NCERT solutions for class 10

1. https://www.entrancei.com/ https://www.entrancei.com/ncert-solutions-class-10-maths

2. https://www.entrancei.com/ INTRODUCTION AND APPLICATIONS OF TRIGONOMETRY 1. Trigonometric ratios of angle A in right triangle ABC right angled at B (i) Side sine A (sin A) opposite = BC to angle A = Hypotenuse AC C (ii) cosine A (cos A) adjacent Side = AB to angle A = Hypotenuse AC (iii) tangent A (tan A) opposite Side = BC to angle A = Side adjacent to angle A A B A B (iv) cosecant A (cosec A)= Hypotenuse AC = Side opposite to angle A BC Hypotenuse AC = (v) secant A (sec A) = Side adjacent to angle A AB Side opposite to angle A AB = (vi) cotangent A (cot A) = Side adjacent to angle A BC Above formulae can be memorized as below: Pandit Har Badri Har Prasad Bole Sona Chandi Tole prepinducu lar( p ) base( b ) prepinducu lar( b p ) sin  =  =  = cos tan hypotenuse ( h ) hypotenuse ( h ) base( ) 2. Reciprocal relation 1 1 1 cosec A = , sec A = cosA , cot A = . sin A tan A For example: 24 = In a ABC, right angled at B, tan A , find the other trigonometric ratios of the angle 7 A. In ABC BC 24 24 = =  tan A 7 AB 7 https://www.entrancei.com/ncert-solutions-class-10-maths

3. https://www.entrancei.com/  BC : AB = 24 : 7 Let the positive real number k be the constant of proportionality. C  BC = 24 k and AB = 7 k Now, using Pythagoras theorem, we have 24 k 2 2 2 2 2 + = + AC = AB BC ( 7 k ) ( 24 k ) 2 2 2 + = = 49 k 576 k 625 k A 7 cm B 2  AC = 625k  AC = 25 k BC 24 k 24 24 = =  = sin A = , i.e., sin A AC 25 k 25 25 AB 7 k 7 7 = = = cos A = , i.e., cos A AC 25 k 25 25 1 25 1 25 1 7 = = = = = cosec A = and , sec A cot A sin A 24 cos A 7 tan A 24 3. Quotient relation of trigonometric ratios sin A cos A A tan A = cot A = , sin A cos 4. Values of all the trigonometric ratios of 0 , 30 , 45 , 60  and 90 . A 0° 30° 45° 60° 90° 1 1 3 0 1 sin A 2 2 2 1 1 3 1 0 cos A 2 2 2 1 0 1 Not defined tan A 3 3 2 Not defined 2 1 cosec A 2 3 2 1 2 Not defined sec A 2 3 1 Not defined 1 0 cot A 3 3 sin for  = 0, 30, 45, 60 and 90.  There is an easy way to remember the values of  0 30 45 60 90 https://www.entrancei.com/ncert-solutions-class-10-maths

4. https://www.entrancei.com/ Write the five numbers in the sequence of 0, 1, 2, 3, 4. Divide by 4 and take their square root. cos  Write the values of sin  by cos  in reverse order 0 1 Increasing order sin  1 1 3 2 2 2 1 0 Decreasing order 1 1 3 2 2 1 2 1 0 Not defined Increasing order tan  Dividing values of sin  by cos sin tan 3  3  =  i.e.,  cos Note: (i) The values of sin  increases from 0 to 1 as  increases from 0 to 90 and value of cos  decreases from 1 to 0 as  increases from 0 to 90. The value of tan  also increases from 0 to a bigger number as  increases from 0 to 90. A  B A tan tan  and sec , cosec cosec B A  A  A A  (ii) If A and B are acute angles such that then sec B  , sin cot sin cot B B , . cos cos B ,  A B , For example: 1 1 2 2 2 2 2 + + + Evaluate: sin 30º cos 45º 4 tan 30º sin 90º cot 60º 2 8 1 1 2 2 2 2 2 + + + sin 30º cos 45º 4tan 30º sin 90º cot 60º 2 8 2 2 2 2         1 1 1 1 1 1 2       1       =  +  + ) 1 (  +  4     2 2 8 2 3 3     1 1 1 1 1 1 1 4 1 =  +  +  +  = + + + 4 1   3       4 2 3 2 8 3 8 3 2 24 + + + 32 12 1 48 = = = 2 24 24 5. Trigonometric ratios of complementary angles. (i) sin (90° - A) = cos A (iii) tan (90° - A) = cot A (v) cosec (90° - A) = sec A (ii) (iv) (vi) cos (90° - A ) = sin A cot (90° - A) = tan A sec (90° - A) = cosec A For example: + B C A = sin cos If A, B and C are the interior angles of a triangle ABC, show that . 2 2 Since A, B and C are angles of a ABC, A + B + C = 180º. + 2 + + 2 −     B C A B C A = Now, sin sin         −     180 º A A A = − = = . sin sin 90 º cos         2 2 2 6. Fundamental trigonometric identities. (Square relation) cos2 A + sin2 A = 1 (i) (ii) sec2 A = 1 + tan2 A https://www.entrancei.com/ncert-solutions-class-10-maths

