Paano ko malulutas ito?

Tulad ng X ay katumbas (5m) mula sa tatlong vertices ng tatsulok # ABC #, X ay ang circumcentre ng # DeltaABC #

Kaya # angleBXC = 2 * angleBAC #

Ngayon

# BC ^ 2 = XB ^ 2 + XC ^ 2-2XB * XC * cosangleBXC #

# => BC ^ 2 = 5 ^ 2 + 5 ^ 2-2 * 5 ^ 2 * cos / _BXC #

# => BC ^ 2 = 2 * 5 ^ 2 (1-cos (2 * / _ BAC) #

# => BC ^ 2 = 2 * 5 ^ 2 * 2sin ^ 2 / _BAC #

# => BC=10sin/_BAC=10sin80 ^ @=9.84m#

Katulad nito

#AB=10sin/_ACB=10sin40 ^ @=6.42m#

At

#AC=10sin/_ABC=10*sin60 ^ @=8.66m#

Sagot:

# AB ~~ 6.43m #

# BC ~~ 9.89m #

# AC ~~ 8.66m #

Paliwanag:

Maaari naming malutas ito gamit ang bilog teorama:

Alam namin iyan # XA = XB = XC = 5m # samakatuwid ang tatlong panig ay ang lahat ng radii ng isang bilog na may isang radius ng # 5m #

Samakatuwid, alam natin:

# 2 / _BCA = / _ BXA #

# 2 / _ABC = / _ AXC #

# 2 / _BAC = / _ BXC #

# / _ BXC = 2 (80) = 160 #

# / _ AXC = 2 (60) = 120 #

# / _ BXA = 2 (40) = 80 #

Gamit ang cosine alam namin na:

# c ^ 2 = a ^ 2 + b ^ 2-2bacosC #

# c = sqrt (a ^ 2 + b ^ 2-2bacosC) #

# AB = sqrt (AX ^ 2 + XB ^ 2-2 (AX) (XB) cos (/ _ AXB)) #

#color (white) (AB) = sqrt (5 ^ 2 + 5 ^ 2-2 (5 ^ 2) cos (80)) #

#color (white) (AB) ~~ 6.43m #

# BC = sqrt (BX ^ 2 + XC ^ 2-2 (BX) (XC) cos (/ _ BXC)) #

#color (white) (BC) = sqrt (5 ^ 2 + 5 ^ 2-2 (5 ^ 2) cos (160)) #

#color (white) (BC) ~~ 9.89m #

# AC = sqrt (AX ^ 2 + XC ^ 2-2 (AX) (XC) cos (/ _ AXC)) #

#color (white) (AC) = sqrt (5 ^ 2 + 5 ^ 2-2 (5 ^ 2) cos (120)) #

#color (white) (AC) ~~ 8.66m #

Mga gilid:

# AB ~~ 6.43m #

# BC ~~ 9.89m #

# AC ~~ 8.66m #