The Sum of the Interior Angles of a Triangle

What does the triangle sum theorem tell us?

The theorem tells us that the sum of the interior angles of any triangle is equal to 180°.

How do we find the third interior angle of a triangle, knowing the other two?

By applying the theorem, we subtract the sum of the two given angles from 180°.

How much must the interior angles of a triangle add up to?

180°.

Task:

Given three angles:

Angle $A$ is equal to $30°$

Angle $B$ is equal to $60°$

Angle $C$ is equal to $90°$

Can these angles form a triangle?

Solution

It is known that the sum of the angles of the triangles must be equal to $180°$

Let's add the value of the angles and see if together they are equal to $180°$

$A+B+C=30+60+90=180$

Answer

Yes

Task:

Given three angles:

Angle $A$ is equal to $60°$

Angle $B$ is equal to $60°$

Angle $C$ is equal to $60°$

Can these angles form a triangle?

Solution

It is known that the sum of the angles of the triangles must be equal to $180°$

Let's add the value of the angles and see if together they are equal to $180°$

$A+B+C=60+60+60=180$

Answer

Yes

Task:

Given three angles:

Angle A is equal to $90°$

Angle B is equal to $115°$

Angle C is equal to $35°$

Can these angles form a triangle?

Solution

We know that the sum of the angles of the triangle must be equal to $180°$

We add the total of the angles to see if together they are equal to $180°$

$A+B+C=90+115+35=240$

We observe that the sum of the three angles are equal to $240°$, that is to say that they cannot form a triangle.

Answer

No

Assignment:

Given the parallel lines.

Find the angle $\alpha$

Solution

The angle beta is equal to $90°$. The adjacent angle is also equal to $90°$ since the sum is equal to $180°$ degrees. The adjacent angle gamma $120°$ and their sum is equal to $180°$, therefore, gamma is equal to $60°$ degrees.

$\alpha+\gamma+\delta=180°$

$\alpha+60°+90°=180°$

$\alpha+150°=180°$

$\alpha=180°-150°$

$\alpha=30°$

Answer

$30°$

$CE$ is parallel to $AD$

What is the value of $X$ if it is given that $ABC$ is isosceles, such that $AB=BC$

Solution

Angles $\sphericalangle UCH$ and angle $\sphericalangle ACE$ are opposite angles.

$\text{AC}E=\text{ICH}=2X$

$\sphericalangle DAC$ and angle $\sphericalangle\text{AC}E$ are collateral angles.

$2x+\text{DAC}=180$

$\text{DAC}=180-2x$

$\sphericalangle FGA$ and angle $\sphericalangle DAB$ are opposite angles.

$\text{FGA}=\text{DAB}=x-10$

$\text{BAC}=\text{DAC}-\text{DAB}=$

$180-2x-(x-10)=$

$190-3x$

The sum of the angles in the triangle is $180$

$\text{ACB}+\text{CAB}+B=180$

$\text{ACB}=180-(190-3x)-(3x-30)=20$

$\text{ACB}=\text{BAC}$

$20=190-3x$

$x=56.67$

Answer

$56.67$