5. https://www.entrancei.com/ (iii) cosec2 A = 1 + cot2 A For example: 2 2 2 2 − − (sin A 2 sin B) (cos B 2 cos A) 2 2 − = = Prove that: (tan A tan B) 2 2 cos Acos B cos B cos A We have 2 2 sin A sin B 2 2 − A− = (tan tan B ) 2 2 cos A cos B 2 2 2 2 − sin A cos B 2 cos A sin B = 2 cos A cos B 2 2 2 2 − − − sin A 1 ( sin B 2 ) 1 ( sin A ) sin B = 2 cos A cos B 2 2 2 2 2 2 − − + sin A sin A sin B 2 sin B sin A sin B = 2 cos A cos B 2 2 − (sin A 2 sin B ) = 2 cos A cos B 2 2 sin A sin B 2 2 − A− Also, = (tan tan B ) 2 2 cos A cos B 2 2 2 2 − sin A cos B 2 cos A sin B = 2 cos A cos B 2 2 2 2 − − − 1 ( cos A ) cos B cos A 1 ( cos B ) = 2 2 cos A cos B 2 2 2 2 2 2 − − + cos B cos A cos B 2 cos A cos A cos B = 2 cos A cos B 2 2 − (cos B 2 cos A ) = 2 cos B cos A 2 2 2 2 − − (sin A 2 sin B ) (cos B 2 cos A ) 2 2 = A− Hence, = (tan tan B ) 2 2 cos A cos B cos B cos A APPLICATIONS OF TRIGONOMETRY 1. Line of sight: The line of sight is the line drawn from the eye of an observer to the point in the object viewed by the observer. https://www.entrancei.com/ncert-solutions-class-10-maths

6. https://www.entrancei.com/ Object 2. Angle of elevation: It is the angle formed by the line of sight with horizontal through the eyes of observer when the object is above the horizontal level. Line of sight Angle of elevation Eye Horizontal level 3. Angle of depression: Eye Horizontal level It is the angle formed by the line of sight with the horizontal when the object is below the horizontal level. Angle of depression Line of sight Object 4. The height of length of an object or the distance between distant objects can be determined with the help of trigonometric ratios. 5. Solving right triangle in applications of trigonometry Step: 1 Make all the possible right angles. Step: 2 Mark all given quantities or angles. If the figure involves more than one triangle always assume the common side (say x) connecting these right triangles to proceed. Step: 3 Write trigonometric ratios involving given quantity, assumed quantity and the quantity to be calculated using given angles. Step 4: Step 5: Solve the simultaneous expressions obtained in step 4 to get the solution. For example: A statue 1.6 m tall stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60º and from the same point, the angle of elevation of the top of the pedestal is 45º. Find the height of the pedestal. https://www.entrancei.com/ncert-solutions-class-10-maths

7. https://www.entrancei.com/ Let BD is a pedestal and AD is a statue with height 1.6 m. BCD = 45º, ACB = 60º BC = x m Height of statue = 1.6 m In ABC, we have AB = º 60 tan A 1.6m D BC h + 6 . 1 h = 3 x 60º + 6 . 1 h 45º = ..... (i) x x C B 3 In BCD, we have BD = tan 45 º BC h = 1 x From (i) and (ii), x = h ..... (ii) + 6 . 1 h = h 3 = + h 3 6 . 1 h − h 6 . 1 = h 3 ) 1 − 6 . 1 = h ( 3 + 6 . 1 3 1 =  h − + 3 1 3 1 + 6 . 1 ( 3 − ) 1 = h 3 . 1 ( 1 732 + 6 . 1 ) 1 = 2 h = 0.8 (2.732) h = 2.19 m (approx) https://www.entrancei.com/ncert-solutions-class-10-maths

8. https://www.entrancei.com/ https://www.entrancei.com/ncert-solutions-class-10-maths