Question 11

The sum of the three angles inside any triangle is always $180^\circ$.

Step 1 — Understanding Triangle Rules

Let the given angle be $A$. Let the second angle be $B$. Let the third angle be $C$. The sum of these three angles must be $180^\circ$. $A + B + C = 180^\circ$ For a triangle to be possible, every angle must be greater than $0^\circ$. So, $A > 0^\circ$, $B > 0^\circ$, and $C > 0^\circ$. From the sum, we know $C = 180^\circ - A - B$. So, $180^\circ - A - B$ must be greater than $0^\circ$. $180^\circ - A - B > 0^\circ$ We can rearrange this inequality. $180^\circ > A + B$ This means the sum of the two angles ($A$ and $B$) must be less than $180^\circ$. Also, the angle $B$ we choose must be greater than $0^\circ$. For a triangle to be not possible, the sum of the two angles must be $180^\circ$ or more. $A + B \ge 180^\circ$

Step 2 — For the angle $30^\circ$

Step 2.1 — Finding angles for a possible triangle

The given angle $A$ is $30^\circ$. We need to find an angle $B$ such that $A + B < 180^\circ$. Let us substitute the value of $A$. $30^\circ + B < 180^\circ$ To find the range for $B$, we subtract $30^\circ$ from both sides. $B < 180^\circ - 30^\circ$ $B < 150^\circ$ So, any angle $B$ less than $150^\circ$ will work.

$\boxed{\text{Another angle } B \text{ must be less than } 150^\circ.}$

Step 2.2 — Giving examples for a possible triangle

We need two different angles that are less than $150^\circ$. Let us choose $B = 60^\circ$. The sum of the two angles is: $30^\circ + 60^\circ = 90^\circ$ This sum is less than $180^\circ$. The third angle $C$ would be: $C = 180^\circ - 90^\circ = 90^\circ$ Since $90^\circ > 0^\circ$, this is a valid triangle. Let us choose $B = 90^\circ$. The sum of the two angles is: $30^\circ + 90^\circ = 120^\circ$ This sum is less than $180^\circ$. The third angle $C$ would be: $C = 180^\circ - 120^\circ = 60^\circ$ Since $60^\circ > 0^\circ$, this is a valid triangle.

Step 2.3 — Finding angles for a not possible triangle

The given angle $A$ is $30^\circ$. We need to find an angle $B$ such that $A + B \ge 180^\circ$. Let us substitute the value of $A$. $30^\circ + B \ge 180^\circ$ To find the range for $B$, we subtract $30^\circ$ from both sides. $B \ge 180^\circ - 30^\circ$ $B \ge 150^\circ$ So, any angle $B$ that is $150^\circ$ or more will make a triangle not possible.

$\boxed{\text{Another angle } B \text{ must be greater than or equal to } 150^\circ.}$

Step 2.4 — Giving examples for a not possible triangle

We need two different angles that are $150^\circ$ or more. Let us choose $B = 170^\circ$. The sum of the two angles is: $30^\circ + 170^\circ = 200^\circ$ This sum is greater than $180^\circ$. So, a third angle cannot be formed. Let us choose $B = 160^\circ$. The sum of the two angles is: $30^\circ + 160^\circ = 190^\circ$ This sum is greater than $180^\circ$. So, a third angle cannot be formed.

Step 3 — For the angle $70^\circ$

Step 3.1 — Finding angles for a possible triangle

The given angle $A$ is $70^\circ$. We need to find an angle $B$ such that $A + B < 180^\circ$. Let us substitute the value of $A$. $70^\circ + B < 180^\circ$ To find the range for $B$, we subtract $70^\circ$ from both sides. $B < 180^\circ - 70^\circ$ $B < 110^\circ$ So, any angle $B$ less than $110^\circ$ will work.

$\boxed{\text{Another angle } B \text{ must be less than } 110^\circ.}$

Step 3.2 — Giving examples for a possible triangle

We need two different angles that are less than $110^\circ$. Let us choose $B = 70^\circ$. The sum of the two angles is: $70^\circ + 70^\circ = 140^\circ$ This sum is less than $180^\circ$. The third angle $C$ would be: $C = 180^\circ - 140^\circ = 40^\circ$ Since $40^\circ > 0^\circ$, this is a valid triangle. Let us choose $B = 40^\circ$. The sum of the two angles is: $70^\circ + 40^\circ = 110^\circ$ This sum is less than $180^\circ$. The third angle $C$ would be: $C = 180^\circ - 110^\circ = 70^\circ$ Since $70^\circ > 0^\circ$, this is a valid triangle.

Step 3.3 — Finding angles for a not possible triangle

The given angle $A$ is $70^\circ$. We need to find an angle $B$ such that $A + B \ge 180^\circ$. Let us substitute the value of $A$. $70^\circ + B \ge 180^\circ$ To find the range for $B$, we subtract $70^\circ$ from both sides. $B \ge 180^\circ - 70^\circ$ $B \ge 110^\circ$ So, any angle $B$ that is $110^\circ$ or more will make a triangle not possible.

$\boxed{\text{Another angle } B \text{ must be greater than or equal to } 110^\circ.}$

Step 3.4 — Giving examples for a not possible triangle

We need two different angles that are $110^\circ$ or more. Let us choose $B = 120^\circ$. The sum of the two angles is: $70^\circ + 120^\circ = 190^\circ$ This sum is greater than $180^\circ$. So, a third angle cannot be formed. Let us choose $B = 150^\circ$. The sum of the two angles is: $70^\circ + 150^\circ = 220^\circ$ This sum is greater than $180^\circ$. So, a third angle cannot be formed.

