{"id":43911,"date":"2021-01-06T15:07:45","date_gmt":"2021-01-06T09:37:45","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=43911"},"modified":"2021-01-06T15:11:20","modified_gmt":"2021-01-06T09:41:20","slug":"top-20-ssc-cgl-trigonometry-questions","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/top-20-ssc-cgl-trigonometry-questions\/","title":{"rendered":"Top-20 SSC CGL Trigonometry Questions"},"content":{"rendered":"

SSC CGL Trigonometry Questions:<\/strong><\/h1>\n

Download Top-20 Trigonometry questions for SSC CGL exam 2020. Most important\u00a0 Trigonometry questions based on asked questions in previous exam papers for SSC CGL.<\/p>\n

Download Top-20 SSC CGL Trigonometry Questions <\/a><\/p>\n

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Question 1:\u00a0<\/b>$5tan\\theta = 4$, then the value of\u00a0$(\\frac{5sin\\theta – 3cos\\theta}{5sin\\theta + 3cos\\theta})$ is<\/p>\n

a)\u00a0$\\frac{1}{7}$<\/p>\n

b)\u00a0$\\frac{2}{7}$<\/p>\n

c)\u00a0$\\frac{5}{7}$<\/p>\n

d)\u00a0$\\frac{2}{5}$<\/p>\n

Question 2:\u00a0<\/b>The least value of $(4sec^2\\theta + 9cosec^2\\theta)$ is<\/p>\n

a)\u00a01<\/p>\n

b)\u00a019<\/p>\n

c)\u00a025<\/p>\n

d)\u00a07<\/p>\n

Question 3:\u00a0<\/b>If $x=cosec\\theta-sin\\theta$ and $y=sec\\theta-cos\\theta$, then the value of $x2<\/sup>y2<\/sup>(x2<\/sup> + y2<\/sup> + 3)$<\/p>\n

a)\u00a00<\/p>\n

b)\u00a01<\/p>\n

c)\u00a02<\/p>\n

d)\u00a03<\/p>\n

Question 4:\u00a0<\/b>If $ 0 \\leq \\theta \\leq \\frac{\\pi}{2}$, $2ycos\\theta=sin\\theta$ and $\\frac{x}{2cosec\\theta}=y$, then the value of $x^2-4y^2$\u00a0is<\/p>\n

a)\u00a01<\/p>\n

b)\u00a02<\/p>\n

c)\u00a03<\/p>\n

d)\u00a04<\/p>\n

Question 5:\u00a0<\/b>The value of $\\frac{1}{cosec\\theta – cot\\theta} – \\frac{1}{sin\\theta}$<\/p>\n

a)\u00a0$cot\\theta$<\/p>\n

b)\u00a0$cosec\\theta$<\/p>\n

c)\u00a0$tan\\theta$<\/p>\n

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

Question 6:\u00a0<\/b>If\u00a0$\\cos\\theta + \\sin\\theta = \\sqrt{2}\\cos\\theta$, then $\\cos\\theta – \\sin\\theta$ is<\/p>\n

a)\u00a0-$\\sqrt{2}\\cos\\theta$<\/p>\n

b)\u00a0-$\\sqrt{2}\\sin\\theta$<\/p>\n

c)\u00a0$\\sqrt{2}\\sin\\theta$<\/p>\n

d)\u00a0$\\sqrt{2}\\tan\\theta$<\/p>\n

Question 7:\u00a0<\/b>If $cos^4\\theta-sin^4\\theta=\\frac{2}{3}$, then the value of $1-2sin^2\\theta$ is,<\/p>\n

a)\u00a00<\/p>\n

b)\u00a0$\\frac{2}{3}$<\/p>\n

c)\u00a0$\\frac{1}{3}$<\/p>\n

d)\u00a0$\\frac{4}{3}$<\/p>\n

Question 8:\u00a0<\/b>The value of $\\frac{1}{1 + tan^2\\theta}$ + $\\frac{1}{1 + cot^2\\theta}$ is<\/p>\n

