Bizzare problem. What are the choices?
The triangle angles add to 180.
4x + 7 + 2x +3 + 6x - 10 = 180
12x = 180
x = 15
Dilation doesn't change the angles so we must have
4x + 7 = 5x - 8 and 2x + 3 = 3x - 12 and 6x - 10 = 7x - 25
Fortunately x=15 is the solution to all of these so it's consistent.
Answer:
The value of x is 15.
Step-by-step explanation:
It is given that in ΔABC, ∠A = 4x + 7, ∠B = 2x + 3, and ∠C = 6x - 10.
According to angle sum property of triangle, the sum of all interior angles of a triangle is 180°.
In ΔABC,
[tex]\angle A+\angle B+\angle C=180[/tex]
[tex](4x+7)+(2x+3)+(6x-10)=180[/tex]
Combine like terms.
[tex](4x+2x+6x)+(7+3-10)=180[/tex]
[tex]12x=180[/tex]
Divide both sides by 12.
[tex]x=15[/tex]
The value of x is 15.
[tex]\angle A=4x+7\Rightarrow 4(15)+7=67^{\circ}[/tex]
[tex]\angle B=2x+3\Rightarrow 2(15)+3=33^{\circ}[/tex]
[tex]\angle C=6x-10\Rightarrow 6(15)-10=80^{\circ}[/tex]
[tex]\angle A'=5x-8\Rightarrow 5(15)-8=67^{\circ}[/tex]
[tex]\angle B'=3x-12\Rightarrow 3(15)-12=33^{\circ}[/tex]
[tex]\angle C'=7x-25\Rightarrow 7(15)-25=80^{\circ}[/tex]
We get
[tex]\angle A\cong \angle A'[/tex]
[tex]\angle B\cong \angle B'[/tex]
Two corresponding angles of triangles are congruent. By AA criterion
[tex]\triangle ABC\cong \triangle A'B'C'[/tex]
Hence proved.
