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Right-Angled Triangles Simplified Revision Notes

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5.1.2 Right-Angled Triangles

Right-angled triangles are triangles in which one of the angles is exactly 90 degrees. This type of triangle has several unique properties and is foundational in trigonometry, geometry, and many practical applications.

Key Components of a Right-Angled Triangle:

infoNote
  1. Hypotenuse:
  • The hypotenuse is the side opposite the right angle. It is always the longest side of the triangle.
  1. Adjacent Side:
  • The adjacent side is the side next to the angle of interest, excluding the hypotenuse.
  1. Opposite Side:
  • The opposite side is the side opposite the angle of interest.

Pythagoras' Theorem:

infoNote

Pythagoras' Theorem is a fundamental relation in a right-angled triangle, stating: Hypotenuse2=Opposite Side2+Adjacent Side2\text{Hypotenuse}^2 = \text{Opposite Side}^2 + \text{Adjacent Side}^2

Or in terms of the sides:

c2=a2+b2c^2 = a^2 + b^2

Where:

  •  c\ c is the length of the hypotenuse,
  •  a and b\ a \ and \ b are the lengths of the other two sides.

Trigonometric Ratios:

infoNote

In a right-angled triangle, the primary trigonometric ratios are defined as follows:

  • Sine (sin): sinθ=OppositeHypotenuse\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}}
  • Cosine (cos): cosθ=AdjacentHypotenuse\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}}
  • Tangent (tan): tanθ=OppositeAdjacent\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}}
infoNote

Example Problem:

Problem: In a right-angled triangle, the length of the hypotenuse is 10 units, and one of the angles is  30\ 30^\circ . Find the lengths of the opposite and adjacent sides.


Solution:

  1. Identify the Sides:
  • Hypotenuse  c=10\ c = 10 units.
  • The angle  θ=30.\ \theta = 30^\circ .
  • We need to find the opposite side  a\ a and adjacent side  b.\ b .
  1. Use the Sine Function to Find the Opposite Side: sin30=a10\sin 30^\circ = \frac{a}{10} Since  sin30=12,\ \sin 30^\circ = \frac{1}{2} , we have: 12=a10\frac{1}{2} = \frac{a}{10} Solving for  a\ a : a=102=5 unitsa = \frac{10}{2} = 5 \text{ units}
  2. Use the Cosine Function to Find the Adjacent Side: cos30=b10\cos 30^\circ = \frac{b}{10} Since  cos30=32\ \cos 30^\circ = \frac{\sqrt{3}}{2} , we have:  32=b10\ \frac{\sqrt{3}}{2} = \frac{b}{10} Solving for  b:\ b : b=10×32=53 unitsb = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3} \text{ units} Final Answer:
  • Opposite side  a=:success[5]\ a = :success[5] units.
  • Adjacent side  b=:success[53]\ b = :success[5\sqrt{3}] units.

Area of a Right-Angled Triangle:

The area of a right-angled triangle can be calculated using:

Area=12×base×height\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}

In the context of trigonometry:

  • The base and height correspond to the two sides forming the right angle (not the hypotenuse).
infoNote

Example Problem for Area:

Problem: Find the area of a right-angled triangle where the lengths of the two shorter sides are 6 units and 8 units.


Solution:

  1. Identify the Base and Height:
  • Base  =6\ = 6 units.
  • Height  =8\ = 8 units.
  1. Calculate the Area: Area=12×6×8=12×48=24 square units\text{Area} = \frac{1}{2} \times 6 \times 8 = \frac{1}{2} \times 48 = 24 \text{ square units} Final Answer:
  • The area is 24 square units.

Summary:

Right-angled triangles are critical in trigonometry and geometry, with their properties governed by Pythagoras' Theorem and the trigonometric ratios. These concepts allow for solving various problems related to angles, side lengths, and areas in both theoretical and applied contexts.

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