Odgovor:
Glejte spodnjo razlago
Pojasnilo:
# 6sinA + 8cosA = 10 #
Razdelitev obeh strani z #10#
# 3 / 5sinA + 4 / 5cosA = 1 #
Let # cosalpha = 3/5 # in # sinalpha = 4/5 #
# cosalpha = cosalpha / sinalpha = (3/5) / (4/5) = 3/4 #
Zato, # sinAcosalpha + sinalphacosA = sin (A + alpha) = 1 #
Torej, # A + alpha = pi / 2 #, #mod 2pi #
# A = pi / 2-alfa #
# tanA = tan (pi / 2-alfa) = cotalpha = 3/4 #
# tanA = 3/4 #
# QED #
Odgovor:
glej spodaj.
Pojasnilo:
# ali, 6sinA - 10 = -8cosA #
#ali, (6sinA -10) ^ 2 = (-8cosA) ^ 2 #
# ali, 36sin ^ 2A- 2 * 6sinA * 10 + 100 = 64cos ^ 2A #
# ali, 36sin ^ 2A - 120sinA + 100 = 64cos ^ 2A #
#ali, 36sin ^ 2A - 120sinA + 100 = 64 (1 - sin ^ 2A) #
#ali, 36sinA - 120sinA +100 = 64 - 64Sin ^ 2A #
# ali, 100 sin ^ 2A - 120SinA + 36 = 0 #
#ali, (10sinA-6) ^ 2 = 0 #
#ali, 10sinA - 6 = 0 #
ali ali SinA = 6/10 #
# ali, SinA = 3/5 = p / h #
Z uporabo Pitagorjevega izreka dobimo
# b ^ 2 = h ^ 2 - p ^ 2 #
#or, b ^ 2 = 5 ^ 2 - 3 ^ 2 #
#ali, b ^ 2 = 25 - 9 #
#ali, b ^ 2 = 16 #
#ali, b = 4 #
# so, TanA = p / b = 3/4 #
Je ta odgovor pravilen?
Odgovor:
glej rešitev
Pojasnilo:
# 6sinA + 8cosA = 10 #
delitev obeh strani z #sqrt (6 ^ 2 + 8 ^ 2) #=#10#
# (6sinA) / 10 + 8cosA / 10 = 10/10 = 1 #
# cosalphasinA + sinalphacosA #=1
kje # tanalpha = 4/3 # ali # alpha = 53degree #
to se spremeni v
#sin (alpha + A) = sin90 #
#alpha + A = 90 #
# A = 90-alpha #
pri sprejemanju # tan #obe strani
# 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 #
# barva (rdeča) (sin ^ 2A + cos ^ 2A = 1) #
# => (3/5) sinA + (4/5) cosA = sinA * sinA + cosA * cosA #
# => sinA = 3/5 in cosA = 4/5 #
Zato #tanA = sinA / cosA = (3/5) / (4/5) = (3/5) × (5/4) = 3/4 #
