Question 261100
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We know that angles in a triangle equal 180 deg.


Then it follows, {{{A+B+C=180^o}}}, Working Eqn.


Let us see the conditions:
Angle B is 3 times angle A: {{{B=3A}}}, condt'n 1


Angle C is 18 less than 6 imes angle A: {{{C=6A-18}}}, condt'n 2

Substitute the conditions in our Working Eqn:
{{{A+red(3A)+red(6A-18)=180^o}}}
{{{10A=180+18=198}}} ---> {{{cross(10)A/cross(10)=cross(198)19.8/cross(10)}}}
{{{red(A=19.8^o)}}}, Answer


Let's check by substituting in the condt'ns.
Condt'n 1: {{{B=3(19.8)=59.4^o}}}
Condt'n 2: {{{C=6(19.8)-18=-118.8-18=100.8^o}}}


Then via Working Eqn:
{{{A+B+C=180^o}}}
{{{19.8+59.4+100.8=180^o}}}
{{{180^o=180^o}}}


Thank you,
Jojo</font>