In Delta ABC,angle A = 2x - 10^{circ},angle B | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The sum of the angles in a triangle is $180^{\circ}$.Therefore,$\angle A + \angle B + \angle C = 180^{\circ}$.Substituting the given values: $(2x

Vedclass · Accepted Answer

$\angle A = 70^{\circ}, \angle B = 50^{\circ}, \angle C = 60^{\circ}$

Vedclass · Answer

(A) The sum of the angles in a triangle is $180^{\circ}$.Therefore,$\angle A + \angle B + \angle C = 180^{\circ}$.Substituting the given values: $(2x - 10^{\circ}) + (x + 10^{\circ}) + (2x - 20^{\circ}) = 180^{\circ}$.Combining like terms: $5x - 20^{\circ} = 180^{\circ}$.$5x = 200^{\circ}$.$x = 40^{\circ}$.Now,calculate each angle:$\angle A = 2(40^{\circ}) - 10^{\circ} = 80^{\circ} - 10^{\circ} = 70^{\circ}$.$\angle B = 40^{\circ} + 10^{\circ} = 50^{\circ}$.$\angle C = 2(40^{\circ}) - 20^{\circ} = 80^{\circ} - 20^{\circ} = 60^{\circ}$.Thus,the angles are $\angle A = 70^{\circ}, \angle B = 50^{\circ}, \angle C = 60^{\circ}$.

In $\Delta ABC$,$\angle A = 2x - 10^{\circ}$,$\angle B = x + 10^{\circ}$,and $\angle C = 2x - 20^{\circ}$. Find the measure of each angle of $\Delta ABC$.

Similar Questions

In the given figure,if $PQ || RS$,$RS || MN$ and $y: z = 7: 11$,then find $x$. (in $^o$)

In $\Delta PQR$,$\angle P = 42^{\circ}$ and $\angle Q = 75^{\circ}$,then find $\angle R$. (in $^{\circ}$)

In the given figure,find the values of $x$ and $y$ and then show that $PQ \parallel XY.$

Prove that two lines that are respectively perpendicular to two intersecting lines intersect each other.

$A$ transversal intersects two parallel lines. Prove that the bisectors of any pair of corresponding angles so formed are parallel.

Vedclass Test Series

Exam Paper Generator

Online Exam Module