One angle of a triangle is $78^{o}$ and the other two angles are in the ratio 7 : 10. Calculate the bigger angle of the triangle.

Open in App

Solution

Two angles of the triangle are in ratio 7 : 10
Let the angles be $7 x$ and $10 x$
Now, sum of angles of triangle = 180
$78 + 7 x + 10 x = 180$
$17 x = 102$
$x = 6$
Thus, angles are $42^{\circ}$ and $60^{\circ}$

Suggest Corrections

Similar questions

Q. In a right angled triangle, one angle is

78^{o} .

Find the other angle.

One angle of a triangle is $60^{\circ}$ and the other two angles are in the ratio 2:3. Find the smallest angle in the triangle.

Q. If one angle of a triangle is

40^{\circ}

and the other two angles are in the ratio 3:4. Then the largest angle in the triangle is

One angle of a triangle is $61^{\circ}$ and the other two angles are in the ratio $(\frac{3}{2}) : (\frac{4}{3})$ . Find the smallest angle in the triangle.

Q. One angle of a triangle is

82^{\circ}

and the other two are in the ratio 3 : 4. Find the value of other two angles.

Join BYJU'S Learning Program

One angle of a triangle is 78o and the other two angles are in the ratio 7 : 10. Calculate the bigger angle of the triangle.

Two angles of the triangle are in ratio 7 : 10Let the angles be 7x and 10xNow, sum of angles of triangle = 18078+7x+10x=18017x=102x=6Thus, angles are 42∘ and 60∘

One angle of a triangle is $78^{o}$ and the other two angles are in the ratio 7 : 10. Calculate the bigger angle of the triangle.

Two angles of the triangle are in ratio 7 : 10
Let the angles be $7 x$ and $10 x$
Now, sum of angles of triangle = 180
$78 + 7 x + 10 x = 180$
$17 x = 102$
$x = 6$
Thus, angles are $42^{\circ}$ and $60^{\circ}$