A Tale of Three Intersecting Lines — NCERT Solutions

Given: Two sides 3 cm and 7 cm with included angle 75°.

Steps of Construction:

Step 1: Draw a line segment $AB = 7$ cm (taking the longer side as base).

Step 2: At point A, construct an angle of $75°$ using a protractor, drawing the other arm of the angle.

Step 3: On this arm, mark point C such that $AC = 3$ cm.

Step 4: Join BC.

$\triangle ABC$ is the required triangle with $AB = 7$ cm, $AC = 3$ cm, and $\angle A = 75°$.

Note: A triangle always exists for any two positive side lengths and any included angle strictly between $0°$ and $180°$, so this triangle is always constructible.

Given: Two sides 6 cm and 3 cm with included angle 25°.

Steps of Construction:

Step 1: Draw a line segment $AB = 6$ cm.

Step 2: At point A, construct an angle of $25°$ using a protractor.

Step 3: On this arm, mark point C such that $AC = 3$ cm.

Step 4: Join BC.

$\triangle ABC$ is the required triangle with $AB = 6$ cm, $AC = 3$ cm, and $\angle A = 25°$.

Given: Two sides 3 cm and 8 cm with included angle 120°.

Steps of Construction:

Step 1: Draw a line segment $AB = 8$ cm.

Step 2: At point A, construct an angle of $120°$ using a protractor.

Step 3: On this arm, mark point C such that $AC = 3$ cm.

Step 4: Join BC.

$\triangle ABC$ is the required triangle with $AB = 8$ cm, $AC = 3$ cm, and $\angle A = 120°$.

Answer: No, a triangle is always possible when two sides and their included angle are given, as long as the angle is strictly between $0°$ and $180°$ (exclusive).

Justification:

• Draw one side as the base, say $AB$.
• At A, draw the included angle. The second side $AC$ is drawn along this angle arm.
• Join B to C. The triangle $ABC$ is always formed.
• The two sides and the included angle uniquely determine the third side BC by the construction.
• There is no condition like the triangle inequality that can fail here, because once two sides and the included angle are fixed, the third side is automatically determined and is always positive (as long as the angle is between $0°$ and $180°$).

Conclusion: For any two positive lengths and any angle strictly between $0°$ and $180°$, a triangle always exists.

Given: $\angle A = 75°$, $AB = 5$ cm (included side), $\angle B = 75°$.

Check existence: $\angle A + \angle B = 75° + 75° = 150° < 180°$ ✓. Triangle exists.

Steps of Construction:

Step 1: Draw base $AB = 5$ cm.

Step 2: At A, construct an angle of $75°$ (above AB).

Step 3: At B, construct an angle of $75°$ (above AB, on the same side).

Step 4: The point C where the two arms intersect is the third vertex.

$\triangle ABC$ is the required triangle with $AB = 5$ cm, $\angle A = 75°$, $\angle B = 75°$.

The third angle: $\angle C = 180° - 75° - 75° = 30°$.

Given: $\angle A = 25°$, $AB = 3$ cm (included side), $\angle B = 60°$.

Check existence: $\angle A + \angle B = 25° + 60° = 85° < 180°$ ✓. Triangle exists.

Steps of Construction:

Step 1: Draw base $AB = 3$ cm.

Step 2: At A, construct an angle of $25°$ above AB.

Step 3: At B, construct an angle of $60°$ above AB.

Step 4: The intersection point of the two arms is vertex C.

$\triangle ABC$ is the required triangle with $AB = 3$ cm, $\angle A = 25°$, $\angle B = 60°$.

The third angle: $\angle C = 180° - 25° - 60° = 95°$.

Given: $\angle A = 120°$, $AB = 6$ cm (included side), $\angle B = 30°$.

Check existence: $\angle A + \angle B = 120° + 30° = 150° < 180°$ ✓. Triangle exists.

Steps of Construction:

Step 1: Draw base $AB = 6$ cm.

Step 2: At A, construct an angle of $120°$ above AB.

Step 3: At B, construct an angle of $30°$ above AB.

Step 4: The intersection point of the two arms is vertex C.

$\triangle ABC$ is the required triangle with $AB = 6$ cm, $\angle A = 120°$, $\angle B = 30°$.

The third angle: $\angle C = 180° - 120° - 30° = 30°$.

Answer: No, a triangle does not always exist for every combination of two angles and an included side.

Condition for existence: A triangle exists if and only if the sum of the two given angles is strictly less than $180°$.

Examples where triangle is NOT possible:

1. $\angle A = 90°$, $\angle B = 90°$: Sum $= 180°$. The two arms from A and B are parallel and never meet. No triangle.

2. $\angle A = 100°$, $\angle B = 100°$: Sum $= 200° > 180°$. No triangle.

3. $\angle A = 40°$, $\angle B = 140°$: Sum $= 180°$. The arms are parallel. No triangle.

4. $\angle A = 60°$, $\angle B = 130°$: Sum $= 190° > 180°$. No triangle.

Rule: A triangle exists when $\angle A + \angle B < 180°$.

