Calculate, with reasons, the angles marked x and y.

Question

ScanSolve AI Tutor · Accepted Answer

Here are the solutions for the geometry problems:

** (Top left triangle, labeled "1.2")**

The sum of angles in a triangle is $180^\circ$ .
$(3x - 15^\circ) + 62^\circ + 34^\circ = 180^\circ$ $3x - 15 + 96 = 180$ $3x + 81 = 180$
Solve for $x$ .
$3x = 180 - 81$ $3x = 99$ $x = \frac{99}{3}$ $x = \boxed{33}$
** (Top right triangle)**
Angles on a straight line sum to $180^\circ$ .
The interior angle $x$ and the exterior angle $130^\circ$ form a linear pair.
$x + 130^\circ = 180^\circ$ $x = 180^\circ - 130^\circ$ $x = \boxed{50^\circ}$
The exterior angle of a triangle is equal to the sum of the two opposite interior angles.
$130^\circ = (y - 20^\circ) + y$ $130 = 2y - 20$
Solve for $y$ .
$130 + 20 = 2y$ $150 = 2y$ $y = \frac{150}{2}$ $y = \boxed{75^\circ}$
** (Middle left figure)**
Lines AB and DC are parallel (indicated by arrows).
BC is a transversal. Consecutive interior angles sum to $180^\circ$ .
$\angle ABC + \angle BCD = 180^\circ$ $y + 90^\circ = 180^\circ$ $y = 180^\circ - 90^\circ$ $y = \boxed{90^\circ}$
In $\triangle BCE$ , the sum of angles is $180^\circ$ .
$\angle BEC = 180^\circ - \angle BCE - \angle CBE$ $\angle BEC = 180^\circ - 90^\circ - 40^\circ$ $\angle BEC = 50^\circ$
Assuming the arrow on DE indicates that DE is parallel to BC.
If DE || BC, then $\angle DEC$ and $\angle BCE$ are consecutive interior angles, so they sum to $180^\circ$ .
$x + 90^\circ = 180^\circ$ $x = 180^\circ - 90^\circ$ $x = \boxed{90^\circ}$
** (Middle right, labeled "10.3")**
The two horizontal lines are parallel.
The angles $6c - 40^\circ$ and $2c - 20^\circ$ are consecutive interior angles, so they sum to $180^\circ$ .
$(6c - 40^\circ) + (2c - 20^\circ) = 180^\circ$ $8c - 60 = 180$
Solve for $c$ .
$8c = 180 + 60$ $8c = 240$ $c = \frac{240}{8}$ $c = \boxed{30}$
Angle $d$ and the angle $6c - 40^\circ$ are vertically opposite angles, so they are equal.
$d = 6c - 40^\circ$ $d = 6(30) - 40$ $d = 180 - 40$ $d = \boxed{140^\circ}$
** (Bottom left, intersecting lines)**
Vertically opposite angles are equal.
$t + 30^\circ = 60^\circ$
Solve for $t$ .
$t = 60 - 30$ $t = \boxed{30}$
** (Bottom left triangle)**
The markings on the sides indicate that all three sides of the triangle are equal, meaning it is an equilateral triangle.
All angles in an equilateral triangle are $60^\circ$ .
$2x - 30^\circ = 60^\circ$
Solve for $x$ .
$2x = 60 + 30$ $2x = 90$ $x = \frac{90}{2}$ $x = \boxed{45}$
** (Bottom middle, angles on a straight line)**
Angles on a straight line sum to $180^\circ$ .
The square symbol indicates a $90^\circ$ angle.
$20^\circ + 90^\circ + a = 180^\circ$ $110^\circ + a = 180^\circ$
Solve for $a$ .
$a = 180^\circ - 110^\circ$ $a = \boxed{70^\circ}$
** (Bottom right triangle, labeled "10.2")**
The interior angle adjacent to the $130^\circ$ exterior angle forms a linear pair, so it is $180^\circ - 130^\circ = 50^\circ$ .
This is the vertex angle of the isosceles triangle.
The markings indicate that the triangle is isosceles, and the two angles marked $b$ are the base angles, which are equal.
The sum of angles in a triangle is $180^\circ$ .
$b + b + 50^\circ = 180^\circ$ $2b + 50 = 180$
Solve for $b$ .
$2b = 180 - 50$ $2b = 130$ $b = \frac{130}{2}$ $b = \boxed{65^\circ}$

Calculate, with reasons, the angles marked x and y.