a)\u00a01<\/p>\n

b)\u00a02<\/p>\n

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

d)\u00a0$\\frac{1}{4}$<\/p>\n

Question 9:\u00a0<\/b>Maximum value of $(2sin\\theta+3 cos\\theta)$ is<\/p>\n

a)\u00a02<\/p>\n

b)\u00a0$\\sqrt{13}$<\/p>\n

c)\u00a0$\\sqrt{15}$<\/p>\n

d)\u00a01<\/p>\n

Question 10:\u00a0<\/b>The value of $cos^2 30^{\\circ} + sin^2 60^{\\circ} + tan^2 45^{\\circ} + sec^2 60^{\\circ} + cos0^{\\circ}$ is<\/p>\n

a)\u00a0$4\\frac{1}{2}$<\/p>\n

b)\u00a0$5\\frac{1}{2}$<\/p>\n

c)\u00a0$6\\frac{1}{2}$<\/p>\n

d)\u00a0$7\\frac{1}{2}$<\/p>\n

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Question 11:\u00a0<\/b>If $cos x + cos^{2} x = 1,$ then $sin^{8} x + 2 sin^{6} x + sin^{4}$ x is equal to<\/p>\n

a)\u00a00<\/p>\n

b)\u00a03<\/p>\n

c)\u00a02<\/p>\n

d)\u00a01<\/p>\n

Question 12:\u00a0<\/b>From an aeroplane just over a straight road, the angles of depression of two consecutive kilometre stones situated at opposite sides of the aeroplane were found to be 60\u00b0 and 30\u00b0 respectively. The height (in km) of the aeroplane from the road at that instant was (Given \u221a3 = 1.732)<\/p>\n

a)\u00a00.433<\/p>\n

b)\u00a08.66<\/p>\n

c)\u00a04.33<\/p>\n

d)\u00a00.866<\/p>\n

Question 13:\u00a0<\/b>In \u0394ABC, \u2220C = 90\u00b0 and AB = c, BC = a, CA = b; then the value of (cosec B – cos A) is<\/p>\n

a)\u00a0$\\frac{c^2}{ab}$<\/p>\n

b)\u00a0$\\frac{b^2}{ca}$<\/p>\n

c)\u00a0$\\frac{a^2}{bc}$<\/p>\n

d)\u00a0$\\frac{bc}{a^{2}}$<\/p>\n

Question 14:\u00a0<\/b>The maximum value of sin \u03b8 + cos \u03b8 is<\/p>\n

a)\u00a0$1$<\/p>\n

b)\u00a0$\\sqrt{2}$<\/p>\n

c)\u00a0$2$<\/p>\n

d)\u00a0$3$<\/p>\n

Question 15:\u00a0<\/b>Find the value of tan 4\u00b0 tan 43\u00b0 tan 47 tan 86\u00b0<\/p>\n

a)\u00a0$\\frac{2}{3}$<\/p>\n

b)\u00a0$1$<\/p>\n

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

d)\u00a0$2$<\/p>\n

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Question 16:\u00a0<\/b>The value of tan1\u00b0tan2\u00b0tan3\u00b0 \u2026\u2026\u2026\u2026\u2026tan89\u00b0 is<\/p>\n

a)\u00a01<\/p>\n

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

c)\u00a00<\/p>\n

d)\u00a0None of the options<\/p>\n

Question 17:\u00a0<\/b>If \u03b8 is an acute angle and $\\tan^2\\theta+\\frac{1}{\\tan^2\\theta}=2$ then the value of \u03b8 is :<\/p>\n

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

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

c)\u00a015\u00b0<\/p>\n

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

Question 18:\u00a0<\/b>If tan \u03b8 + cot \u03b8 = 5, then $tan^2 \u03b8 + cot^2 \u03b8$ is<\/p>\n

a)\u00a023<\/p>\n

b)\u00a025<\/p>\n

c)\u00a026<\/p>\n

d)\u00a024<\/p>\n

Question 19:\u00a0<\/b>A person of height 6ft. wants to pluck a fruit which is on a 26\/3 ft. high tree. If the person is standing 8\/\u221a3 ft. away from the base of the tree, then at what angle should he throw a stone so that it hits the fruit ?<\/p>\n

a)\u00a075\u00b0<\/p>\n

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

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

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

Question 20:\u00a0<\/b>The angle of elevation of a tower from a distance of 100 metre from its foot is 30\u00b0. Then the height of the tower is<\/p>\n

