Answer :

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.


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