Trigonometry, Quantitative Aptitude Subject

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Aptitude TrigonometryPage 1

In a △ABC right angled at B if AB = 12, and BC = 5 find sin A.

(a)5 / 13

(b)5 / 12

(d)12 / 5

Answer!

Answer is: A

AC = √(AB)² + (BC)²
AC = √(12)² + (5)²
= √144 + 25
= √169 = 13
When we consider t-ratios of ∠A we have
Base AB = 12
Perpendicular = BC = 5
Hypotenuse = AC = 13
Sin A = (Perpendicular / Hypotenuse) = 5 / 13

Find the value of 2sin² 30⁰ tan 60⁰ - 3cos² 60⁰ sec² 30⁰

(a)(√3 - 2) / 2

(b)(√3 - 2) / 3

(c)(√3 - 1) / 2

(d)(√3 - 2) / 5

Answer!

Answer is: A2(1 / 2)² x √3 - 3(1 / 2)² x (2 / √3)²
= 2 x (1 / 4) x √3 - 3 x (1 / 4) x (4 / 3)
= √3 / 2 - 1 = (√3 - 2) / 2

Find the value of X.
tan 3X = sin 45^o cos 45^o + sin 30^o

(a)10^o

(b)15^o

(d)30^o

Answer!

Answer is: Btan 3X = sin 45^o cos 45^o + sin 30^o
tan 3X = (1 / √) x (1 / √2) + (1 / 2)
tan 3X = 1 / 2 + 1 / 2 = 1
tan 3X = 1
tan 3X = tan 45^o
3X = 45^o
X = 15^o

If X is an acute angle tan X + cot X = 2 find the value of tan⁷X + cot⁷X.

(a)5

(b)2

(d)4

Answer!

Answer is: Btan X + cot X = 2
tan X + 1 / tan X = 2
tan²X + 1 = 2tan X
tan²X - 2tan X + 1 = 0
(tan X - 1)² = 0
tan X = 1
tan X = tan 45^o
X = 45^o
Now, tan⁷X + cot⁷X
= tan⁷45^o + cot⁷45^o
= 1 + 1 = 2

Find the value of
sin 36^o / cos 54^o - sin 54^o / cos 36^o

(a)0

(b)1

(d)3

Answer!

Answer is: Asin 36^o / cos 54^o - sin 54^o / cos 36^o
sin (90 - 54)^o / cos 54^o - sin (90 - 36)^o / cos 36^o
cos 54^o / cos 54^o - cos 36^o / cos 36^o
1 - 1 = 0

Evaluate the cot 12^o cot 38^o cot 52^o cot 60^o cot 78^o.

(a)√3

(b)3√3

(d)3 / √3

Answer!

Answer is: CWe have
cot 12^o cot 38^o cot 52^o cot 60^o cot 78^o
= (cot 12^o cot 78^o) (cot 38^o cot 52^o) cot 60^o
= [cot 12^o cot (90^o - 12^o)] [cot 38^o cot (90^o - 38^o)] cot 60^o
= [cot 12^o tan 12^o] [cot 38^o tan 38^o] cot 60^o
= 1 x 1 x 1 / √3 = 1 / √3

If tan 2X = cot (X + 6^o), where 2X and X + 6^o are acute angles.
Find the value of X.

(a)14^o

(b)25^o

(d)32^o

Answer!

Answer is: CWe have,
tan 2X = cot (X + 6^o)
cot (90^o - 2X) = cot (X + 6^o)
90 – 2X = X + 6^o
3X = 84^o
X = 28^o

Find the value of [(1 + cot X) - cosec X] [1 + tan X + sec X]

(a)0

(b)2

(d)8

Answer!

Answer is: B[(1 + cot X) - cosec X] [1 + tan X + sec X]
[1 + (cos X / sin X) - (1 / sin X)][1 + (sin X / cos X) + (1 / cos X)]
= [(sin X + cos X - 1) / sin X][(cos X + sin X - 1) / cos X]
= [(sin X + cos X)² - 1 / sin X cos X]
(sin ² X + cos² X + 2 sin X cos X - 1) / sin X cos X
(1 + 2 sin X cos X - 1) / sin X cos X = 2

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Aptitude TrigonometryPage 1

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