a)\u00a0$50\\sqrt{3}$ metre<\/p>\n

b)\u00a0$100\\sqrt{3}$ metre<\/p>\n

c)\u00a0$\\frac{50}{\\sqrt{3}}$ metre<\/p>\n

d)\u00a0$\\frac{100}{\\sqrt{3}}$ metre<\/p>\n

More SSC CGL Important Questions and Answers PDF<\/a><\/p>\n

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

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

Taking $cos\\theta$ outside in numerator and in denominator and making $tan\\theta$
\nhence eq will be\u00a0\u00a0$(\\frac{5tan\\theta – 3}{5tan\\theta + 3})$
\nAs it is given that $5tan\\theta$ = 4
\nafter putting values and solving we will get the equation reduced to 1\/7.<\/p>\n

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

$4sec^2\\theta+9cosec^2\\theta$
\nor $4+4tan^2\\theta+9+9cot^2\\theta$
\nor $13+4tan^2\\theta+9cot^2\\theta$
\nor $ 13+4tan^2\\theta+\\frac{9}{tan^2\\theta} $
\nor $ \u00a013-12+(2tan\\theta+\\frac{3}{tan\\theta})^2 $ \u00a0 \u00a0(eq. (1) )
\nor now above expression to be minimum, equation $(2tan\\theta+\\frac{3}{tan\\theta})^2$ should be minimum.
\nSo applying $A.M.\\geq G.M. $
\n$\\frac{(2tan\\theta +\\frac{3}{tan\\theta})}{2} \\geq \\sqrt{6}$
\nor\u00a0${(2tan\\theta+\\frac{3}{tan\\theta})}=2\\sqrt{6}$ ( for value to be minimum)
\nAfter putting above value in eq.(1) , we will get least value of expression as 25.<\/p>\n

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

$x=cosec\\theta – sin\\theta=\\frac{cos^2\\theta}{sin\\theta}=cot\\theta cos\\theta$
\nSimilarly $y=tan\\theta sin\\theta$
\n$xy=sin\\theta cos\\theta$
\n$x^2+y^2+3=(sec^2\\theta +cosec^2\\theta )$
\nNow putting above values in given equation, and after solving it will be reduced to 1<\/p>\n

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

$2y=tan\\theta$
\n$x=2ycosec\\theta$
\nHence value of $x^2 – 4y^2 $ = $4y^2(cosec^2\\theta – 1)$
\nor $tan^2\\theta cot^2\\theta$ = 1<\/p>\n

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

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

or $\\frac{cos\\theta – cos^2\\theta}{(1-cos\\theta)sin\\theta}$ = $cot\\theta$<\/p>\n

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

$\\sin^2 \\theta + \\cos^2 \\theta = 1$
\nSo, $\\sin^2 \\theta + \\cos^2 \\theta + 2\\sin\\theta * \\cos \\theta = 2 \\cos^2\\theta$
\nHence, $\\cos^2 \\theta – \\sin^2 \\theta = 2 \\sin\\theta*\\cos\\theta$
\nSo, $\\cos\\theta – \\sin\\theta = \\sqrt{2}\\sin\\theta$<\/p>\n

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

$cos^4\\theta-sin^4\\theta=(cos^2\\theta-sin^2\\theta)(cos^2\\theta+sin^2\\theta)=cos^2\\theta-sin^2\\theta=\\frac{2}{3}$
\n$cos^2\\theta-sin^2\\theta =1-2sin^2\\theta=\\frac{2}{3}$<\/p>\n

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

$1 + \\tan ^2 \\theta = \\sec ^2 \\theta$
\n$1 + \\cot ^2 \\theta = \\csc ^2 \\theta$
\nSo, the given fraction becomes,<\/p>\n

$\\frac{1}{\\sec ^2 \\theta} + \\frac{1}{\\csc ^2 \\theta} = \\sin^2\\theta + \\cos^2 \\theta = 1$<\/p>\n

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

$\\because$ Maximum Value of $a\\sin{\\theta}+b\\cos{\\theta}=\\sqrt{a^{2}+b^{2}}$
\n$\\therefore$ Maximum Value of $2\\sin{\\theta}+3\\cos{\\theta}=\\sqrt{2^{2}+3^{2}}$
\n$=\\sqrt{13}$
\nHence, Correct option is B.<\/p>\n

