Question 19

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

Step 1 — Equilateral Triangle Angles

An equilateral triangle has all three sides equal. So, all its three angles are also equal. Let us call each angle $x$.

The sum of angles in a triangle is $180^\circ$. So, we write:

$x + x + x = 180^\circ$

$3x = 180^\circ$

$x = \frac{180^\circ}{3}$

$\boxed{\text{Each angle} = \mathbf{60^\circ}}$

Step 2 — Equilateral Right-angled Triangle

A right-angled triangle has one angle of $90^\circ$. We found that each angle in an equilateral triangle is $60^\circ$. Since $60^\circ$ is not equal to $90^\circ$, an equilateral triangle cannot be right-angled.

Step 3 — Equilateral Obtuse-angled Triangle

An obtuse-angled triangle has one angle greater than $90^\circ$. We know each angle in an equilateral triangle is $60^\circ$. No angle is greater than $90^\circ$. So, an equilateral triangle cannot be obtuse-angled.

Step 4 — Isosceles Right-angled Triangle

An isosceles triangle has two equal sides. It also has two equal angles. Let us try to make one angle $90^\circ$. The other two angles must be equal. The sum of angles in a triangle is $180^\circ$. So, we write:

$\text{Equal angle} + \text{Equal angle} + 90^\circ = 180^\circ$

$2 \times \text{Equal angle} = 180^\circ - 90^\circ$

$2 \times \text{Equal angle} = 90^\circ$

$\text{Equal angle} = \frac{90^\circ}{2}$

$\boxed{\text{Each equal angle} = \mathbf{45^\circ}}$

So, a triangle with angles $90^\circ, 45^\circ, 45^\circ$ is possible. This is an isosceles right-angled triangle. To construct it:

1. Draw a line segment, say AB.
2. At point A, draw a line perpendicular to AB. This creates a $90^\circ$ angle.
3. From A, measure equal lengths along AB and the perpendicular line. Let these points be B and C.
4. Connect points B and C.
5. Triangle ABC is an isosceles right-angled triangle.

Step 5 — Isosceles Obtuse-angled Triangle

An isosceles triangle has two equal angles. Let us try to make one angle obtuse. An obtuse angle is greater than $90^\circ$.

Case 1: The obtuse angle is one of the equal angles. Let the equal angles be $y$. Let $y > 90^\circ$. The sum of angles is $180^\circ$. So, $y + y + \text{third angle} = 180^\circ$. This means $2y + \text{third angle} = 180^\circ$. If $y > 90^\circ$, then $2y > 180^\circ$. This would make the total sum greater than $180^\circ$. This is not possible for a triangle. So, the equal angles cannot be obtuse.

Case 2: The obtuse angle is the unique angle. Let the unique angle be $z$. Let $z > 90^\circ$. The two other angles are equal. Let them be $y$. So, we write:

$y + y + z = 180^\circ$

$2y = 180^\circ - z$

If we choose $z = 100^\circ$, then:

$2y = 180^\circ - 100^\circ$

$2y = 80^\circ$

$y = \frac{80^\circ}{2}$

$\boxed{\text{Each equal angle} = \mathbf{40^\circ}}$

So, a triangle with angles $100^\circ, 40^\circ, 40^\circ$ is possible. This is an isosceles obtuse-angled triangle. To construct it:

1. Draw a line segment, say PQ.
2. At point P, draw a line PR. Make angle QPR obtuse (e.g., $100^\circ$).
3. Measure equal lengths along PQ and PR from point P. Let these points be Q and R.
4. Connect points Q and R.
5. Triangle PQR is an isosceles obtuse-angled triangle.

Answer

(i) An equilateral triangle cannot be right-angled. (ii) An equilateral triangle cannot be obtuse-angled. (i) An isosceles triangle can be right-angled. (ii) An isosceles triangle can be obtuse-angled.