Given: $\angle A = 40°$ (acute angle at A).

For the line from B to not meet the line $l$ (the arm from A), the line from B must be parallel to $l$ or inclined further to the right (away from $l$).

When the line from B is parallel to $l$:

• AB is the transversal cutting two parallel lines.
• Co-interior angles (same-side interior angles) add up to $180°$.
• So $\angle A + \angle B = 180°$, giving $\angle B = 180° - 40° = 140°$.

For the lines to not meet, $\angle B \geq 140°$.

A possible value: $\angle B = 150°$ (the lines will not meet).

Given: $\angle A = 40°$.

The smallest value of $\angle B$ for which the lines do not meet is when the line from B is parallel to the arm from A.

Using the property of parallel lines (co-interior angles sum to $180°$):
$$\angle A + \angle B = 180°$$
$$40° + \angle B = 180°$$
$$\angle B = 140°$$

Conclusion: The smallest value of $\angle B$ for the lines to not meet is $\mathbf{140°}$. For $\angle B \geq 140°$, the lines are parallel or diverge, and no triangle is formed.

Rule: Given one angle $\alpha$, a triangle is possible with another angle $\beta$ if $\alpha + \beta < 180°$, i.e., $\beta < 180° - \alpha$. A triangle is not possible if $\alpha + \beta \geq 180°$.

(a) Given angle: 30°
For triangle to be possible: $\beta < 180° - 30° = 150°$

• Possible: $\beta = 60°$ (since $30° + 60° = 90° < 180°$); $\beta = 100°$ (since $30° + 100° = 130° < 180°$)
• Not possible: $\beta = 150°$ (since $30° + 150° = 180°$); $\beta = 160°$ (since $30° + 160° = 190° > 180°$)

(b) Given angle: 70°
For triangle to be possible: $\beta < 180° - 70° = 110°$

• Possible: $\beta = 50°$ (since $70° + 50° = 120° < 180°$); $\beta = 80°$ (since $70° + 80° = 150° < 180°$)
• Not possible: $\beta = 110°$ (since $70° + 110° = 180°$); $\beta = 120°$ (since $70° + 120° = 190° > 180°$)

(c) Given angle: 54°
For triangle to be possible: $\beta < 180° - 54° = 126°$

• Possible: $\beta = 70°$ (since $54° + 70° = 124° < 180°$); $\beta = 90°$ (since $54° + 90° = 144° < 180°$)
• Not possible: $\beta = 126°$ (since $54° + 126° = 180°$); $\beta = 140°$ (since $54° + 140° = 194° > 180°$)

(d) Given angle: 144°
For triangle to be possible: $\beta < 180° - 144° = 36°$

• Possible: $\beta = 20°$ (since $144° + 20° = 164° < 180°$); $\beta = 30°$ (since $144° + 30° = 174° < 180°$)
• Not possible: $\beta = 36°$ (since $144° + 36° = 180°$); $\beta = 50°$ (since $144° + 50° = 194° > 180°$)

Rule: Two angles $\alpha$ and $\beta$ can be angles of a triangle if and only if $\alpha + \beta < 180°$ (so that the third angle $= 180° - \alpha - \beta > 0°$).

(a) 35°, 150°:
$$35° + 150° = 185° > 180°$$
Third angle would be $180° - 185° = -5°$ (negative). Cannot be angles of a triangle.

(b) 70°, 30°:
$$70° + 30° = 100° < 180°$$
Third angle $= 180° - 100° = 80° > 0°$. Can be angles of a triangle.

(c) 90°, 85°:
$$90° + 85° = 175° < 180°$$
Third angle $= 180° - 175° = 5° > 0°$. Can be angles of a triangle.

(d) 50°, 150°:
$$50° + 150° = 200° > 180°$$
Third angle would be $180° - 200° = -20°$ (negative). Cannot be angles of a triangle.

Concept (Angle Sum Property): The sum of all three angles of a triangle is $180°$.
$$\angle A + \angle B + \angle C = 180°$$

(a) 36°, 72°:
$$\text{Third angle} = 180° - 36° - 72° = 180° - 108° = 72°$$

(b) 150°, 15°:
$$\text{Third angle} = 180° - 150° - 15° = 180° - 165° = 15°$$

(c) 90°, 30°:
$$\text{Third angle} = 180° - 90° - 30° = 180° - 120° = 60°$$

(d) 75°, 45°:
$$\text{Third angle} = 180° - 75° - 45° = 180° - 120° = 60°$$

If all three angles are 70°:
$$70° + 70° + 70° = 210° \neq 180°$$
This violates the angle sum property. A triangle with all angles equal to 70° cannot be constructed.

If two angles are 70°:
$$\text{Third angle} = 180° - 70° - 70° = 40°$$
So the third angle would be $40°$.

If all three angles are equal:
Let each angle $= x$.
$$x + x + x = 180°$$
$$3x = 180°$$
$$x = 60°$$

Conclusion: For all three angles to be equal, each must be $60°$. Such a triangle is an equilateral triangle. All equilateral triangles have each angle equal to $60°$.