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

Substituting values of angles, we get,
\n3\/4 + 3\/4+ 1 + 4 + 1 = 7.5. Option D is the right answer.<\/p>\n

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

$cos x + cos^2 x = 1$<\/p>\n

=> $cos x = 1 – cos^2 x$<\/p>\n

=> $cos x = sin^2 x$<\/p>\n

$\\therefore$ $sin^{8} x + 2 sin^{6} x + sin^{4} x$<\/p>\n

= $(sin^4 x + sin^2 x)^2$<\/span><\/p>\n

= $((cos x)^2 + sin^2 x)^2$<\/span><\/p>\n

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

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

<\/p>\n

OC = height of plane = $h$<\/p>\n

$\\angle$OAC = $\\angle$DOA = 60\u00b0<\/p>\n

$\\angle$OBC = $\\angle$BOE = 30\u00b0<\/p>\n

AB = 2 and let AC = $x$<\/p>\n

=> BC = $(2-x)$<\/p>\n

From, $\\triangle$OAC<\/p>\n

$tan60^{\\circ} = \\frac{OC}{AC}$<\/p>\n

=> $\\sqrt{3} = \\frac{h}{x}$<\/p>\n

=> $x = \\frac{h}{\\sqrt{3}}$ ————Eqn(1)<\/p>\n

From, $\\triangle$OBC<\/p>\n

$tan30^{\\circ} = \\frac{OC}{BC}$<\/p>\n

=> $\\frac{1}{\\sqrt{3}} = \\frac{h}{2-x}$<\/p>\n

=> $\\sqrt{3}h = 2 – \\frac{h}{\\sqrt{3}}$ [From eqn(1)]<\/p>\n

=> $\\frac{3h+h}{\\sqrt{3}} = 2$<\/p>\n

=> $h = \\frac{2\\sqrt{3}}{4} = \\frac{\\sqrt{3}}{2}$<\/p>\n

= $\\frac{1.732}{2}$ = 0.866<\/strong><\/p>\n

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

<\/p>\n

In $\\triangle$$ABC, AB^2 = AC^2 + BC^2$<\/p>\n

=> $c^2 = a^2 + b^2 => c^2 – b^2 = a^2$<\/p>\n

$cosecB = \\frac{AB}{BC} = \\frac{c}{b}$<\/p>\n

$cosA = \\frac{AC}{AB} = \\frac{b}{c}$<\/p>\n

$\\therefore cosecB – cosA = \\frac{c}{b} – \\frac{b}{c}$<\/p>\n

= $\\frac{c^2-b^2}{bc} = \\frac{a^2}{bc}$<\/p>\n

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

for\u00a0\u00a0asin \u03b8 + bcos \u03b8 + c,<\/p>\n

maximum value = $c+\\sqrt{a^{2}+b^{2}}$<\/p>\n

minimum value =\u00a0\u00a0$c-\\sqrt{a^{2}+b^{2}}$<\/p>\n

for\u00a0\u00a0sin \u03b8 + cos \u03b8 , a = 1, b = 1, c = 0<\/p>\n

maximum value = $c+\\sqrt{a^{2}+b^{2}}=0+\\sqrt{1^{2}+1^{2}}=\\sqrt{2}$<\/p>\n

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

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

Expression : tan 4\u00b0 tan 43\u00b0 tan 47 tan 86\u00b0<\/p>\n

$\\because$ $tan(90-\\theta) = cot\\theta$<\/p>\n

=> $tan 4^{\\circ} = tan(90^{\\circ}-86^{\\circ}) = cot 86^{\\circ}$<\/p>\n

Similarly, $tan 43^{\\circ} = cot 47^{\\circ}$<\/p>\n

=> $(cot 86^{\\circ} \\times tan 86^{\\circ}) * (tan 47^{\\circ} \\times cot 47^{\\circ})$<\/p>\n

Using, $tan\\theta cot\\theta$ = 1<\/p>\n

=> 1*1 = 1<\/p>\n

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

Expression : tan1\u00b0tan2\u00b0tan3\u00b0 \u2026\u2026\u2026\u2026\u2026tan88\u00b0tan89\u00b0<\/p>\n

$\\because$ $tan(90^{\\circ}-\\theta) = cot\\theta$<\/p>\n

=> tan 89\u00b0 = tan(90\u00b0-1) = cot 1\u00b0<\/p>\n

Similarly, tan 88\u00b0 = cot 2\u00b0 and so on till tan 46\u00b0 = cot 44\u00b0<\/p>\n

