NCERT Class 10th solution of Exercise 10.2 NCERT Class 10th Maths Projects Exercise 10.1 Q1. How many tangents can a circle have? Answer: Infinitely many. Q2. Fill in the blanks: i) A tangents to a circle intersects it in _____ point (s). ii) A line intersecting a circle in two points is called a ______. iii) A circle can have ______ parallel tangents at the most. iv) The common point of a tangent to a circle and the circle is called ______. Answer: i) `1`     ii) secant     iii) `2`     iv) point of contact. Q3. A tangent `PQ` at a point `P` of a circle of radius `5 cm` meets a line through the centre `O` at a point `Q` so that `OQ = 12 cm`. Length `PQ` is : A) `12 cm`     B) `13 cm`     C) `8.5 cm`     D) `sqrt119 cm` Answer: D) `sqrt119 cm` Q4. Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle. Sol. : Step of Construction:- 1) Draw a circle of centre `O` and...

Chapter 8 Introduction to Trigonometry NCERT Class 10th solution of Exercise 8.1 NCERT Class 10th solution of Exercise 8.2 NCERT Class 10th solution of Exercise 8.3 NCERT Class 10th Maths Projects Exercise 8.4 Q1. Express the trigonometric ratios `sin A, sec A` and `tan A` in terms of `cot A`. Sol. : `sin A = 1/(cosec A) = 1/(sqrt (cosec^2 A)) = 1/sqrt(1 + cot^2 A)` `sec A = 1/(cos A) = (1/sin A)/(cos A/sin A) = (cosec A)/(cot A)` `sec A = sqrt(cosec^2 A)/(Cot A) = sqrt(1 + cot^2 A)/cot A` `tan A = 1/cot A` Answer: `sin A = 1/sqrt(1 + cot^2 A)`, `sec A = sqrt(1 + cot^2 A)/cot A` and `tan A = 1/cot A` ☝  Like, Share,  and  Subscribe.  Q2. Write all the other trigonometric ratios of `angle A` in terms of `sec A`. Sol. : Let `triangle ABC` is right-angled at `angleB` and `sec A = x` `sec A = x = (AC)/(AB)` By Pythagoras theorem `BC^2 = AC^2 - AB^2` `BC^2 = x^2 - 1` `BC = sqrt(x^2 -1)` `sin A = (BC)/(AC) = sqrt(x^2 - 1)/x = sqrt(sec^2 - 1)/(sec A)` `cos A = (AB)/(AC) = 1...

Chapter 8 Introduction to Trigonometry NCERT Class 10th solution of Exercise 8.1 NCERT Class 10th solution of Exercise 8.2 NCERT Class 10th solution of Exercise 8.4 NCERT Class 10th Maths Projects Exercise 8.3 Q1. Evaluate: i) `sin 18^circ/cos 72^circ`   Sol. : `= sin 18^circ/cos 72^circ` `= sin 18^circ/cos(90^circ - 18^circ)` `= sin 18^circ/sin 18^circ` `= 1` Answer: `= 1`.    ii) `tan 26^circ/cot 64^circ` Sol. : `= tan 26^circ/cot 64^circ` `= tan 26^circ/cot(90^circ - 26^circ)` `= tan 26^circ/tan 26^circ` `= 1` Answer: `= 1`      iii) `cos 48^circ - sin 42^circ` Sol. : `= cos 48^circ - sin 42^circ` `= cos(90^circ - 42^circ) - sin 42^circ` `= sin 42^circ - sin 42^circ` `= 0` Answer: `= 0` iv) `cosec 31^circ - sec 59^circ` Sol. : `= cosec 31^circ - sec 59^circ` `= cosec(90^circ - 59^circ) - sec 59^circ` `= sec 59^circ - sec 59^circ` `= 0` Answer: `= 0`  ☝  Like, Share, and Subscribe  Q2. Show that: i) `tan 48^circ tan23^circ tan42^circ tan 6...