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Introduction to Trigonometry - Evaluate trigonometric ratios at standard angles 0 degree, 30 degree, 45 degree, 60 degree, and 90 degree

Grade 10CBSE

Review the key concepts, formulae, and examples before starting your quiz.

πŸ”‘Concepts

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The trigonometric ratios of 45∘45^\circ are derived using an isosceles right-angled triangle where the two legs are equal. If each leg is aa, the hypotenuse is a2a\sqrt{2} by Pythagoras theorem. Thus, sin⁑45∘=aa2=12\sin 45^\circ = \frac{a}{a\sqrt{2}} = \frac{1}{\sqrt{2}} and cos⁑45∘=aa2=12\cos 45^\circ = \frac{a}{a\sqrt{2}} = \frac{1}{\sqrt{2}}.

Isosceles right triangle with sides a, a and hypotenuse a*sqrt(2) showing 45 degree angle.
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For angles 30∘30^\circ and 60∘60^\circ, we use an equilateral triangle with side 2a2a. Dropping a perpendicular from one vertex bisects the base and creates a 30βˆ˜βˆ’60βˆ˜βˆ’90∘30^\circ-60^\circ-90^\circ triangle with sides aa, a3a\sqrt{3}, and 2a2a.

Equilateral triangle bisected to show 30 and 60 degree angles with side ratios.
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As θ\theta increases from 0∘0^\circ to 90∘90^\circ, sin⁑θ\sin \theta increases from 00 to 11, while cos⁑θ\cos \theta decreases from 11 to 00. The value of tan⁑θ\tan \theta increases from 00 to infinity (undefined at 90∘90^\circ).

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Reciprocal relationships are essential for evaluation: sec⁑θ\sec \theta is the inverse of cos⁑θ\cos \theta, cosec θ\text{cosec } \theta is the inverse of sin⁑θ\sin \theta, and cot⁑θ\cot \theta is the inverse of tan⁑θ\tan \theta.

πŸ“Formulae

sin⁑0∘=0,sin⁑30∘=12,sin⁑45∘=12,sin⁑60∘=32,sin⁑90∘=1\sin 0^\circ = 0, \sin 30^\circ = \frac{1}{2}, \sin 45^\circ = \frac{1}{\sqrt{2}}, \sin 60^\circ = \frac{\sqrt{3}}{2}, \sin 90^\circ = 1

cos⁑0∘=1,cos⁑30∘=32,cos⁑45∘=12,cos⁑60∘=12,cos⁑90∘=0\cos 0^\circ = 1, \cos 30^\circ = \frac{\sqrt{3}}{2}, \cos 45^\circ = \frac{1}{\sqrt{2}}, \cos 60^\circ = \frac{1}{2}, \cos 90^\circ = 0

tan⁑0∘=0,tan⁑30∘=13,tan⁑45∘=1,tan⁑60∘=3,tan⁑90∘=Not Defined\tan 0^\circ = 0, \tan 30^\circ = \frac{1}{\sqrt{3}}, \tan 45^\circ = 1, \tan 60^\circ = \sqrt{3}, \tan 90^\circ = \text{Not Defined}

cosec θ=1sin⁑θ\text{cosec } \theta = \frac{1}{\sin \theta}

sec⁑θ=1cos⁑θ\sec \theta = \frac{1}{\cos \theta}

cot⁑θ=1tan⁑θ=cos⁑θsin⁑θ\cot \theta = \frac{1}{\tan \theta} = \frac{\cos \theta}{\sin \theta}

πŸ’‘Examples

Problem 1:

Evaluate: sin⁑60∘cos⁑30∘+sin⁑30∘cos⁑60∘\sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ

Solution:

Step 1: Substitute the values of trigonometric ratios. sin⁑60∘=32\sin 60^\circ = \frac{\sqrt{3}}{2} cos⁑30∘=32\cos 30^\circ = \frac{\sqrt{3}}{2} sin⁑30∘=12\sin 30^\circ = \frac{1}{2} cos⁑60∘=12\cos 60^\circ = \frac{1}{2}

Step 2: Place them into the expression: =(32Γ—32)+(12Γ—12)= (\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}) + (\frac{1}{2} \times \frac{1}{2})

Step 3: Simplify the terms: =34+14= \frac{3}{4} + \frac{1}{4}

Step 4: Add the fractions: =3+14=44=1= \frac{3+1}{4} = \frac{4}{4} = 1

Explanation:

This problem uses the standard values of trigonometric ratios for 30∘30^\circ and 60∘60^\circ. It also demonstrates the identity sin⁑(A+B)=sin⁑Acos⁑B+cos⁑Asin⁑B\sin(A+B) = \sin A \cos B + \cos A \sin B where A=60∘A=60^\circ and B=30∘B=30^\circ.

