Tiêu đề Advanced series 9th class Physics - Solutions.pdf ✅

Class 9 – Physics | A - 9 Basic Mathematics for Physics 1 1. Basic Mathematics for Physics Solutions Trigonometry LEVEL-I 1. Correct option: (C) We know, from the definition of tan A , that sin tan cos A A A = . Therefore, sin tan cos A A A = 2. Correct option: (D) We know, from the definition of cot A , that Adjacent side of cot Opposite side of A A A  =  . Therefore, Adjacent side of cot Opposite side of A A A  =  3. Correct option: (B) We know, 1 csc sin A A = Given, Opposite side of sin Hypotenuse A A  = Hene, 1 csc sin Hypotenuse csc Opposite side of A A A A = =  Therefore, if Opposite side of sin Hypotenuse A A  = , then Hypotenuse csc Opposite side of A A =  . 4. Correct option: (C) Given, in a triangle PQR, right angled at R, PQ = 29 units, QR = 21 units and  = PQR  . We know, from Pythagoras theorem, 2 2 2 c a b = + where c is the length of the hypotenuse of the right-angled triangle, and ab, are the length of its other two sides. Applying this to the given triangle PQR, we have :
Class 9 – Physics | A - 9 Basic Mathematics for Physics 2 2 2 2 2 2 2 2 2 PQ PR QR PR PQ QR PR PQ QR = + = − = − Substituting the given values PQ = 29 units, QR = 21 units, we get : PR = √29 2 − 21 2 PR = √29 2 − 21 2 PR = √400 PR = 20 We know, Opposite side of sin Hypotenuse A A  = and Adjacent side of cos Hypotenuse A A  = Applying this to the given triangle, we get: Opposite side of sin Hypotenuse sin Hypotenuse 20 sin 29 Q PR     = = = and Adjacent side of cos Hypotenuse cos Hypotenuse 21 cos 29 Q QR     = = = Therefore, 2 2 2 2 2 2 2 2 2 2 2 2 2 20 21 sin cos 29 29 20 21 sin cos 29 841 sin cos 841 sin cos         =      + = +           + + =       + =     + Therefore, if in a triangle PQR, right angled at R, PQ = 29 units, QR = 21 units and  = PQR  , then 2 2 sin cos   =  + . OR We can also solve this question from the identity that 2 2 sin cos   =  + .