Complementary angles are two angles whose values add up to 90o.
Since 90o measures the right angle, it can be concluded that if two angles are complementary, they form a right angle together. For instance, in the above diagram, the ray has divided a right angle into a pair of complementary angles of 30o and 60o.
Complementary angles may or may not share a common ray and vertex.
For instance, in the above diagram, one angle is 25o and the other is 65o. When added, the sum of both angles is 90o,which makes it a right angle.
For instance, in the above complementary angle, the value of θ can be calculated as –
Θ = 90o – 53o
= 37o
Complementary angles are naturally present in a right triangle. In a right-angle triangle, the two acute angles are complementary. As the sum of all interior angles in a triangle is 180o since the right angle is 90o, the sum of the remaining two acute angles will be 90o.
The above figure is a right-angle triangle with ∠ABC = 90o and ∠ACB = 40o. Since the sum of all interior angles of a triangle is 180o,
∠BAC = 180o – (∠ABC + ∠ACB)
Thus,
∠BAC =180o – (90o + 40o)
= 180o – 130o
= 50o
Also,
∠BAC + ∠ACB = 50o + 40o = 90o
Thus, it can be observed that the sum of the values of the two acute angles in the right-angle triangle is equal to that of the right angle. This proves that the angles ∠BAC and ∠ACB are complementary.
The arms of a clock
Complementary angles on arches.
In pizza slices.
The beams in the bridges are welded to form complementary angles for better balance.
In sports such as basketball and football, there are plays where the angle at which the ball is tackled is integral to the strategy.
Areas of differences | Complementary Angles | Supplementary Angles |
The sum of the angles | The sum of two complementary angles is 90o. | The sum of two supplementary angles is 180o. |
Diagram | ||
Nature of angles | The complementary angles are always acute angles with a measure of less than 90o. | Supplementary angles will always have one acute and one obtuse angle. That is, one angle >90o and one angle <90o but >180o. |
The sum of the angles | The sum of two complementary angles is 90o. | The sum of two supplementary angles is 180o. |
Linear angle relation | It does not apply to a pair of linear angles. | It applies to a pair of linear angles. |
In Euclidean geometry, a quadrilateral is a four-sided 2D figure whose sum of internal angles is 360°. There are five types of quadrilaterals –
Complementary angles are observed in quadrilaterals and other polygons. Rectangles and squares have right angles and any line passing diagonally to join two opposite ends will create complementary angles on either side.
In the above diagram of the square, ∠ADBC and ∠CBA are divided by a diagonal ray DB. Since both angles are right angles, the ray DB creates 2 pairs of complementary angles –
Squares and rectangles have right angles, thus, when a diagonal connects the opposite angles, they will have adjacent complementary angles that measure 45o.
Figures such as the trapezium, parallelogram and rhombus have supplementary angles as they have obtuse angles with at least one angle measuring more than 90o.
Base (b) or side adjacent to ∠C |
The sine and cosine of angles are ratios of the pairs of sides in right-angled triangles. In a right-angled triangle –
Thus, in the ΔABC, ∠B is the right angle. Since the sum of the measures of the three angles must be 180o,
m∠A + m∠B + m∠C = 180o
Since, m∠B = 90o
m∠A + m∠C = 90o
Thus, in ΔABC, the trigonometric ratios of complementary angles are –
for reference angle θ | for reference angle (90° − θ) |
sin θ = AB/ ACcos θ = AB/BCtan θ = BC/ABcosec θ = AC/ABsec θ = AC/BCcot θ = BC/AB | sin (90° − θ) = cos θcos (90° − θ) = sin θtan (90° − θ) = cot θcot (90° − θ) = tan θsec (90° − θ) = cosec θcosec (90° − θ) = sec θ |
From the above equations, it can be observed that -
In the above triangle –
sin C = cos A = AB/ AC
sin A = cos C = AB/ AC
tan A = cot C = AB/BC
tan C = cot A = BC/AB
sec C = cosec A = AC/BC
sec A = cosec C= AC/AB
Q1. If sec 4θ = cosec (θ – 40o), where 4θ is an acute angle, find the value of θ.
Solution1:
sec 4θ = cosec (θ – 40o)
⇒ cosec (90° - 4θ) = cosec (θ - 40°),
[Since sec θ = c0sec (90° - θ)]
⇒ (90° - 4θ) = (θ - 40°)
⇒ -4θ - θ = -40° - 90°
⇒ -5θ = -130°
⇒ θ = 26°
Therefore, θ = 26°
Q2. In any right ΔABC, prove that
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