Problem 2:

In β–³ABC\triangle ABC, right-angled at BB, AB=5AB = 5 cm and ∠ACB=30∘\angle ACB = 30^\circ. Determine the length of the side BCBC.

Solution:

Step 1: Identify the given information and the required side. Given: Perpendicular (ABAB) = 55 cm, θ=30∘\theta = 30^\circ. To find: Base (BCBC).

Step 2: Choose the trigonometric ratio that relates Perpendicular and Base. tan⁑θ=PerpendicularBase\tan \theta = \frac{\text{Perpendicular}}{\text{Base}} tan⁑30∘=ABBC\tan 30^\circ = \frac{AB}{BC}

Step 3: Substitute the known values. 13=5BC\frac{1}{\sqrt{3}} = \frac{5}{BC}

Step 4: Solve for BCBC. BC=53BC = 5\sqrt{3} cm

Explanation:

To find a missing side when an angle and one side are given, identify which trigonometric ratio (sin, cos, or tan) connects the given side and the side to be found. Here, tangent is used because we are dealing with the opposite (perpendicular) and adjacent (base) sides.

Problem 3:

In a right triangle PQRPQR, right-angled at QQ, PQ=3PQ = 3 cm and PR=6PR = 6 cm. Find ∠QPR\angle QPR and ∠PRQ\angle PRQ.

Right triangle PQR with PQ=3 and PR=6.

Solution:

  1. To find ∠PRQ\angle PRQ (let it be θ\theta): sin⁑θ=OppositeHypotenuse=PQPR=36=12\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{PQ}{PR} = \frac{3}{6} = \frac{1}{2}
  2. Since sin⁑30∘=12\sin 30^\circ = \frac{1}{2}, we have ∠PRQ=30∘\angle PRQ = 30^\circ.
  3. To find ∠QPR\angle QPR: We know ∠P+∠Q+∠R=180∘\angle P + \angle Q + \angle R = 180^\circ ∠QPR+90∘+30∘=180∘\angle QPR + 90^\circ + 30^\circ = 180^\circ ∠QPR=60∘\angle QPR = 60^\circ.

Explanation:

We use the definition of the sine ratio for the known sides to identify the standard angle. Then, the angle sum property of a triangle is used to find the third angle.

Problem 4:

Evaluate the following expression: 2tan⁑245∘+cos⁑230βˆ˜βˆ’sin⁑260∘2 \tan^2 45^\circ + \cos^2 30^\circ - \sin^2 60^\circ

A right-angled isosceles triangle ABC showing sides of length 1 and an angle of 45 degrees to illustrate tan 45 = 1.

Solution:

We know the values of the trigonometric ratios at standard angles: tan⁑45∘=1\tan 45^\circ = 1 cos⁑30∘=32\cos 30^\circ = \frac{\sqrt{3}}{2} sin⁑60∘=32\sin 60^\circ = \frac{\sqrt{3}}{2}

Substituting these values into the expression: 2(1)2+(32)2βˆ’(32)22(1)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 - \left(\frac{\sqrt{3}}{2}\right)^2

Calculating the squares: 2(1)+34βˆ’342(1) + \frac{3}{4} - \frac{3}{4}

Simplifying the terms: 2+0=22 + 0 = 2

Explanation:

To solve this problem, we identify the values of the specific trigonometric functions at the given angles. Notice that cos⁑30∘\cos 30^\circ and sin⁑60∘\sin 60^\circ are equal (32\frac{\sqrt{3}}{2}), so their squares cancel each other out, leaving only the term containing tan⁑45∘\tan 45^\circ.

Evaluate trigonometric ratios at standard angles 0 degree, 30 degree, 45 degree, 60 degree, and 90…