It is given that [tex]$\angle[/tex] [tex]A+\angle B=1080$[/tex]
We know that the sum of all the angles in a triangle i[tex]s $180 \%$[/tex]
So we can write it as
[tex]\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180 \circ[/tex]
By substituting[tex]$\angle A+\angle B=108 \circ$[/tex]
in the above equation
[tex]$108 \circ+\angle C=180 \circ$[/tex]
On further calculation
[tex]$\angle \mathrm{C}=180 \text { o-108० }$[/tex]
By subtraction
[tex]$\angle \mathrm{C}=720$[/tex]
It is given that [tex]$\angle B+\angle C=130$ o[/tex]
By substituting the value of [tex]$\angle C$$\angle B+72 o=130 \%$[/tex]
It is given that [tex]$\angle B+\angle C=130$ o[/tex]
By substituting the value of [tex]$\angle C$\angle B+72 \circ=130 \circ$On further calculation$\angle B=130 \circ-72 \circ$[/tex]
By subtraction
[tex]$B=580$[/tex]
By substituting [tex]$\angle B=58 \circ$[/tex]
in equation (1)
So we get $[tex]\angle A+\angle B=108$ o[/tex]
[tex]$\angle A+58{ }^{\circ}=108 \circ$[/tex]
On further calculation [tex]$\angle A=108 \mathrm{o}-58 \mathrm{o}$[/tex]
By subtraction [tex]$\angle A=50$ 。[/tex]
Therefore, [tex]$\angle A=50 \circ, \angle B=58 \circ$ and $\angle C=72^{\circ}$.[/tex]
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