Find Angles of Similar Triangles
Find Angles of Similar Triangles

Complementary angles in Trigonometry

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Complementary angles in trigonometry : Two angles are said to be complementary, if their sum is 900.
It follows from the above definition that θ and ( 90 - θ ) are complementary angles in trigonometry for an acute angle θ
In ΔABC, ∠B = 900
∴ ∠A + ∠C = 900
∠C =900 - ∠A
For the sake of easiness in this derivation, we will write ∠C and ∠A as C and A respectively
Thus
C = 900 - A
sin A = BC / AC cosec A = AC / BC
cos A = AB / AC sec A = AC / AB
tan A = BC / AB cot A = AB / BC
and
sin C = sin (900 - A ) = AB / AC; cosec C = cosec (900 - A) = AC / AB
cos C = cos (900 - A) = BC / AC; sec C = sec (900 - A) = AC / BC
tan C = tan (900 - A) = AB / BC; cot C = cot (900 - A) = BC /AB
sin (900 - A) = cos A tan(900 - A) = cot A sec(900 - A) = cosec A
cos (900 - A) = sin A cot(900 - A) = tan A cosec (900 - A) = sec A

This means, for example
sin 700 = cos 200
The cofunction of the sine is the cosine. And 20° is the complement of 70°.
Some Solved Examples on complementary angles in trigonometry :
1) Evaluate : cos 370 / sin 530
Solution :
cos 370/sin 530 = cos( 90 – 53 )/sin 53 = sin 530/sin 530 = 1
________________________________________________________________
2) Show that : ( cos 700) / (sin 200) + (cos 590) / sin 310 - 8 sin2 300 = 0
Solution : Consider
( cos 700) / (sin 200) + (cos 590) / sin 310 - 8 sin2 300
= [ cos ( 900 - 200)] / [sin 200] + [ cos(900 - 310)] / sin 310 - 8 x(1/2) 2
= sin 200 / sin 200 + sin 310 / sin 310 - 8 x 1/ 4
= 1 + 1 – 2
= 0
∴ ( cos 700) / (sin 200) + (cos 590) / sin 310 - 8 sin2 300 = 0



Trigonometry

SOHCAHTOA -Introduction to Trigonometry
Trigonometric ratios and their Relation
Trigonometry for specific angles
Complementary angles in Trigonometry
Trigonometric Equations

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