Dictionary Rank of a Word | Permutations & Combinations

PERMUTATIONS & COMBINATIONS Rank of the word or Dictionary order of the English words like COMPUTER, COLLEGE, SUCCESS, SOCCER, RAIN, FATHER, etc. Dictionary Rank of a Word Method of finding the Rank (Dictionary Order) of the word  “R A I N” Given word: R A I N Total letters = 4 Letters in alphabetical order: A, I, N, R No. of words formed starting with A = 3! = 6 No. of words formed starting with I = 3! = 6 No. of words formed starting with N = 3! = 6 After N there is R which is required R ----- Required A ---- Required I ---- Required N ---- Required RAIN ----- 1 word   RANK OF THE WORD “R A I N” A….. = 3! = 6 I……. = 3! = 6 N….. = 3! = 6 R…A…I…N = 1 word 6 6 6 1 TOTAL 19 Rank of “R A I N” is 19 Method of finding the Rank (Dictionary Order) of the word  “F A T H E R” Given word is :  "F A T H E R" In alphabetical order: A, E, F, H, R, T Words beginni

Math Assignment Class XI Ch-3 | Trigonometric Functions

Math Assignment / Class XI / Chapter 3 /  Trigonometric Functions

Extra questions of chapter 3 class 11 Trigonometric Functions with answer and  hints to the difficult questions. Important and useful math. assignment for the students of class 11

For better results

• Students should learn all the basic points of Trigonometry up to 11th standard
• Student should revise NCERT book thoroughly with examples.
• Now revise this assignment. This assignment integrate the knowledge of the students.