Step-by-step solution

The sum of angles in a triangle is $180^\circ$ .

Solve for $x$ .

Angles on a straight line sum to $180^\circ$ .

The exterior angle of a triangle is equal to the sum of the two opposite interior angles.

Solve for $y$ .

Lines AB and DC are parallel (indicated by arrows).

In $\triangle BCE$ , the sum of angles is $180^\circ$ .

Assuming the arrow on DE indicates that DE is parallel to BC.

The two horizontal lines are parallel.

Solve for $c$ .

Angle $d$ and the angle $6c - 40^\circ$ are vertically opposite angles, so they are equal.

Vertically opposite angles are equal.

Solve for $t$ .

The markings on the sides indicate that all three sides of the triangle are equal, meaning it is an equilateral triangle.

Solve for $x$ .

Angles on a straight line sum to $180^\circ$ .

Solve for $a$ .

The interior angle adjacent to the $130^\circ$ exterior angle forms a linear pair, so it is $180^\circ - 130^\circ = 50^\circ$ .

The markings indicate that the triangle is isosceles, and the two angles marked $b$ are the base angles, which are equal.

Solve for $b$ .

Still stuck on this one?

More Mathematics Questions

Calculate, with reasons, the angles marked x and y.

Step-by-step solution

The sum of angles in a triangle is 180∘180^\circ180∘.

Solve for xxx.

Angles on a straight line sum to 180∘180^\circ180∘.

The exterior angle of a triangle is equal to the sum of the two opposite interior angles.

Solve for yyy.

Lines AB and DC are parallel (indicated by arrows).

In △BCE\triangle BCE△BCE, the sum of angles is 180∘180^\circ180∘.

Assuming the arrow on DE indicates that DE is parallel to BC.

The two horizontal lines are parallel.

Solve for ccc.

Angle ddd and the angle 6c−40∘6c - 40^\circ6c−40∘ are vertically opposite angles, so they are equal.

Vertically opposite angles are equal.

Solve for ttt.

The markings on the sides indicate that all three sides of the triangle are equal, meaning it is an equilateral triangle.

Solve for xxx.

Angles on a straight line sum to 180∘180^\circ180∘.

Solve for aaa.

The interior angle adjacent to the 130∘130^\circ130∘ exterior angle forms a linear pair, so it is 180∘−130∘=50∘180^\circ - 130^\circ = 50^\circ180∘−130∘=50∘.

The markings indicate that the triangle is isosceles, and the two angles marked bbb are the base angles, which are equal.

Solve for bbb.

Still stuck on this one?

More Mathematics Questions

The sum of angles in a triangle is $180^\circ$ .

Solve for $x$ .

Angles on a straight line sum to $180^\circ$ .

Solve for $y$ .

In $\triangle BCE$ , the sum of angles is $180^\circ$ .

Solve for $c$ .

Angle $d$ and the angle $6c - 40^\circ$ are vertically opposite angles, so they are equal.

Solve for $t$ .

Solve for $x$ .

Angles on a straight line sum to $180^\circ$ .

Solve for $a$ .

The interior angle adjacent to the $130^\circ$ exterior angle forms a linear pair, so it is $180^\circ - 130^\circ = 50^\circ$ .

The markings indicate that the triangle is isosceles, and the two angles marked $b$ are the base angles, which are equal.

Solve for $b$ .