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50 questions · timed · auto-graded

Question 21 Mark
Prove that:
sec(70° -θ) = cosec(20° +θ)
Answer
sec(70° -θ) = sec[90° - (20° +θ)] = cosec(20° +θ)
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Question 31 Mark
Prove that: tan (55° + x) = cot (35° - x)
Answer
tan (55° + x) = tan [90° - (35° - x)] = cot(35° - x)
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Question 41 Mark
Evaluate
$2\left(\frac{\tan 35^{\circ}}{\cot 55^{\circ}}\right)+\left(\frac{\cot 55^{\circ}}{\tan 35^{\circ}}\right)-3\left(\frac{\sec 40^{\circ}}{\cos e c 50^{\circ}}\right)$
Answer
$2\left(\frac{\tan 35^{\circ}}{\cot 55^{\circ}}\right)+\left(\frac{\cot 55^{\circ}}{\tan 35^{\circ}}\right)-3\left(\frac{\sec 40^{\circ}}{\cos e c 50^{\circ}}\right) $
$=2\left(\frac{\tan \left(90^{\circ}-55^{\circ}\right)}{\cot 55^{\circ}}\right)+\left(\frac{\cot \left(90^{\circ}-35^{\circ}\right)}{\tan 35^{\circ}}\right)-3\left(\frac{\sec \left(90^{\circ}-50^{\circ}\right)}{\cos e c 50^{\circ}}\right)$
$ =2\left(\frac{\cot 55^{\circ}}{\cot 55^{\circ}}\right)+\left(\frac{\tan 35^{\circ}}{\tan 35^{\circ}}\right)-3\left(\frac{\cos e c 50^{\circ}}{\cos e c 50^{\circ}}\right) $
$ =2(1)^2+1^2+-3 $
$=2+1-3$
$=0$
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Question 51 Mark
Find A, if 0° ≤ A ≤ 90° and sin 3A - 1 = 0
Answer
sin 3A - 1 = 0
⇒ sin3A = 1
We know sin 90° = 1
∴ 3A = 90°
Hence, A = 30°
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Question 61 Mark
Use tables to find the acute angle θ, if the value of tan θ is0.2419
Answer
From the tables, it is clear that tan 13° 36’ = 0.2419
Hence, θ = 13° 36’
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Question 71 Mark
Use tables to find the acute angle θ, if the value of cos θ is0.9848
Answer
From the tables, it is clear that cos 10° = 0.9848
Hence, θ = 10°
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Question 81 Mark
Use tables to find the acute angle θ, if the value of sin θ is 0.3827
Answer
From the tables, it is clear that sin 22° 30' = 0.3827
Hence, θ = 22° 30
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Question 91 Mark
Use tables to find the acute angle θ, if the value of sin θ is 0.4848
Answer
From the tables, it is clear that sin 29° = 0.4848
Hence, θ = 29°
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Question 101 Mark
Use trigonometrical tables to find tangent of17° 27'
Answer
tan 17° 27' = tan (17 24' + 3') = 0.3134 + 0.0010 = 0.3144
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Question 131 Mark
Use tables to find cosine of9° 23’ + 15° 54’
Answer
cos (9° 23’ + 15° 54’) = cos 24° 77’ = cos 25° 17’ = cos (25° 12’ + 5’) = 0.9048 − 0.0006 = 0.9042
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Question 141 Mark
Use tables to find cosine of 65° 41’
Answer
cos 65° 41’ = cos (65° 36’ + 5’) = 0.4131 − 0.0013 = 0.4118
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Question 151 Mark
Use tables to find cosine of 26° 32’
Answer
cos 26° 32’ = cos (26° 30’ + 2’) = 0.8949 − 0.0003 = 0.8946
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Question 181 Mark
Use tables to find sine of 10° 20' + 20° 45'
Answer
sin (10° 20' + 20° 45') = sin 30° 65' = sin 31° 5' = 0.5150 + 0.0012 = 0.5162
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Question 191 Mark
Use tables to find sine of 62° 57'
Answer
sin 62° 57' = sin ( 62° 54' + 3') = 0.8902 + 0.0004 = 0.8906
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Question 231 Mark
Evaluate:
$14 \sin 30^\circ + 6 \cos 60^\circ - 5 \tan 45^\circ$
Answer
$14 \sin 30^{\circ}+6 \cos 60^{\circ}-5 \tan 45^{\circ}$
$=14\left(\frac{1}{2}\right)+6\left(\frac{1}{2}\right)-5(1)$
$=7+3-5=5$
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Question 241 Mark
Prove that:
tan (55° - A) - cot (35° + A)
Answer
tan (55° - A) - cot (35° + A)
= tan [90° - (55° - A)] - cot (35° + A)
= cot (90° - 55° + A) - cot (35° + A)
= cot (35° + A) - cot (35° + A)
= 0
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Question 251 Mark
Express the following in terms of angles between 0° and 45°:
cos74° + sec67°
Answer
cos74° + sec67°
= cos(90 - 16)° + sec(90 - 23)°
= sin16° + cosec23°
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Question 261 Mark
Express the following in terms of angles between 0° and 45°:
cosec68° + cot72°
Answer
cosec68° + cot72°
= cosec(90 - 22)° + cot(90 -18)°
= sec22° + tan18°
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Question 271 Mark
Express the following in terms of angle between 0° and 45°:
sin 59° + tan 63°
Answer
sin 59° + tan 63°
= sin(90- 31)° + tan(90 - 27)°
= cos 31° + cot 27°
