{"id":34063,"date":"2019-08-27T15:04:07","date_gmt":"2019-08-27T09:34:07","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=34063"},"modified":"2019-08-27T15:04:07","modified_gmt":"2019-08-27T09:34:07","slug":"trigonometry-questions-for-rrb-ntpc-set-2-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/trigonometry-questions-for-rrb-ntpc-set-2-pdf\/","title":{"rendered":"Trigonometry Questions for RRB NTPC Set-2 PDF"},"content":{"rendered":"

Trigonometry Questions for RRB NTPC Set-2 PDF<\/strong><\/span><\/h2>\n

Download RRB NTPC Trigonometry Questions and Answers PDF. Top 15 RRB NTPC Maths questions based on asked questions in previous exam papers very important for the Railway NTPC exam.<\/p>\n

Download Trigonometry Questions for RRB NTPC Set-2 PDF<\/a><\/p>\n\n

Take a free mock test for RRB NTPC<\/a><\/p>\n

Download RRB NTPC Previous Papers PDF<\/a><\/p>\n

Question 1:\u00a0<\/b>Find cosec\u03b8, if Cot\u03b8 = 40\/9 ?<\/p>\n

a)\u00a09\/41<\/p>\n

b)\u00a041\/9<\/p>\n

c)\u00a041\/40<\/p>\n

d)\u00a09\/40<\/p>\n

Question 2:\u00a0<\/b>Find the area of a triangle, whose inradius is 2 cm and ex-radii are 4cm, 8cm, 4 cm ?<\/p>\n

a)\u00a016 sq.cm<\/p>\n

b)\u00a017 sq.cm<\/p>\n

c)\u00a018 sq.cm<\/p>\n

d)\u00a019 sq.cm<\/p>\n

Question 3:\u00a0<\/b>$1+Tan^2A$ = ?<\/p>\n

a)\u00a0$Cosec^2A$<\/p>\n

b)\u00a0$Sin^2A$<\/p>\n

c)\u00a0$Sec^2A$<\/p>\n

d)\u00a0$Cos^2A$<\/p>\n

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FREE RRB NTPC YOUTUBE VIDEOS<\/a><\/p>\n

Question 4:\u00a0<\/b>Find sec\u03b8, if sin\u03b8 = 21\/29 ?<\/p>\n

a)\u00a021\/20<\/p>\n

b)\u00a029\/21<\/p>\n

c)\u00a020\/21<\/p>\n

d)\u00a029\/20<\/p>\n

Question 5:\u00a0<\/b>Find cos\u03b8-sec\u03b8, if cot\u03b8 = 15\/8 ?<\/p>\n

a)\u00a0-64\/255<\/p>\n

b)\u00a064\/255<\/p>\n

c)\u00a0-64\/289<\/p>\n

d)\u00a064\/289<\/p>\n

Question 6:\u00a0<\/b>Find the angle of elevation of the Sun if the length of shadow of a pole of height $70\\sqrt3 cm$ is 70 cm ?<\/p>\n

a)\u00a090\u00b0<\/p>\n

b)\u00a060\u00b0<\/p>\n

c)\u00a045\u00b0<\/p>\n

d)\u00a030\u00b0<\/p>\n

Question 7:\u00a0<\/b>Find $sec\\theta-tan\\theta$, if $sec\\theta+tan\\theta=2$ ?<\/p>\n

a)\u00a01\/2<\/p>\n

b)\u00a0-2<\/p>\n

c)\u00a01<\/p>\n

d)\u00a0-1<\/p>\n

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Question 8:\u00a0<\/b>Find $cosec\\theta$, if $cot\\theta = 12\/5$ ?<\/p>\n

a)\u00a05\/13<\/p>\n

b)\u00a013\/5<\/p>\n

c)\u00a012\/13<\/p>\n

d)\u00a013\/12<\/p>\n

Question 9:\u00a0<\/b>Find $\\frac{1-cos\\frac{\\theta}{2}}{sin\\frac{\\theta}{2}}$ = ?<\/p>\n

a)\u00a0$tan\\frac{\\theta}{4}$<\/p>\n

b)\u00a0$cot\\frac{\\theta}{2}$<\/p>\n

c)\u00a0$tan\\frac{\\theta}{2}$<\/p>\n

d)\u00a0$cot\\frac{\\theta}{4}$<\/p>\n

Question 10:\u00a0<\/b>If $\\tan A = \\frac{5}{12}$, find the value of $\\cosec A$. (Angle A is an acute angle)<\/p>\n