Given: $\angle A = 50°$, $\angle B = \angle C$.

Using Angle Sum Property:
$$\angle A + \angle B + \angle C = 180°$$
$$50° + \angle B + \angle C = 180°$$

Since $\angle B = \angle C$:
$$50° + 2\angle B = 180°$$
$$2\angle B = 180° - 50° = 130°$$
$$\angle B = \frac{130°}{2} = 65°$$

Therefore, $\angle C = 65°$.

Answer: $\angle B = \angle C = 65°$.

Given: $BC = 5$ cm, $AB = 6$ cm, $CA = 5$ cm.

Check Triangle Inequality:
$5 + 5 = 10 > 6$ ✓; $5 + 6 = 11 > 5$ ✓. Triangle exists. (It is isosceles with $BC = CA = 5$ cm.)

Steps of Construction:

Step 1: Draw base $BC = 5$ cm.

Step 2: With B as centre, draw an arc of radius 6 cm.

Step 3: With C as centre, draw an arc of radius 5 cm intersecting the first arc at A.

Step 4: Join AB and AC. $\triangle ABC$ is formed.

Constructing Altitude from A to BC:

Step 5: With A as centre, draw an arc that cuts BC (or its extension) at two points, say P and Q.

Step 6: With P and Q as centres, draw arcs of equal radius (more than half PQ) on the opposite side of A. Let them intersect at point D.

Step 7: Join AD. The foot of the perpendicular from A to BC is the point where AD meets BC. This segment is the altitude from A.

Note: Since the triangle is isosceles ($BC = CA$), the altitude from A bisects BC.

Given: $RY = 4$ cm, $TR = 7$ cm, $\angle R = 140°$ (included angle between TR and RY).

Steps of Construction:

Step 1: Draw base $RY = 4$ cm.

Step 2: At R, construct an angle of $140°$ using a protractor.

Step 3: On this arm, mark point T such that $RT = 7$ cm.

Step 4: Join TY. $\triangle TRY$ is formed.

Constructing Altitude from T to RY:

Step 5: Since $\angle R = 140°$ is obtuse, the altitude from T will fall outside the triangle (on the extension of RY beyond R).

Step 6: Extend RY beyond R. From T, draw a perpendicular to the line RY (extended). Use a set square or compass construction to drop a perpendicular from T to line RY.

Step 7: The foot of the perpendicular (let's call it H) lies on the extension of RY. TH is the required altitude from T.

Given: $\angle B = 90°$, $AC = 5$ cm (hypotenuse).

Analysis: We know one angle ($\angle B = 90°$) and the hypotenuse ($AC = 5$ cm). The other two angles $\angle A$ and $\angle C$ must satisfy:
$$\angle A + \angle C = 90°$$

For any value of $\angle A$ strictly between $0°$ and $90°$, we get a valid $\angle C = 90° - \angle A$, also strictly between $0°$ and $90°$. Each such choice gives a different triangle.

Steps of Construction (for one example, say ∠A = 30°, ∠C = 60°):

Step 1: Draw $AC = 5$ cm as the base.

Step 2: At A, construct $\angle A = 30°$.

Step 3: At C, construct $\angle C = 60°$.

Step 4: The intersection of the two arms is B, where $\angle B = 90°$.

Answer: There are infinitely many different triangles with $\angle B = 90°$ and $AC = 5$ cm, because $\angle A$ can take any value strictly between $0°$ and $90°$, giving a different triangle each time.

Equilateral Triangle:
In an equilateral triangle, all three angles are equal to $60°$.

(i) Equilateral and right-angled:
A right-angled triangle has one angle of $90°$. But in an equilateral triangle, all angles are $60°$. Since $60° \neq 90°$, an equilateral triangle cannot be right-angled.

(ii) Equilateral and obtuse-angled:
An obtuse-angled triangle has one angle greater than $90°$. But all angles in an equilateral triangle are $60° < 90°$. So an equilateral triangle cannot be obtuse-angled.

Conclusion: An equilateral triangle is always acute-angled (all angles = 60°). It can be neither right-angled nor obtuse-angled.

Isosceles Triangle:

(i) Isosceles and right-angled:
This is possible. Example: An isosceles right-angled triangle has angles $90°, 45°, 45°$.

Construction: Draw $AB = BC = 5$ cm with $\angle B = 90°$. Join AC. $\triangle ABC$ is isosceles ($AB = BC$) and right-angled ($\angle B = 90°$).

(ii) Isosceles and obtuse-angled:
This is possible. Example: An isosceles triangle with angles $120°, 30°, 30°$.

Construction: Draw base $BC = 4$ cm. At B and C, construct angles of $30°$ each. The apex angle at A will be $180° - 30° - 30° = 120°$ (obtuse). Since $\angle B = \angle C = 30°$, we have $AB = AC$ (sides opposite equal angles are equal), making it isosceles.

Conclusion: An isosceles triangle can be both right-angled and obtuse-angled.