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Question 281 Mark
Prove.$\frac{\sec A-1}{\sec A+1}=\frac{1-\cos A}{1+\cos A}$
Answer
$\text { LHS }=\frac{\sec A-1}{\sec A+1}=\frac{\frac{1}{\cos A}-1}{\frac{1}{\cos A}+1} $
$ =\frac{1-\cos A}{1+\cos A}=\text { RHS }$
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Question 301 Mark
Prove that:
sec(70° -θ) = cosec(20° +θ)
Answer
sec(70° -θ) = sec[90° - (20° +θ)] = cosec(20° +θ)
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Question 311 Mark
Prove that: tan (55° + x) = cot (35° - x)
Answer
tan (55° + x) = tan [90° - (35° - x)] = cot(35° - x)
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Question 321 Mark
Evaluate
3 cos80° cosec10°+ 2 cos59° cosec31°
Answer
3 cos80° cosec10°+ 2 cos59° cosec31°
= 3 cos(90° - 10°) cosec10° + 2 cos(90° - 31°) cosec31°
= 3 sin10° cosec10° + 2 sin31° cosec31°
= 3 + 2 = 5
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Question 331 Mark
Evaluate
sin27° sin63° - cos63° cos27°
Answer
sin27° sin63° - cos63° cos27°
= sin(90° - 63°) sin63° - cos63° cos(90° - 63°)
= cos63° sin63° - cos63° sin63°
= 0
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Question 341 Mark
Evaluate
cos40° cosec50° + sin50° sec40°
Answer
cos40° cosec50° + sin50° sec40°
= cos(90°-50°) cosec50° + sin(90° - 40°) sec40°
= sin50° cosec50° + cos40° sec40°
= 1+ 1 = 2
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Question 351 Mark
Use tables to find the acute angle θ, if the value of tan θ is0.7391
Answer
From the tables, it is clear that tan 36° 24’ = 0.7373
tan θ − tan 36° 24’ = 0.7391 − 0.7373 = 0.0018
From the tables, diff of 4’ = 0.0018
Hence, θ = 36° 24’ + 4’ = 36° 28’
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Question 361 Mark
Use tables to find the acute angle θ, if the value of tan θ is0.4741
Answer
From the tables, it is clear that tan 25° 18’ = 0.4727
tan θ − tan 25° 18’ = 0.4741 − 0.4727 = 0.0014
From the tables, diff of 4’ = 0.0014
Hence, θ = 25° 18’ + 4’ = 25° 22’
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Question 371 Mark
Use tables to find the acute angle θ, if the value of tan θ is0.2419
Answer
From the tables, it is clear that tan 13° 36’ = 0.2419
Hence, θ = 13° 36’
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Question 381 Mark
Use tables to find the acute angle θ, if the value of cos θ is0.6885
Answer
From the tables, it is clear that cos 46° 30’ = 0.6884
cos q − cos 46° 30’ = 0.6885 − 0.6884 = 0.0001
From the tables, diff of 1’ = 0.0002
Hence, θ = 46° 30’ − 1’ = 46° 29’
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Question 391 Mark
Use tables to find the acute angle θ, if the value of cos θ is0.9574
Answer
From the tables, it is clear that cos 16° 48’ = 0.9573
cos θ − cos 16° 48’ = 0.9574 − 0.9573 = 0.0001
From the tables, diff of 1’ = 0.0001
Hence, θ = 16° 48’ − 1’ = 16° 47’
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Question 401 Mark
Use tables to find the acute angle θ, if the value of cos θ is0.9848
Answer
From the tables, it is clear that cos 10° = 0.9848
Hence, θ = 10°
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Question 411 Mark
Use tables to find the acute angle θ, if the value of sin θ is0.6525
Answer
From the tables, it is clear that sin 40° 42' = 0.6521
sin θ − sin 40° 42' = 0.6525 −; 0.6521 = 0.0004
From the tables, diff of 2' = 0.0004
Hence, θ = 40° 42' + 2' = 40° 44
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Question 421 Mark
Use tables to find the acute angle θ, if the value of sin θ is0.3827
Answer
From the tables, it is clear that sin 22° 30' = 0.3827
Hence, θ = 22° 30
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Question 431 Mark
Use tables to find the acute angle θ, if the value of sin θ is0.4848
Answer
From the tables, it is clear that sin 29° = 0.4848
Hence, θ = 29°
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Question 441 Mark
Use trigonometrical tables to find tangent of17° 27'
Answer
tan 17° 27' = tan (17 24' + 3') = 0.3134 + 0.0010 = 0.3144
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Question 471 Mark
Use tables to find cosine of9° 23’ + 15° 54’
Answer
cos (9° 23’ + 15° 54’) = cos 24° 77’ = cos 25° 17’ = cos (25° 12’ + 5’) = 0.9048 − 0.0006 = 0.9042
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Question 481 Mark
Use tables to find cosine of65° 41’
Answer
cos 65° 41’ = cos (65° 36’ + 5’) = 0.4131 − 0.0013 = 0.4118
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Question 491 Mark
Use tables to find cosine of26° 32’
Answer
cos 26° 32’ = cos (26° 30’ + 2’) = 0.8949 − 0.0003 = 0.8946
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[1 Mark Question Answer] - Mathematics STD 10 Questions - Vidyadip