Question

In a triangle, the sum of two angles is ${118}^{\circ }$ and their difference is ${32}^{\circ }$; find each angle of the triangle.

• A. ${75}^{\circ },{43}^{\circ }$ and ${62}^{\circ }$
• B. ${70}^{\circ },{48}^{\circ }$ and ${62}^{\circ }$
• C. ${85}^{\circ },{43}^{\circ }$ and ${52}^{\circ }$
• D. ${75}^{\circ },{53}^{\circ }$ and ${52}^{\circ }$

Answer 1

The correct option is A ${75}^{\circ },{43}^{\circ }$ and ${62}^{\circ }$

Let the angles of the triangle be $a,b,c$
We know, $a+b+c={180}^o$
Given, sum of two angles, $a+b={118}^o$...(I)
Difference between two angles, $a-b={32}^o$...(II)
Adding I and II, we get
$2a={150}^o$
$\therefore a={75}^{\circ }$
Put $a={75}^o$ in equation I,
$b={118}^o-{75}^o$
$\therefore b={43}^{\circ }$
Sum of the angles of a triangle = 180
$a+b+c={180}^o$
${75}^o+{43}^o+c={180}^o$
$c={180}^o-{118}^o$
$c={62}^{\circ }$