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The angles in a triangle are in a ratio of 19 : 10 : 7. What is the sum of thrice the smallest angle and the twice the largest angle ?
- A275°
- B295°
- C280°
- D273°
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0
CCommunity
Solution
Let the angles of the triangle be $19x , 10x , 7x$
=> Sum of angles = $19x + 10x + 7x = 180^{\circ}$
=> $36x = 180^{\circ}$
=> $x = \frac{180}{36} = 5^{\circ}$
=> Angles are = $95^{\circ} , 50^{\circ} , 35^{\circ}$
$\therefore$ Sum of thrice the smallest angle and the twice the largest angle
= $(3 \times 35^{\circ}) + (2 \times 95^{\circ})$
= $105 + 190 = 295^{\circ}$
Let the angles of the triangle be $19x , 10x , 7x$
=> Sum of angles = $19x + 10x + 7x = 180^{\circ}$
=> $36x = 180^{\circ}$
=> $x = \frac{180}{36} = 5^{\circ}$
=> Angles are = $95^{\circ} , 50^{\circ} , 35^{\circ}$
$\therefore$ Sum of thrice the smallest angle and the twice the largest angle
= $(3 \times 35^{\circ}) + (2 \times 95^{\circ})$
= $105 + 190 = 295^{\circ}$
0
CCommunity
Solution
Let the angles of the triangle be $19x , 10x , 7x$
=> Sum of angles = $19x + 10x + 7x = 180^{\circ}$
=> $36x = 180^{\circ}$
=> $x = \frac{180}{36} = 5^{\circ}$
=> Angles are = $95^{\circ} , 50^{\circ} , 35^{\circ}$
$\therefore$ Sum of thrice the smallest angle and the twice the largest angle
= $(3 \times 35^{\circ}) + (2 \times 95^{\circ})$
= $105 + 190 = 295^{\circ}$
Let the angles of the triangle be $19x , 10x , 7x$
=> Sum of angles = $19x + 10x + 7x = 180^{\circ}$
=> $36x = 180^{\circ}$
=> $x = \frac{180}{36} = 5^{\circ}$
=> Angles are = $95^{\circ} , 50^{\circ} , 35^{\circ}$
$\therefore$ Sum of thrice the smallest angle and the twice the largest angle
= $(3 \times 35^{\circ}) + (2 \times 95^{\circ})$
= $105 + 190 = 295^{\circ}$