The lengths of two sides of a triangle are 8 centimetres and 10 centimetres and the angle between them is 40°. Calculate its area. What is the area of the triangle with sides of the same length, but angle between them 140°?

Consider a triangle ABC with AB = 10 cm, AC = 8 cm and

∠ BAC = 40°

Draw perpendicular from C on AB, say CH.

From right angles triangle ABC,

CH = CA × sin 40°

⇒ CH = 8 × sin 40°

(From the table, sin 40° = 0.642 )

⇒ CH = 5.136 cm

Area (Δ ABC) = 25.68 cm²

If given angle is 140° i.e. ∠ BAC = 140°

Draw perpendicular from C on AB and extend AB to meet it at H.

∴ CH ⊥ HB

HB is a straight line.

⇒ ∠ HAB = ∠ HAC + ∠ BAC = 180°

⇒ ∠ HAC + 140° = 180°

⇒ ∠ HAC = 40°

In right triangle CHA,

∠ HAC = 40° and AC = 8 cm

CH = CA × sin 40°

⇒ CH = 8 × sin 40°

(From the table, sin 40° = 0.642 )

⇒ CH = 5.136 cm

Area (Δ ABC) = 25.68 cm²

Hence, Area is same only the position of perpendicular line was changed if the angle is considered as 140°.

Area of triangle (given angle is 40°) = 25.68 cm² and area of triangle ( angle is taken as 140° ) = 25.68 cm².