Step 4 — For the angle $54^\circ$

Step 4.1 — Finding angles for a possible triangle

The given angle $A$ is $54^\circ$. We need to find an angle $B$ such that $A + B < 180^\circ$. Let us substitute the value of $A$. $54^\circ + B < 180^\circ$ To find the range for $B$, we subtract $54^\circ$ from both sides. $B < 180^\circ - 54^\circ$ $B < 126^\circ$ So, any angle $B$ less than $126^\circ$ will work.

$\boxed{\text{Another angle } B \text{ must be less than } 126^\circ.}$

Step 4.2 — Giving examples for a possible triangle

We need two different angles that are less than $126^\circ$. Let us choose $B = 72^\circ$. The sum of the two angles is: $54^\circ + 72^\circ = 126^\circ$ This sum is less than $180^\circ$. The third angle $C$ would be: $C = 180^\circ - 126^\circ = 54^\circ$ Since $54^\circ > 0^\circ$, this is a valid triangle. Let us choose $B = 54^\circ$. The sum of the two angles is: $54^\circ + 54^\circ = 108^\circ$ This sum is less than $180^\circ$. The third angle $C$ would be: $C = 180^\circ - 108^\circ = 72^\circ$ Since $72^\circ > 0^\circ$, this is a valid triangle.

Step 4.3 — Finding angles for a not possible triangle

The given angle $A$ is $54^\circ$. We need to find an angle $B$ such that $A + B \ge 180^\circ$. Let us substitute the value of $A$. $54^\circ + B \ge 180^\circ$ To find the range for $B$, we subtract $54^\circ$ from both sides. $B \ge 180^\circ - 54^\circ$ $B \ge 126^\circ$ So, any angle $B$ that is $126^\circ$ or more will make a triangle not possible.

$\boxed{\text{Another angle } B \text{ must be greater than or equal to } 126^\circ.}$

Step 4.4 — Giving examples for a not possible triangle

We need two different angles that are $126^\circ$ or more. Let us choose $B = 140^\circ$. The sum of the two angles is: $54^\circ + 140^\circ = 194^\circ$ This sum is greater than $180^\circ$. So, a third angle cannot be formed. Let us choose $B = 130^\circ$. The sum of the two angles is: $54^\circ + 130^\circ = 184^\circ$ This sum is greater than $180^\circ$. So, a third angle cannot be formed.

Step 5 — For the angle $144^\circ$

Step 5.1 — Finding angles for a possible triangle

The given angle $A$ is $144^\circ$. We need to find an angle $B$ such that $A + B < 180^\circ$. Let us substitute the value of $A$. $144^\circ + B < 180^\circ$ To find the range for $B$, we subtract $144^\circ$ from both sides. $B < 180^\circ - 144^\circ$ $B < 36^\circ$ So, any angle $B$ less than $36^\circ$ will work.

$\boxed{\text{Another angle } B \text{ must be less than } 36^\circ.}$

Step 5.2 — Giving examples for a possible triangle

We need two different angles that are less than $36^\circ$. Let us choose $B = 10^\circ$. The sum of the two angles is: $144^\circ + 10^\circ = 154^\circ$ This sum is less than $180^\circ$. The third angle $C$ would be: $C = 180^\circ - 154^\circ = 26^\circ$ Since $26^\circ > 0^\circ$, this is a valid triangle. Let us choose $B = 26^\circ$. The sum of the two angles is: $144^\circ + 26^\circ = 170^\circ$ This sum is less than $180^\circ$. The third angle $C$ would be: $C = 180^\circ - 170^\circ = 10^\circ$ Since $10^\circ > 0^\circ$, this is a valid triangle.

Step 5.3 — Finding angles for a not possible triangle

The given angle $A$ is $144^\circ$. We need to find an angle $B$ such that $A + B \ge 180^\circ$. Let us substitute the value of $A$. $144^\circ + B \ge 180^\circ$ To find the range for $B$, we subtract $144^\circ$ from both sides. $B \ge 180^\circ - 144^\circ$ $B \ge 36^\circ$ So, any angle $B$ that is $36^\circ$ or more will make a triangle not possible.

$\boxed{\text{Another angle } B \text{ must be greater than or equal to } 36^\circ.}$

Step 5.4 — Giving examples for a not possible triangle

We need two different angles that are $36^\circ$ or more. Let us choose $B = 40^\circ$. The sum of the two angles is: $144^\circ + 40^\circ = 184^\circ$ This sum is greater than $180^\circ$. So, a third angle cannot be formed. Let us choose $B = 50^\circ$. The sum of the two angles is: $144^\circ + 50^\circ = 194^\circ$ This sum is greater than $180^\circ$. So, a third angle cannot be formed.

Answer

(a) Another angle for which a triangle is possible will be any angle less than 150 degrees. Two different angles are 60 degrees, 90 degrees. Another angle for which a triangle is not possible will be any angle greater than or equal to 150 degrees. Two different angles are 170 degrees, 160 degrees. (b) Another angle for which a triangle is possible will be any angle less than 110 degrees. Two different angles are 70 degrees, 40 degrees. Another angle for which a triangle is not possible will be any angle greater than or equal to 110 degrees. Two different angles are 120 degrees, 150 degrees. (c) Another angle for which a triangle is possible will be any angle less than 126 degrees. Two different angles are 72 degrees, 54 degrees. Another angle for which a triangle is not possible will be any angle greater than or equal to 126 degrees. Two different angles are 140 degrees, 130 degrees. (d) Another angle for which a triangle is possible will be any angle less than 36 degrees. Two different angles are 10 degrees, 26 degrees. Another angle for which a triangle is not possible will be any angle greater than or equal to 36 degrees. At least two different angles are 40 degrees, 50 degrees.