #
Antwoord:
zie hieronder.
Uitleg:
# of, 6sinA - 10 = -8cosA #
#of, (6sinA -10) ^ 2 = (-8cosA) ^ 2 #
#of, 36sin ^ 2A- 2 * 6sinA * 10 + 100 = 64cos ^ 2A #
#of, 36sin ^ 2A - 120sinA + 100 = 64cos ^ 2A #
#of, 36sin ^ 2A - 120sinA + 100 = 64 (1 - sin ^ 2A) #
#of, 36sinA - 120sinA +100 = 64 - 64Sin ^ 2A #
#of, 100 sin ^ 2A - 120SinA + 36 = 0 #
#of, (10sinA-6) ^ 2 = 0 #
#of, 10sinA - 6 = 0 #
#of, SinA = 6/10 #
# of, SinA = 3/5 = p / h #
Met behulp van de stelling van Pythagoras, krijgen we
# b ^ 2 = h ^ 2 - p ^ 2 #
#of, b ^ 2 = 5 ^ 2 - 3 ^ 2 #
#of, b ^ 2 = 25 - 9 #
#of, b ^ 2 = 16 #
#of, b = 4 #
# so, TanA = p / b = 3/4 #
Is dit antwoord juist?
Antwoord:
zie oplossing
Uitleg:
# 6sinA + 8cosA = 10 #
door beide kanten te verdelen #sqrt (6 + 2 ^ 8 ^ 2) #=#10#
# (6sinA) / 10 + 8cosA / 10 = 10/10 = 1 #
# CosalphasinA + sinalphacosA #=1
waar # Tanalpha = 4/3 # of # Alpha = 53degree #
dit transformeert naar
#sin (alfa + A) = sin90 #
#alpha + A = 90 #
# A = 90-alpha #
nemen #bruinen#beide kanten
# Tana = tan (90-alfa) #
# Tana = cotalpha #
# Tana = 3/4 #
# 6sinA + 8cosA = 10 #
# => 3sinA + 4cosA = 5 #
# => (3/5) sinA + (4/5) cosA = 1 #
# => (3/5) sinA + (4/5) cosA = (sinA) ^ 2 + (cosA) ^ 2 #
# kleur (rood) (sin ^ 2A + cos ^ 2A = 1) #
# => (3/5) sinA + (4/5) cosA = sinA * sinA + cosA * cosA #
# => sinA = 3/5 en cosA = 4/5 #
Vandaar, #tanA = sinA / cosA = (3/5) / (4/5) = (3/5) × (5/4) = 3/4 #