=> (tan1\u00b0tan2\u00b0tan3\u00b0…….tan45\u00b0……cot3\u00b0cot2\u00b0cot1\u00b0)<\/p>\n

Using, $tan\\theta cot\\theta$ = 1 and $tan45^{\\circ}$ = 1<\/p>\n

=> 1*1 = 1<\/p>\n

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

Expression : $\\tan^2\\theta+\\frac{1}{\\tan^2\\theta}=2$<\/p>\n

=> $(tan\\theta + \\frac{1}{tan\\theta})^2 – 2 = 2$<\/p>\n

=> $(tan\\theta + \\frac{1}{tan\\theta})^2 = 4$<\/p>\n

=> $tan\\theta + \\frac{1}{tan\\theta} = 2$<\/p>\n

[It can’t be -2 as $\\theta$ is in 1st quadrant, and $tan\\theta$ is positive in 1st quadrant.]<\/p>\n

=> $tan^2\\theta + 1 = 2tan\\theta$<\/p>\n

=> $(tan\\theta – 1)^2 = 0$<\/p>\n

=> $tan\\theta = 1$<\/p>\n

=> $\\theta = 45^{\\circ}$<\/p>\n

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

Expression : $tan\\theta + cot\\theta = 5$<\/p>\n

Squaring both sides, we get :<\/p>\n

=> $tan^2\\theta + cot^2\\theta + 2tan\\theta cot\\theta = 25$<\/p>\n

We know that, $tan\\theta cot\\theta = 1$<\/p>\n

=> $tan^2\\theta + cot^2\\theta = 25-2 = 23$<\/p>\n

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

<\/p>\n

Height of person = CD = 6 ft<\/p>\n

Height of tree = AB = $\\frac{26}{3}$ ft<\/p>\n

Distance between them = BD = $\\frac{8}{\\sqrt{3}}$ ft<\/p>\n

To find : $\\angle$ACE = $\\theta$ = ?<\/p>\n

Solution : AE = AB – BE = $\\frac{26}{3}$ – 6<\/p>\n

=> AE = $\\frac{8}{3} ft$<\/p>\n

and BD = CE = $\\frac{8}{\\sqrt{3}}$ ft<\/p>\n

Now, in $\\triangle$AEC<\/p>\n

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

=> $tan\\theta$ = $\\frac{\\frac{8}{3}}{\\frac{8}{\\sqrt{3}}}$<\/span><\/p>\n

=> $tan\\theta$ = $\\frac{1}{\\sqrt{3}}$<\/span><\/span><\/p>\n

=> $\\theta$ = 30\u00b0<\/span><\/span><\/span><\/p>\n

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

<\/p>\n

Height of tower = AB<\/p>\n

In $\\triangle$ABC<\/p>\n

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

=> $tan30^{\\circ} = \\frac{AB}{100}$<\/p>\n

=> $\\frac{1}{\\sqrt{3}} = \\frac{AB}{100}$<\/p>\n

=> $AB = \\frac{100}{\\sqrt{3}}$<\/p>\n

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 <\/p>\n","protected":false},"excerpt":{"rendered":"

SSC CGL Trigonometry Questions: Download Top-20 Trigonometry questions for SSC CGL exam 2020. Most important\u00a0 Trigonometry questions based on asked questions in previous exam papers for SSC CGL. Take a free SSC CGL Tier-1 mock test Download SSC CGL Tier-1 Previous Papers PDF Question 1:\u00a0$5tan\\theta = 4$, then the value of\u00a0$(\\frac{5sin\\theta – 3cos\\theta}{5sin\\theta + 3cos\\theta})$ […]<\/p>\n","protected":false},"author":53,"featured_media":43913,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[9,504],"tags":[1625,462,4349],"class_list":{"0":"post-43911","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-ssc","8":"category-ssc-cgl","9":"tag-maths","10":"tag-ssc-cgl","11":"tag-trigonometry"},"better_featured_image":{"id":43913,"alt_text":"top 20 ssc cgl trigonometry questions","caption":"top 20 ssc cgl trigonometry questions","description":"top 20 ssc cgl trigonometry 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