a)\u00a01.2<\/p>\n

b)\u00a02.6<\/p>\n

c)\u00a0$\\frac{24}{5}$<\/p>\n

d)\u00a02.4<\/p>\n

Question 11:\u00a0<\/b>If tan 330\u00b0 = x, then the value of x is<\/p>\n

a)\u00a0$\\frac{-1}{\\sqrt{3}}$<\/p>\n

b)\u00a0${-\\sqrt{3}}$<\/p>\n

c)\u00a0$\\frac{-1}{2}$<\/p>\n

d)\u00a0$\\frac{-1}{\\sqrt{2}}$<\/p>\n

Download General Science Notes PDF<\/a><\/p>\n\n

Question 12:\u00a0<\/b>In which of the following quadrilaterals only one pair of opposite angles is supplementary?<\/p>\n

a)\u00a0Isosceles Trapezium<\/p>\n

b)\u00a0Parallelogram<\/p>\n

c)\u00a0Cyclic quadrilateral<\/p>\n

d)\u00a0Rectangle<\/p>\n

Question 13:\u00a0<\/b>If \u221a$(1 – cos^2A)$ = x, then the value of x is<\/p>\n

a)\u00a0cosecA<\/p>\n

b)\u00a0sinA<\/p>\n

c)\u00a0tanA<\/p>\n

d)\u00a0secA<\/p>\n

Question 14:\u00a0<\/b>If \u221a$(sec^2A-1)\/secA$ = x, then the value of x is<\/p>\n

a)\u00a0cosecA<\/p>\n

b)\u00a0cosA<\/p>\n

c)\u00a0sinA<\/p>\n

d)\u00a0tanA<\/p>\n

Question 15:\u00a0<\/b>What is the value of $(sinA-2sin^3A)\/(2cos^3A-cosA)$?<\/p>\n

a)\u00a0cotA<\/p>\n

b)\u00a0cosecA<\/p>\n

c)\u00a0secA<\/p>\n

d)\u00a0tanA<\/p>\n

RRB NTPC Free Mock Test<\/a><\/p>\n

RRB NTPC Previous Papers (Download PDF)<\/a><\/p>\n

Answers & Solutions:<\/strong><\/span><\/p>\n

1)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Cot\u03b8 = $\\frac{adj}{opp}$ = 40\/9<\/p>\n

Cosec\u03b8 = $\\frac{hyp}{opp}$ = 41\/9\u00a0 ($\\because 40^2+9^2 = 41^2$)<\/p>\n

So the answer is option B.<\/p>\n

2)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

If $ r$ is the inradius and $r_1,r_2,r_3$ are the radii of excircles, then area of triangle = $\\sqrt{rr_1r_2r_3} = \\sqrt{2*4*8*4} = \\sqrt{256} = 16$<\/p>\n

So the answer is option A.<\/p>\n

3)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

$1+Tan^2A$ =\u00a0$Sec^2A$\u00a0 \u00a0($\\because Sec^2A-Tan^2A = 1$)<\/p>\n

or<\/p>\n

$1+Tan^2A$\u00a0= $1+\\frac{Sin^2A}{Cos^2A}$ = $\\frac{Cos^2A+Sin^2A}{Cos^2A}$ = $\\frac{1}{Cos^2A}$ = $Sec^2A$<\/p>\n

So the answer is option C.<\/p>\n

4)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

Sin\u03b8 = $\\frac{opp}{hyp}$ = 21\/29<\/p>\n

Cos\u03b8 = $\\frac{adj}{hyp}$ = 20\/29\u00a0($\\because 20^2+21^2 = 29^2$)<\/p>\n

sec\u03b8 = 29\/20<\/p>\n

So the answer is option D.<\/p>\n

5)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Cot\u03b8 = $\\frac{adj}{opp}$ = 15\/8<\/p>\n

Cos\u03b8 = $\\frac{adj}{hyp}$ = 15\/17\u00a0 ($\\because 15^2+8^2 = 17^2$)<\/p>\n

cos\u03b8-sec\u03b8 = 15\/17 – 17\/15 = (225-289)\/(255) = -64\/255<\/p>\n

So the answer is option A.<\/p>\n

6)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

<\/figure>\n

From\u00a0 the diagram, Tan\u03b8 = $\\frac{70\\sqrt3}{70} = \\sqrt3$ = Tan60\u00b0<\/p>\n

==>\u00a0\u03b8 = 60\u00b0<\/p>\n

So the answer is option B.<\/p>\n

7)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

We Know That, $(sec^2(\\theta)-tan^2(\\theta)) = 1$<\/p>\n

Applying square on both sides,<\/p>\n

$(sec^2(\\theta)-tan^2(\\theta))^2 = 1$<\/p>\n

$[sec\\theta-tan\\theta][sec\\theta+tan\\theta] = 1$<\/p>\n

$[sec\\theta-tan\\theta][2] = 1$<\/p>\n

$[sec\\theta-tan\\theta] = 1\/2$<\/p>\n

So the answer is option A.<\/p>\n

8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

$cot\\theta = 12\/5$<\/p>\n

<\/figure>\n

from the diagram,<\/p>\n

$Cosec\\theta$ = $\\frac{ hypotenuse}{ opposite side} = \\frac{13}{5}$<\/p>\n

So the answer is option B.<\/p>\n

9)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

$\\frac{1-cos\\frac{\\theta}{2}}{sin\\frac{\\theta}{2}}$<\/p>\n

=\u00a0$\\frac{2sin^2\\frac{\\theta}{4}}{2sin\\frac{\\theta}{4}cos\\frac{\\theta}{4}}$<\/p>\n

