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Complementary Angles

By JAIN College
Complementary Angles
Table of Contents

Introduction

Complementary angles are two angles whose values add up to 90o.

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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.

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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.

Properties of Complementary Angles

  • The sum of complementary angles in 90o.
  • Since by definition acute angles are less than 90o, the pair of complementary angles will individually be acute angles.
  • Complementary angles can be adjacent or nonadjacent angles. When two angles share a common vertex and side, it is called an adjacent angle. Likewise, when two angles do not share a common vertex and side they are called non-adjacent angles.
  • To find the value of a missing angle in complementary angles, the value of the known angle is subtracted from 90o.
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For instance, in the above complementary angle, the value of θ can be calculated as –

Θ = 90o – 53o

                = 37o

  • In trigonometry, the sine of one angle is equal to the cosine of its complementary.
  • The product of the sine and cosine of complementary angles is always 1.

Complementary Angles in the Right Triangle

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.

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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.

Practical Examples of Complementary Angles

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The arms of a clock

  • A close-up of a window
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Complementary angles on arches.

  • A slice of pizza with pepperoni on top
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In pizza slices.

  • A bridge over water with trees and a mountain in the background
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The beams in the bridges are welded to form complementary angles for better balance.

  • A diagram of a light source
Description automatically generated with medium confidence A football field with a goal
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In sports such as basketball and football, there are plays where the angle at which the ball is tackled is integral to the strategy.

Complementary vs. Supplementary Angles

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 A diagram of a variety of angles
Description automatically generated with medium confidence A diagram of a variety of angles
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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.

Complementary angles in Quadrilaterals and Polygons

In Euclidean geometry, a quadrilateral is a four-sided 2D figure whose sum of internal angles is 360°.  There are five types of quadrilaterals –

  1. Rectangle 
  2. Square
  3. Parallelogram 
  4. Rhombus
  5. Trapezium

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 –

  • ∠ADB and ∠BDC
  • ∠ABD and ∠DBC

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.

Complementary Angles in Trigonometry

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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 –

  • The sine of an angle is the ratio of the opposite angle to the hypotenuse.
  • The cosine of an angle is the ratio of the side adjacent to the angle to the hypotenuse.

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 -

  • Sine of an angle = cosine of its complementary angle
  • Cosine of an angle = sine of its complementary angle
  • Tangent of an angle = cotangent of its complementary angle

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

Trigonometric Problems with Complementary Angles

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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About the Author
Mekhala Joshi

JAIN College

JAIN PU College, a part of the renowned JGI Group, is committed to empowering students with quality education. Beyond academics, the college ensures its online content reflects the same standard of excellence. Every blog and article is meticulously vetted and proofread by subject matter experts to ensure accuracy, relevance, and clarity. From insightful educational topics to engaging discussions, JAIN PU College's content is crafted to inform, inspire, and add value to its readers, reflecting the institution's commitment to intellectual growth and innovation.

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