ASSIGNMENT FOR XI  STANDARD TRIGONOMETRY

 Question 1Find the degree measure for the following radian measure$i)\:\; \; \frac{2\pi }{15}Radian\; \; \; \; \; ............. \; \; [Ans\; \; 24^{o}]$$ii)\:\; \; \frac{11 }{16}Radian\; \; \; \; \; ............. \; \; [Ans\; \; 39^{o}\: 22'\: 30'']$$iii)\:\; \; (-2)Radian\; \; \; \; \; ............. \; \; [Ans\; \; -(114^{o}\: 32'\: 44'')]$ Question 2Find the radian measure for the following degree measure$i)\:\; \; -37^{o}30'\; \; \; \; \; ............. \; \; [Ans\; \; -\left ( \frac{5\pi }{24} \right )]$$ii)\:\; \; 40^{o}\: 20'\; \; \; \; \; ............. \; \; [Ans\; \; \left ( \frac{121\pi }{540} \right )]$$iii)\:\; \; 5^{o}\: 37'\:30'' \; \; \; \; \; ............. \; \; [Ans\; \; \left ( \frac{\pi }{32} \right )]$ Question 3Find the magnitude, radian and degree, of the interior angles of a regular$i)\:\; \; Pentagon \; \; \; \; \; ............. \; \; [Ans\; \; \left ( \frac{3\pi }{5}\; :\; 108^{o} \right )]$$ii)\:\; \; Octagon \; \; \; \; \; ............. \; \; [Ans\; \; \left ( \frac{3\pi }{4}\; :\; 135^{o} \right )]$$iii)\; \; \; Heptagon \; \; \; \; \; ............. \; \; Ans\; \; \left ( \frac{5\pi }{7}\; :\; 128^{o}\; 34'\; 17'' \right )$ Question 4If sin θ =12/13 and θ lie in the second quadrant, then find the value ofsec θ + tan θ.   ………………    Ans. [- 5] Question 5Prove the followings  i)   cos24o + cos 55o + cos 125o + cos 204o + cos 300o = 1/2ii)  sin 600o tan(-690o) + sec 840o cot(-945o) = 3/2iii) $sin^{2}\left \{ \frac{\pi }{8} \right \} +sin^{2}\left \{ \frac{3\pi }{8} \right \}+sin^{2}\left \{ \frac{5\pi }{8} \right \}+sin^{2}\left \{ \frac{7\pi }{8} \right \}=2$ Question 6:   Simplify  $tan\left ( \frac{\pi }{20} \right )tan\left ( \frac{3\pi }{20} \right )tan\left ( \frac{5\pi }{20} \right )tan\left ( \frac{7\pi }{20} \right )tan\left ( \frac{9\pi }{20} \right )$ Solution$tan\left ( \frac{\pi }{20} \right )tan\left ( \frac{3\pi }{20} \right )tan\left ( \frac{\pi }{4} \right )tan\left ( \frac{10\pi-3\pi }{20} \right )tan\left ( \frac{10\pi-\pi }{20} \right )$$=tan\left ( \frac{\pi }{20} \right )tan\left ( \frac{3\pi }{20} \right ).1.tan\left ( \frac{\pi}{2}-\frac{3\pi}{20} \right )tan\left ( \frac{\pi}{2}-\frac{\pi}{20} \right )$$=tan\left ( \frac{\pi }{20} \right )tan\left ( \frac{3\pi }{20} \right )cot\left ( \frac{3\pi}{20} \right )cot\left (\frac{\pi}{20} \right ) = 1$ Question 7Evaluate the following :$i) \; \; tan105^{o}\; \; ......\; \; Ans \left [ \frac{\sqrt{3}+1}{1-\sqrt{3}} \right ]$$ii) \; \; cos105^{o}+sin105^{o}\; \; ......\; \; Ans \left [ \frac{1}{\sqrt{2}} \right ]$ Question 8Prove that : tan70o = tan20o + 2tan50oSolution Hint:$70^{o}=50^{o}+20^{o}\]$tan\: 70^{o}=tan(50^{o}+20^{o})$$tan\: 70^{o}=\frac{tan50^{o}+tan20^{o}}{1-tan50^{o}\; tan20^{o}}$Now cross-multiply and simplify the above fraction we get the required result. Question 9$Prove \; that: \frac{cos9^{o}+sin9^{o}}{cos9^{o}-sin9^{o}}=tan54^{o}$ Question 10 If tanA = ktanB, then show that: $sin(A+B)=\left (\frac{k+1}{k-1} \right )sin(A-B)$Solution Hint: tanA = ktanB$\frac{sinA}{cosA}=k\; \frac{sinB}{cosB}$$\Rightarrow \frac{sinA\: cosB}{cosA\: sinB}=\frac{k}{1}$ Now apply componendo and dividendo and using trigonometric formulas. Question 11If tan(A + B) = p, tan(A – B) = q, then show that : $tan\: 2A = \frac{p+q}{1-pq}$Solution Hint: Start from RHS and then putting the value of p and q then simplify and get the result. Question 12An angle α is divided into two parts such that the ratio of the tangents of the two parts = k and difference of two parts = x then show that: $sin\: x=\frac{k-1}{k+1}\: sin\alpha$Solution hint:Let two parts of α are p and q, ThenATQ : p + q = α, p – q = x, $\frac{tanp}{tanq}=\frac{k}{1}$Now applying componendo and dividendo and simplify we get the required result. Question 13Prove that: $sec\left ( \frac{\pi }{4}+\theta \right )sec\left ( \frac{\pi }{4}-\theta \right )=2\: sec\: 2\theta$ Question 14 Prove that: $\frac{sin11AsinA+sin7Asin3A}{cos11AsinA+cos7Asin3A}=tan8A$ Question 15Prove That : $tan20^{o}\; tan40^{o}\; tan80^{o}=tan60^{o}$Solution Hint : Use tan θ = sin θ/cos θ, then using AB and CD formulas Question 16If cos(θ+2 α) = n cos θ , then prove that : $cot\; \alpha =\frac{1+n}{1-n}\; tan(\theta +\alpha )$Solution Hint: $\frac{cos\: \theta }{cos(\theta +2\alpha) }=\frac{1}{n}$Applying componendo and dividendo then applying CD formulas then simplify the fractions we get the required result. Question 17 Prove that : sin20o sin40o sin60o sin80o = 3/16 Question 18 Prove that : cos20o cos40o cos60o cos80o = 1/16 Question 19Find the value of sin18o $......\; \; Ans\; \left [sin18^{o}=\frac{\sqrt{5}-1}{4} \right ]$Solution:θ = 18o ⇒ 5 θ = 90o ⇒ 2 θ + 3 θ = 90o ⇒ 2 θ = 90o - 3 θNow taking sin on both side we getSin (2 θ) = Sin (90o - 3 θ) Sin (2 θ) = Cos (3 θ)2Sin θ cos θ = 4Cos3 θ – 3cos θ 2Sin θ = 4Cos2 θ – 32Sin θ = 4(1-sin2 θ) – 34 sin2 θ + 2sin θ – 1 = 0Now using quadratic formula here and find the value of sin θ Question 20Prove that : $\frac{sec8A-1}{sec4A-1}=\frac{tan8A}{tan2A}$Solution Hint: $\frac{sec8A-1}{sec4A-1}=\frac{1-cos8A}{cos8A}\times \frac{cos4A}{1-cos4A}$$=\frac{2sin^{2}4A.\: cos4A}{cos8A.\: 2sin^{2}2A}=\frac{(2sin4A\: cos4A)sin4A}{cos8A.\: 2sin^{2}2A}\$$=\left (\frac{sin8A}{cos8A} \right ).\frac{2sin2Acos2A}{2sin^{2}A}=\frac{tan8A}{tan4A}$  Question 21i) Evaluate: $sin\left ( 22\frac{1}{2} \right )^{o}\: \: ...\: \: Ans\left [ \frac{\sqrt{2-\sqrt{2}}}{2} \right ]$Solution Hint: Using: $Sin\theta =\sqrt{\frac{1-cos2\theta }{2}},\; \; where\; \theta =\frac{45}{2}$ii) Evaluate: $tan\left ( 22\frac{1}{2} \right )^{o}\: \: ...\: \: Ans\left [ \sqrt{2}-1 \right ]$ Question 22Solve: $cos3\theta +8cos^{3}\theta =0$$Ans\left [ \theta =(2n+1)\frac{\pi }{2} \; or\; \; n\pi \pm \frac{\pi }{3}\right ]$ Question 23 Solve: $cos\theta + cos2\theta + cos3\theta =0$$Ans\left [ \theta =(2n+1)\frac{\pi }{4}, \; \; or \; \; 2n\pi \pm \frac{2\pi }{3} \right ]$ Question 24Solve: $\sqrt{2}sec\; \theta +tan\; \theta =1$Solution Hint: $\sqrt{2}sec\; \theta +tan\; \theta =1$$\sqrt{2}\; \frac{1}{cos\theta } +\frac{sin\theta }{cos\theta } =1\Rightarrow \sqrt{2}+sin\theta =cos\theta$ $cos\theta -sin\theta =\sqrt{2}$ $\frac{1}{\sqrt{2}}cos\theta-\frac{1}{\sqrt{2}}sin\theta=1$$cos\frac{\pi }{4}\; cos\theta-sin\frac{\pi }{4}\; sin\theta=1$$cos\left ( \theta +\frac{\pi }{4} \right )=1=cos\: 0$$\theta +\frac{\pi }{4}=2n\pi \pm 0\; \Rightarrow \; \theta =2n\pi -\frac{\pi }{4}$ Question 25Solve: $\sqrt{3}\; cosx-sinx=1$$Ans\left [ x=2n\pi +\frac{\pi }{6}\; or\; 2n\pi -\frac{\pi }{2} \right ]$Solution Hint: $\sqrt{3}\; cosx-sinx=1$ Dividing on both side by $\sqrt{(\sqrt{3})^{2}+(-1)^{2}}$ i.e. by 2 $\frac{\sqrt{3}}{2}\; cosx-\frac{1}{2}\; sinx=\frac{1}{2}$$cos\frac{\pi }{6}\; cosx-sin\frac{\pi }{6}\; sinx=\frac{1}{2}$ Now proceed this question as previous one. Question 26Prove that : $cos\theta\; cos\frac{\theta }{2}-cos3\theta\; cos\frac{9\theta }{2} = sin4\theta \; sin\frac{7\theta}{2}$Solution : $cos\theta.\; cos\frac{\theta }{2}-cos3\theta.\; cos\frac{9\theta }{2}$$=\frac{1}{2}\left [2cos\theta.\; cos\frac{\theta }{2}-2cos3\theta.\; cos\frac{9\theta }{2} \right ]$$=\frac{1}{2}\left [ cos\left ( \theta +\frac{\theta }{2} \right )+cos\left ( \theta -\frac{\theta }{2} \right ) \right ]- \frac{1}{2}\left [ cos\left ( 3\theta +\frac{9\theta }{2} \right )+cos\left ( 3\theta -\frac{9\theta }{2} \right ) \right ]\$$\[=\frac{1}{2}\left [ cos\left ( \theta +\frac{\theta }{2} \right )+cos\left ( \theta -\frac{\theta }{2} \right ) \right ]-\\ \frac{1}{2}\left [ cos\left ( 3\theta +\frac{9\theta }{2} \right )+cos\left ( 3\theta -\frac{9\theta }{2} \right ) \right ]\$$=\frac{1}{2}\left [ cos\left (\frac{3\theta }{2} \right )+cos\left (\frac{\theta }{2} \right )-cos\left (\frac{15\theta }{2} \right )-cos\left (\frac{3\theta }{2} \right ) \right ]\$$=\frac{1}{2}\left [ cos\left (\frac{\theta }{2} \right )-cos\left (\frac{15\theta }{2} \right ) \right ]$$=\frac{1}{2}\left [ -2sin\frac{1}{2}\left ( \frac{\theta +15\theta }{2} \right ).\; sin\frac{1}{2}\left ( \frac{\theta -15\theta }{2} \right ) \right ]$$=\frac{1}{2}\left [ -2sin4\theta .\; sin\left ( \frac{-7\theta }{2} \right ) \right ]$$=sin4\theta .\; sin\left ( \frac{7\theta }{2} \right )$ Question 27Evaluate: $If\; cos(\alpha +\beta )=\frac{4}{5}\; and\; sin(\alpha -\beta )=\frac{5}{13}$ and $0<\alpha <\frac{\pi }{4}, \; \; Find \; \; tan2\alpha$Solution:$cos(\alpha +\beta )=\frac{4}{5}\Rightarrow sin(\alpha +\beta )=\frac{2}{5}$$sin(\alpha -\beta )=\frac{5}{13}\Rightarrow cos(\alpha -\beta )=\frac{12}{13}$$\[cos[(\alpha +\beta )+(\alpha -\beta )]=cos(\alpha +\beta)cos(\alpha -\beta)-sin(\alpha +\beta)sin(\alpha -\beta)$$$cos(2\alpha )=\frac{4}{5}\times \frac{12}{13}-\frac{3}{4}\times \frac{5}{13}=\frac{33}{52}$$sin[(\alpha +\beta )+(\alpha -\beta )]=sin(\alpha +\beta)cos(\alpha -\beta)+cos(\alpha +\beta)sin(\alpha -\beta)$$sin(2\alpha )=\frac{3}{5}\times \frac{12}{13}+\frac{4}{5}\times \frac{5}{13}=\frac{56}{65}$$tan(2\alpha )=\frac{sin(2\alpha )}{cos(2\alpha )}=\frac{56/65}{33/65}=\frac{56}{33}$ Note:  Each interior angle of a regular polygon is given by : $\frac{(n-2)}{n}\times 180$ Where n is the number of sides.

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