Solve triangle ABC. Show prosedure?

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1 Answer
Apr 27, 2018

#" "#
#/_BAC = 53^@, /_ABC=90^@ and /_ACB=37^@#

#bar(AB)="12 Units", bar(BC) = "16 Units", bar(AC)= "20 Units"#

Explanation:

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Sum of the internal angles of a triangle add to #color(blue)(180^@#

#color(green)("Step 1")#

Hence,

#(4x+1)^@ + (7x-1)^@ + (3x-2)^@ = 180^@#

#4x+1+7x-1+3x-2 = 180#

Combine the like terms:

#(4x+7x+3x)+cancel 1-cancel1-2=180#

#14x+cancel 1-cancel1-2=180#

#14x-2=180#

Add 2 to both sides of the equation.

#14x-cancel 2+cancel 2=180+2#

#14x=182#

Divide both sides of the equation by #14.#

#(14x)/14=182/14#

#(cancel 14x)/cancel14=cancel 182^color(red)(13)/cancel 14#

Hence #color(blue)(x = 13#

Substitute #color(blue)(x = 13# in

#color(green)((4x+1)^@ + (7x-1)^@ + (3x-2)^@ = 180^@# to obtain angles #A, B, C#

#A=[4(13)+1)]^@=(52+1)^@ = 53^@#

#B=[7(13)-1)]^@=(91-1)^@ = 90^@#

#C=[3(13)-2)]^@=(39-2)^@ = 37^@#

Hence, the angles are: #color(red)(A= 53^@, B= 90^@, C=37^@#, triangle #ABC# is a right-triangle.

#color(green)("Step 2")#

We have,
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Since, #color(blue)("Angle B = 90"^@#, we can use the Pythagoras Theorem to find the magnitudes.

#AC# is the Hypotenuse, as it is the opposite side of #B#.

Hence,

#AC^2=AB^2+BC^2#

#(3y-1)^2 = (y+5)^2+(2y+2)^2" "# #[1]#

Use the Identities:

#(a+b)^2 -= a^2+2ab+b^2#

#(a-b)^2 -= a^2-2ab+b^2#

#[1]# becomes

#(9y^2-6y+1)=(y^2+10y+25)+(4y^2+8y+4)#

#9y^2-6y+1=y^2+10y+25+4y^2+8y+4#

Combine like terms:

#9y^2-y^2-4y^2-6y-10y-8y+1-25-4=0#

#4y^2-24y-28=0#

Divide both sides by 4.

#(cancel 4y^2)/cancel 4-(cancel 24^color(red)(6)y)/cancel 4-cancel 28^color(red)(7)/cancel 4=0#

#y^2-6y-7=0#

#y^2-7y+1y-7=0#

#y(y-7)+1(y-7)=0#

#(y-7)(y+1)=0#

#y=7, y=-1#

Ignore #y=-1#, as it would make the side #2y+2=0#

Consider #y=7.#

#AB=y+5=7+5=12#

#BC=2y+2=2*7+2=16#

#AC=3y-1=3*7-1=20#

Results:

#/_BAC = 53^@, /_ABC=90^@ and /_ACB=37^@#

#bar(AB)="12 Units", bar(BC) = "16 Units", bar(AC)= "20 Units"#