As per the given in question , we draw a figure triangle ABC

∠B + ∠C = 60° + 40° = 100°
∴ ∠A = 180° – 100° = 80°
∴ ∠BAD = 40°
In ∆ ABE,
∠AEB = 90°

Correct Option: B

As per the given in question , we draw a figure triangle ABC

∠B + ∠C = 60° + 40° = 100°
∴ ∠A = 180° – 100° = 80°
∴ ∠BAD = 40°
In ∆ ABE,
∠AEB = 90°
∴ ∠BAE = 180° – 90° – 60° = 30°
∴ ∠EAD = ∠BAD – ∠BAE = 40° – 30° = 10°

We draw a figure triangle ABC whose the side BC is produced to D ,

Exterior angle of a triangle is equal to the sum of remaining two interior angles.
From triangle ,
∴ ∠A + ∠B = ∠ACD = 112°

Correct Option: B

We draw a figure triangle ABC whose the side BC is produced to D ,

Exterior angle of a triangle is equal to the sum of remaining two interior angles.
From triangle ,
∴ ∠A + ∠B = ∠ACD = 112°

Given that , ∠A + ∠C = 140° and ∠A + ∠B = 180°
In a ∆ ABC,
We know that , ∠A + ∠B + ∠C = 180°
⇒ ∠B + 140° = 180°
⇒ ∠B = 180° – 140° = 40°
∴ ∠A + 3 ∠B = 180°

Correct Option: C

Given that , ∠A + ∠C = 140° and ∠A + ∠B = 180°
In a ∆ ABC,
We know that , ∠A + ∠B + ∠C = 180°
⇒ ∠B + 140° = 180°
⇒ ∠B = 180° – 140° = 40°
∴ ∠A + 3 ∠B = 180°
⇒ ∠A + 3 × 40° = 180°
⇒ ∠A = 180° – 120° = 60°.

According to question ,
We know that the sum of two sides of a triangle is greater than the third side.
We have , the set of three sides are 5 cm, 8 cm, 15 cm .

Correct Option: B

According to question ,
We know that the sum of two sides of a triangle is greater than the third side.
We have , the set of three sides are 5 cm, 8 cm, 15 cm .
∴ (5 + 8) < 15
(5 + 15) > 8
(15 + 8) > 5

We can say that the point of intersection of altitudes from the vertices of a triangle to opposite sides is called orthocentre.

Correct Option: B

We can say that the point of intersection of altitudes from the vertices of a triangle to opposite sides is called orthocentre. Hence , option B is correct answer .