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Chapters For Class X- CBSE


Trigonometric Ratios

Trigonometry is the study of relationships between the sides and angles of a triangle.

How to remember: (Say in one sentance) sin cos tan P B P H H B
You can replace the alphabet with words such as Pandit Badri Prasad Har Har Bol , or any thing as per your convience


Conside a ∆ABC, right angle at B. Now, if we take a point P on the hypotenuse AC or a point Q on AC extended, of the right triangle ABC and draw PM perpendicular to AB and QN perpendicular to AB extended (see Fig.), how will the trigonometric ratios of ∠ A in ∆ PAM differ from those of ∠ A in ∆ CAB or from those of ∠ A in ∆ QAN? A C B M N P Q

Explaination
Since ∆ PAM is similar to ∆ CAB , therefore
${AM}/{AB}={AP}/{AC}={MP}/{BC}$

i.e ${MP}/{AP}={BC}/{AC}=$sin A.

Similarly, ${AM}/{AP}={AB}/{AC}=cos A$

${MP}/{AM}={BC}/{AB}=tan A$

This shows that the trigonometric ratios of angle A in ∆ PAM not differ from those of angle A in ∆ CAB.

From our observations, it is now clear that the values of the trigonometric ratios of an angle do not vary with the lengths of the sides of the triangle, if the angle remains the same.

If in a right triangle ABC, sin A=$1/3$, this mean that ${BC}/{AC}=1/3$ , i.e., the lengths of the sides BC and AC of the ∆ ABC are in the ratio 1 : 3. So if BC is equal to k, then AC will be 3k, where k is any positive number. To determine other trigonometric ratios for the angle A, we need to find the length of the third side AB.

Since, $AB^2 = AC^2 – BC^2$ ,therefore $AB^2=(3k)^2 – (k)^2 = 8k^2 = (2√2k)^2 $

Therefore, AB = ±2√2k

Now, cos A =${AB}/{AC}={2√2k}/{3k}={2√2}/3$

Similarly, you can obtain the other trigonometric ratios of the angle A.

Remark : Since the hypotenuse is the longest side in a right triangle, the value of $sinA$ or $cosA$ is always less than 1 (or, in particular, equal to 1). 

Example:  Given  tan  A  = $4/3$,  find  the  other trigonometric ratios of the angle A.

Solution  : Draw  a  right  ∆  ABC
Since $tanA={BC}/{AB}=4/3$,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 =  5k
		
	Also, sin A =${BC}/{AC}={4k}/{5k}=4/5$
		cos A=${AB}/{AC}={3k}/{5k}=3/5$
		cot A=$1/{tan A}=3/4$
		cosec A=$1/{sin A}=5/4$
		sec A=$1/{cos A}=5/3$

Example :  If  ∠  B  and  ∠  Q  are acute angles such that sin B = sin Q, then prove that ∠ B = ∠ Q.
Solution : Let consider two right triangles  ABC  and  PQR  where sin B = sin Q 

Therefore, sin B=${AC}/{AB}$

and         sin Q=${PR}/{PQ}$

hence,      ${AC}/{AB}$= ${PR}/{PQ}$

Therefore, ${AC}/{PR}={AB}/{PQ}=k ..........(1)$

Now, using Pythagoras theorem,      
			
			$BC=√{AB^2-AC^2}$
			
	and     $QR=√{PQ^2-PR^2}$
	
	So, ${BC}/{QR}={√{AB^2-AC^2}}/{√{PQ^2-PR^2}}$
	
	From equation (1) , $AB=kPQ$ and $AC=kPR$
	
	     ${BC}/{QR}={√{k^2PQ^2-k^2PR^2}}/{√{PQ^2-PR^2}}$=${k({√{PQ^2-PR^2}})}/{√{PQ^2-PR^2}} ..........(2)$
		 
	From (1) and (2), we have
	
	${AC}/{PR}={AB}/{PQ}={BC}/{QR}$
	
Then, by using Theorem 6.4, ∆ ACB ~ ∆ PRQ and therefore, ∠ B = ∠ Q.
	


Example: Consider ∆ ACB, right-angled at C, in which AB = 29 units, BC = 21 units and ∠ ABC = θ.
 Determine the values of 
	(i)  cos2  θ + sin2  θ, 
	 (ii)  cos2  θ – sin2  θ.
	 
Solution : In ∆ ACB, we have $AC=√{AB^2-BC^2}$
		
							=$√{29^2-21^2}$ =$√{(29-21)(29+21)}$=$√{(8)(50)}$=$√400$=20 (Since,${(a^2-b^2)=(a-b)(a+b)}$)
							
	Since, sin θ =${AC}/{AB}=20/29$  and  cos θ =${BC}/{AB}=21/29$
	
	Now, (i) cos2  θ + sin2  θ = $(21/29)^2 + (20/29)^2$=${21^2+20^2}/29^2$=${441+400}/841$ =$1$
	
	(ii) cos2  θ – sin2  θ= $(21/29)^2 - (20/29)^2$=${(21-20)(21+20)}/29^2$ =$41/841$
	

	
Example: In a right triangle ABC, right-angled at B, if tan A = 1, then verify that
				2. sin A. cos A = 1.
				
Solution :  In ∆ ABC, tan A =${BC}/{AB}=1$
                       i.e., BC =  AB
			Let AB = BC = k, where k is a positive number.
					  
			Since, $AC=√{AB^2+BC^2}$
					  =$√({k^2+k^2}) =k√2$
					  
			Therefore, sin A=${BC}/{AC}={BC}/{AC}=1/√2$
			
			       cos A = ${AB}/{AC}=1/√2$
				   
			So, 2. sin A. cos A = 2.$1/√2$.$1/√2$=1.
And: Hence verified.



Example:  In  ∆   OPQ,  right-angled  at  P, OP = 7 cm and OQ – PQ = 1 cm. Determine the values of sin Q and cos Q.

Solution : 
		In ∆ OPQ, $OQ^2  =  OP^2   +  PQ^2$
		
		Given : OP = 7 cm and OQ – PQ = 1 cm
		
		Therefore, $(1+PQ)^2=OP^2 + PQ^2$, (Since $(a+b)^2=a^2+2ab+b^2$);
		
		i.e $1 + PQ^2  + 2PQ =  OP^2   +  PQ^2$
		
		$1 + 2PQ = OP^2$
		
		Since OP=7; Therefoer $2PQ=48$ ,$PQ=24$
		
		and $QR=1+PQ=1+24=25$
		
Ans:		So, sinQ= $7/25$ and cosQ= $24/25$