8.2 Trigonometric Ratios
Let us take a right triangle ABC as shown in Fig. 8.4.
Here, ∠ CAB (or, in brief, angle A) is an acute angle. Note the position of the side BC with respect to angle A. It faces ∠ A. We call it the side opposite to angle A.
AC is the hypotenuse of the right triangle and the side AB is a part of ∠ A. So, we call it the side adjacent to angle A.the position of sides change when you consider angle C in place of A (see Fig. 8.5).
The trigonometric ratios of the angle A in right triangle ABC (see Fig. 8.4) are defined as follows :Sine of $$\angle A=\frac{side\:L\: opposite\:\: to \:\:angle A}{hypotenuse}=\frac{BC}{AC}$$
Cosine of $$\angle A=\frac{side \:\:adjacent\:\: to \:\:angle A}{hypotenuse}=\frac{AB}{AC}$$
Tangent of $$\angle A=\frac{side\:\: opposite\:\: to \:\:angle A}{ side \:\:adjacent\:\: to\:\: angle A }=\frac{BC}{AB}$$
Cosecant of $$\angle A=\frac{1}{sine \:\:of\:\: \angle A}=\frac{hypotenuse}{side\:\: opposite\:\: to \:\:angle A}=\frac{AC}{BC}$$
Secant of $$\angle A=\frac{1}{cosine \:\:of \angle A}=\frac{hypotenuse}{side\:\: adjacent\:\: to \:\:angle A} =\frac{AC}{AB}$$
Cotangent of $$\angle A=\frac{1}{tangent \angle A}=\frac{side\:\: adjacent \:\:to\:\: angle A}{side opposite\:\: to \:\:angle A}=\frac{AB}{BC}$$
The ratios defined above are abbreviated as sin A, cos A, tan A, cosec A, sec A and cot A respectively. Note that the ratios cosec A, sec A and cot A are respectively, the reciprocals of the ratios sin A, cos A and tan A.
Also , observe that tan $$A=\frac{BC}{AB}=\frac{\frac{BC}{AC}}{\frac{AB}{AC}}=\frac{sin A}{cos A} and cot A =\frac{cos A}{sin A}$$
So, the trigonometric ratios of an acute angle in a right triangle express the relationship between the angle and the length of its sides.
Example 1 : Given tan $$A =\frac{4}{3}$$, find the other trigonometric ration of the angle A.
Solution : Let us first draw a right ∆ ABC (see Fig 8.8).Now , we know that tan $$A=\frac{BC}{AB}=\frac{4}{3}$$.
Therefore, if BC = 4k, then AB = 3k, where k is a positive number.
Now, by using the Pythagoras Theorem, we have
$$AC^{2} = AB^{2} + BC^{2} = (4k )^{2} + (3k)^{2} = 25k^{2}$$
So, AC = 5 k
Now, we can write all the trigonometric ratios using their definitions.
$$Sin A=\frac{BC}{AC}=\frac{4k}{5k}=\frac{4}{5}$$
$$Cos A=\frac{AB}{AC}=\frac{3k}{5k}=\frac{3}{5}$$
Therefore , $$cot A=\frac{1}{tan A}=\frac{3}{4},\:\: cosec A=\frac{1}{sin A}=\frac{5}{4}$$ and sec $$A=\frac{1}{cos A}=\frac{5}{3}$$.
Example 2 : If ∠ B and ∠ Q are acute angles such that sin B = sin Q, then prove that ∠ B = ∠ Q.
Solution : Let us consider two right triangles ABC and PQR where sin B = sin Q.We have $$sin B=\frac{AC}{AB}$$
And $$sin Q=\frac{PR}{PQ}$$
Then $$\frac{AC}{AB}=\frac{PR}{PQ}$$
Therefore, $$\frac{AC}{PR}=\frac{AB}{PQ}=k$$, say……………(1)
Now , using Pythagoras theorem,
$$BC=\sqrt{AB^{2}-AC^{2}}$$
And $$QR=\sqrt{PQ^{2}-PR^{2}}$$
So, $$\frac{BC}{QR}=\frac{\sqrt{AB^{2}-AC^{2}}}{\sqrt{PQ^{2}-PR^{2}}}=\frac{k^{2}PQ^{2}-k^{2}PR^{2}}{\sqrt{PQ^{2}-PR^{2}}}=\frac{k\sqrt{PQ^{2}-PR^{2}}}{\sqrt{PQ^{2}-PR^{2}}}=k$$…………(2)
From (1) and (2), we have
$$\frac{AC}{PR}=\frac{AB}{PQ}=\frac{BC}{QR}$$
Then, by using Theorem 6.4, ∆ ACB ~ ∆ PRQ and therefore, ∠ B = ∠ Q.
Example 3 : Consider ∆ ACB, right-angled at C, in which AB = 29 units, BC = 21 units and ∠ ABC = θ (see Fig. 8.10). Determine the values of
(i) $$cos^{2} θ + sin^{2} θ,$$
(ii) $$cos^{2} θ – sin^{2} θ$$.
Solution : In ∆ ACB, we have
$$AC=\sqrt{AB^{2}-BC^{2}}=\sqrt{29^{2}-21^{2}}$$
=$$\sqrt{(29-21)(29+21)}=\sqrt{8X50}=\sqrt{400}=20$$ units
So, $$sin\theta = \frac{AC}{AB}=\frac{20}{29}, cos\theta =\frac{BC}{AB}=\frac{21}{29}.$$
i)$$cos^{2}\theta +sin^{2}\theta$$ =$$(\frac{20}{29})^{2}+(\frac{21}{29})^{2}=\frac{20^{2}+21^{2}}{29^{2}}=\frac{400+441}{841}=1$$
ii) $$cos^{2}\theta - sin^{2}\theta$$ =$$(\frac{20}{29})^{2}- (\frac{21}{29})^{2} =\frac{21+20)(21-20)}{29^{2}}=\frac{41}{841}$$
Example 4 : In a right triangle ABC, right-angled at B, if tan A = 1, then verify that 2 sin A cos A = 1.
Solution : In $$\Delta ABC,\:\: tan A=\frac{BC}{AB}=1$$
i.e., BC=AB
let AB=BC=k, where k is a positive number.
Now, $$AC=\sqrt{AB^{2}+BC^{2}}$$
=$$\sqrt{k^{2}+k^{2}}=k\sqrt{2}$$
Therefore, $$sin A=\frac{BC}{AC}=\frac{1}{\sqrt{2}} and cos A=\frac{AB}{AC}=\frac{1}{\sqrt{2}}$$
So, $$2sin A cos A=2(\frac{1}{\sqrt{2}})(\frac{1}{\sqrt{2}})=1$$, which is the required value.
Example 5 : In ∆ OPQ, right-angled at P, OP = 7 cm and OQ – PQ = 1 cm (see Fig. 8.12). Determine the values of sin Q and cos Q.
Solution : In ∆ OPQ, we have
$$OQ^{2}=OP^{2}+PQ^{2}$$
i.e., $$(1+PQ)^{2}=OP^{2}+PQ^{2}$$
i.e., $$1+PQ^{2}+2PQ=OP^{2}+PQ^{2}$$
i.e., $$1+2PQ=7^{2}$$
i.e., PQ=24 cm and OQ=1+PQ=25 cm
so, $$ sin Q=\frac{7}{25} and Q=\frac{24}{25}.$$
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