# Angles of a Triangle

I'm redoing some high school math. I'm having trouble thinking through this question.

Question:

The second angle of a triangle is three times as large as the first. The measure of the third angle is 25° greater than that of the first angle. How large are the angles?

What would be a good way to figure this out?

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Thank you both for the great answers. I will accept one and give both +1 when reputation score and time permits. Cheers –  BAQU Aug 11 at 19:41

First angle: $A_1=x$

Second angle: $A_2=3x$

Third angle: $A_3=A_1+25^\circ=x+25^\circ$

Note that the three angles in a triangle must add up to $180^\circ$. You have $$A_1+A_2+A_3=180^\circ$$ or

\begin{align} (x)+(3x)+(x+25^\circ)&=180^\circ \\ 5x&=155^\circ \\ x&=31^\circ \end{align}

Substitute the value of $x$ into your angle expressions and you will get:

$$\boxed{A_1=31^\circ,A_2=93^\circ,A_3=56^\circ}$$

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There's no point in naming things twice. It would probably be simpler to just name them $A_i$ or $x$. –  Eul Can Aug 11 at 19:35
its not twice, but 3 times. –  MonK Aug 11 at 19:39
MonK means that $A_2=3x$ not $2x$. –  Eul Can Aug 11 at 19:40
x+(3x)+(x+25)=180 sould evaluate to 5x = 155 and in turn x = 31. Therefor will angle 1 will be 31°, angle 2 with 93° and angle 3 with 56°? Am I off? –  BAQU Aug 11 at 20:17
Ok. Thank you for the clarification @dlev. –  BAQU Aug 11 at 20:19

$$x + \underbrace{\Big(3x\Big)}_{\text{second angle}} + \underbrace{\Big(x+25\Big)}_{\text{third angle}} = 180.$$ That is an algebraic equation that can be solved for $x$.

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