Trigonometric Ratios of an Acute Angle

Trigonometric ratios are the foundation of trigonometry. They relate the angles of a right-angled triangle to the lengths of its sides. An acute angle in a right-angled triangle is an angle that is less than 90 degrees. These ratios provide a way to calculate unknown side lengths or angles when some information is known. The basic trigonometric ratios are sine (sin), cosine (cos), and tangent (tan). Their reciprocals are cosecant (csc or cosec), secant (sec), and cotangent (cot).

Formulae

Consider a right-angled triangle with angle θ. Let:

  • Opposite be the side opposite to angle θ.
  • Adjacent be the side adjacent to angle θ (not the hypotenuse).
  • Hypotenuse be the side opposite the right angle (the longest side).

Then:

  • Sine: $sin(\theta) = \frac{Opposite}{Hypotenuse}$
  • Cosine: $cos(\theta) = \frac{Adjacent}{Hypotenuse}$
  • Tangent: $tan(\theta) = \frac{Opposite}{Adjacent}$
  • Cosecant: $csc(\theta) = \frac{1}{sin(\theta)} = \frac{Hypotenuse}{Opposite}$
  • Secant: $sec(\theta) = \frac{1}{cos(\theta)} = \frac{Hypotenuse}{Adjacent}$
  • Cotangent: $cot(\theta) = \frac{1}{tan(\theta)} = \frac{Adjacent}{Opposite}$

Examples

Example-1: Consider a right-angled triangle ABC, where angle B is 90 degrees, angle A is θ, AB = 3 cm, BC = 4 cm, and AC = 5 cm.

Then, with respect to angle θ (angle A):

  • Opposite = BC = 4 cm
  • Adjacent = AB = 3 cm
  • Hypotenuse = AC = 5 cm

Therefore:

  • $sin(\theta) = \frac{4}{5}$
  • $cos(\theta) = \frac{3}{5}$
  • $tan(\theta) = \frac{4}{3}$
  • $csc(\theta) = \frac{5}{4}$
  • $sec(\theta) = \frac{5}{3}$
  • $cot(\theta) = \frac{3}{4}$

Example-2: Given a right-angled triangle with an acute angle θ, if $sin(\theta) = \frac{5}{13}$, find $cos(\theta)$ and $tan(\theta)$.

Since $sin(\theta) = \frac{Opposite}{Hypotenuse} = \frac{5}{13}$, let Opposite = 5 and Hypotenuse = 13. Using the Pythagorean theorem ($Adjacent^2 + Opposite^2 = Hypotenuse^2$), we can find the Adjacent:

$Adjacent = \sqrt{Hypotenuse^2 – Opposite^2} = \sqrt{13^2 – 5^2} = \sqrt{169 – 25} = \sqrt{144} = 12$

Now we can calculate:

  • $cos(\theta) = \frac{Adjacent}{Hypotenuse} = \frac{12}{13}$
  • $tan(\theta) = \frac{Opposite}{Adjacent} = \frac{5}{12}$

Real Life Application

Trigonometric ratios are used extensively in various real-life applications:

  • Navigation: Calculating distances and directions, used in GPS systems and mapping.
  • Engineering and Architecture: Designing structures, calculating angles and lengths for bridges, buildings, and other constructions.
  • Surveying: Measuring distances and elevations of land.
  • Physics: Analyzing motion, forces, and waves.
  • Astronomy: Calculating distances and positions of celestial objects.

Fun Fact

The word “sine” comes from the Latin word “sinus,” which means “fold” or “curve.” This term originated from the Arabic word “jiba,” which itself was derived from the Sanskrit word “jya,” meaning “chord.”

Recommended YouTube Videos for Deeper Understanding

Q.1 If in a right-angled triangle ABC, right-angled at B, AB = 3 cm and BC = 4 cm, then what is the value of $sin A$?
Check Solution

Ans: B

Using Pythagoras theorem, $AC = \sqrt{AB^2 + BC^2} = \sqrt{3^2 + 4^2} = 5$. Therefore, $sin A = \frac{opposite}{hypotenuse} = \frac{BC}{AC} = \frac{4}{5}$.

Q.2 In a right-angled triangle PQR, right-angled at Q, if $tan R = \frac{1}{\sqrt{3}}$, then what is the value of $cos R$?
Check Solution

Ans: B

If $tan R = \frac{1}{\sqrt{3}}$, then the ratio of opposite side (PQ) to adjacent side (QR) is $1:\sqrt{3}$. Let PQ = 1 and QR = $\sqrt{3}$. Using Pythagoras theorem, $PR = \sqrt{1^2 + (\sqrt{3})^2} = 2$. Then, $cos R = \frac{adjacent}{hypotenuse} = \frac{QR}{PR} = \frac{\sqrt{3}}{2}$.

Q.3 If $cosec \theta = 2$, then what is the value of $cot \theta$?
Check Solution

Ans: C

Since $cosec \theta = 2$, $sin \theta = \frac{1}{2}$. Using the identity $sin^2 \theta + cos^2 \theta = 1$, we get $cos^2 \theta = 1 – \frac{1}{4} = \frac{3}{4}$, therefore $cos \theta = \frac{\sqrt{3}}{2}$. Hence, $cot \theta = \frac{cos \theta}{sin \theta} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3}$.

Q.4 The value of $sec 60^\circ$ is equal to:
Check Solution

Ans: B

We know that $cos 60^\circ = \frac{1}{2}$. Since $sec \theta = \frac{1}{cos \theta}$, then $sec 60^\circ = \frac{1}{\frac{1}{2}} = 2$.

Q.5 In a right-angled triangle XYZ, right-angled at Y, if $XZ = 10$ cm and $XY = 5$ cm, then what is the value of $cos Z$?
Check Solution

Ans: A

$cos Z = \frac{adjacent}{hypotenuse} = \frac{YZ}{XZ}$. Using Pythagoras theorem, $YZ = \sqrt{XZ^2 – XY^2} = \sqrt{10^2 – 5^2} = \sqrt{75} = 5\sqrt{3}$. Then $cos Z = \frac{XY}{XZ} = \frac{5}{10} = \frac{1}{2}$.

Next Topic: Relationships between Trigonometric Ratios

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