= $\\frac{sin\\frac{\\theta}{4}}{cos\\frac{\\theta}{4}}$<\/p>\n

= $tan\\frac{\\theta}{4}$<\/p>\n

So the answer is option A.<\/p>\n

10)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Given, $\\tan A = \\frac{5}{12}$.
\nAlso, A is an cute angle.
\nWe can consider a right triangle with the legs 5x and 12x; and
\nHypotenuse = $\\sqrt{(5x)^2 + (12x)^2}$ = 13x.<\/p>\n

$\\cosec A = \\frac{13}{5} = 2.6$<\/p>\n

Hence, option B is the correct answer.<\/p>\n

11)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Expression\u00a0:\u00a0tan 330\u00b0 = x<\/p>\n

= $tan(360-30)$<\/p>\n

= $-tan(30)$<\/p>\n

= $\\frac{-1}{\\sqrt{3}}$<\/p>\n

=> Ans – (A)<\/p>\n

12)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

In rectangle, all angles are 90\u00b0 and in a parallelogram, adjacent angles are supplementary while in a cyclic quadrilateral\u00a0only one pair of opposite angles is supplementary.<\/p>\n

=> Ans – (C)<\/p>\n

13)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Expression\u00a0: $\\sqrt{1-cos^2A}$<\/p>\n

$\\because (sin^2A+cos^2A=1)$<\/p>\n

= $\\sqrt{sin^2A}=sinA$<\/p>\n

=> Ans – (B)<\/p>\n

14)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Expression\u00a0: $\\frac{\\sqrt{sec^2A-1}}{secA}$<\/p>\n

$\\because$ $(sec^2A-tan^2A=1)$<\/p>\n

= $\\frac{\\sqrt{tan^2A}}{secA} = \\frac{tanA}{secA}$<\/p>\n

= $(\\frac{sinA}{cosA})\\div(\\frac{1}{cosA})$<\/p>\n

= $(\\frac{sinA}{cosA})\\times(cosA) = sinA$<\/p>\n

=> Ans – (C)<\/p>\n

15)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

Expression\u00a0:\u00a0$(sinA-2sin^3A)\/(2cos^3A-cosA)$<\/p>\n

= $\\frac{sinA}{cosA} \\times \\frac{1-2sin^2A}{2cos^2A-1}$<\/p>\n

$\\because$ $(sin^2A+cos^2A=1)$<\/p>\n

= $tanA \\times \\frac{(sin^2A+cos^2A)-2sin^2A}{2cos^2A-(sin^2A+cos^2A)}$<\/p>\n

= $tanA \\times \\frac{cos^2A-sin^2A}{cos^2A-sin^2A}$<\/p>\n

= $tanA \\times 1 = tanA$<\/p>\n

=> Ans – (D)<\/p>\n\n

DOWNLOAD APP FOR RRB FREE MOCKS<\/a><\/p>\n

We hope this Trigonometry Questions set-2 pdf\u00a0 for RRB NTPC Exam will be highly useful for your Preparation.<\/p>\n","protected":false},"excerpt":{"rendered":"

Trigonometry Questions for RRB NTPC Set-2 PDF Download RRB NTPC Trigonometry Questions and Answers PDF. Top 15 RRB NTPC Maths questions based on asked questions in previous exam papers very important for the Railway NTPC exam. Take a free mock test for RRB NTPC Download RRB NTPC Previous Papers PDF Question 1:\u00a0Find cosec\u03b8, if Cot\u03b8 […]<\/p>\n","protected":false},"author":41,"featured_media":34065,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[169,125,31,1603,1601],"tags":[1637,2052,1647,1619],"class_list":{"0":"post-34063","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-downloads","8":"category-featured","9":"category-railways","10":"category-rrb-ntpc","11":"category-rrb-ntpc-je","12":"tag-rrb-exam","13":"tag-rrb-maths","14":"tag-rrb-mocks","15":"tag-rrb-ntpc-2019"},"better_featured_image":{"id":34065,"alt_text":"Trigonometry Questions for RRB NTPC Set-2 PDF","caption":"Trigonometry Questions for RRB NTPC Set-2 PDF","description":"Trigonometry Questions for RRB NTPC Set-2 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