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Trigonometry Aptitude Questions

Quantitative Aptitude Questions and Answers section on “Trigonometry” with solution and explanation for competitive examinations such as CAT, MBA, SSC, Bank PO, Bank Clerical and other examinations.

1.

The circular measure of an angle of an isoceles triangle is 5π/9. Circular measure of one of the other angles must be
[A]\frac{4\pi }{9}
[B]\frac{2\pi }{9}
[C]\frac{5\pi }{9}
[D]\frac{5\pi }{18}

\frac{2 \pi}{9}
Sum of remaining two angles = \pi -\frac{5\pi }{9}=\frac{4\pi}{9}
∴ Each angle = \frac{1}{2}\times \frac{4\pi }{9}=\frac{2\pi }{9}
Hence option [B] is the right answer.

2.

\left ( \frac{3\pi }{5} \right ) radians is equals to :
[A]120°
[B]180°
[C]108°
[D]100°

108°
\because \pi  radian=180°
\therefore \frac{3\pi }{5} radian = \frac{180}{\pi}\times\frac{3\pi }{5}
=108°
Hence option [C] is the right answer.

3.

If 0\leq \theta \leq \frac{\pi}{2} and \sec ^{2}\theta +\tan^{2}\theta=7, then Θ is:
[A]\frac{\pi}{5} Radian
[B]\frac{\pi}{6} Radian
[C]\frac{5\pi}{12} Radian
[D]\frac{\pi}{3} Radian

\frac{\pi}{3} Radian
Given Expression, \sec ^{2}\theta +\tan ^{2}\theta =7
=>1+\tan ^{2}\theta+\tan ^{2}\theta=7
=>2 \tan ^{2}\theta=7-1=6
=>\tan ^{2}\theta=3
=>\tan \theta=\sqrt{3}
\because \tan 60°=\sqrt{3}
\therefore \theta=60°
\because 180°=\pi Radian
\therefore 60°= \frac{\pi }{180}\times60=\frac{\pi }{3} Radian
Hence option [D] is the right answer.

4.

In circular measure, the value of the angle 11°15′ is :
[A]\frac{\pi^{c}}{16}
[B]\frac{\pi^{c}}{8}
[C]\frac{\pi^{c}}{4}
[D]\frac{\pi^{c}}{12}

\frac{\pi^{c}}{16}
11°15′
=> 11\textdegree + \frac{15\textdegree}{60}
=> 11\textdegree + \frac{1}{4} = \frac{45\textdegree}{4}
=> [180° = π Radian]
\therefore \frac{45^{\circ}}{4} = \frac{\pi}{180}\times \frac{45}{4}=\frac{\pi^{c}}{16}
Hence option [A] is correct answer.

5.

In a triangle ABC, \angle ABC=75\textdegree and \angle ACB=\frac{\pi^{c}}{4}. The circular measure of \angle BAC is:
[A]\frac{\pi}{6}Radian
[B]\frac{\pi}{2}Radian
[C]\frac{5\pi}{12}Radian
[D]\frac{\pi}{3}Radian

\frac{\pi}{3}Radian
\angle ABC = 75^{\circ}
[\because 180 \textdegree = \pi radian]
75^{\circ} = \frac{\pi}{180}\times 75 = \frac{5 \pi}{12} Radian
\therefore \angle BAC = \pi - \frac{\pi}{4} - \frac{5 \pi}{12}
=> \frac{12\pi-3\pi -5\pi}{12} = \frac{4 \pi}{12}
=> \frac{\pi}{3}Radian
Hence option [D] is the right answer.

6.

The degree measure of 1 radian is :
[A]57\textdegree32'16''
[B]57\textdegree61'22''
[C]57\textdegree16'22''
[D]57\textdegree22'16''

\mathbf{57\textdegree16'22''}
\therefore 1 radian  = \frac{180\textdegree}{\pi}
= \frac{180\times7\textdegree}{22}
= \frac{630}{11} = 57\frac{3}{11}\textdegree
= 57\textdegree\frac{3}{11}\times 60' = 57\textdegree\frac{180'}{11}
= 57\textdegree16'\frac{4}{11}\times 60'' = 57\textdegree16'22''
Hence option [C] is the right answer.

7.

In the sum of two angles is 135\textdegree and their difference is \frac{\pi}{12}. Then the circular measure of the greater angle is :
[A]\frac{\pi}{3}
[B]\frac{5\pi}{12}
[C]\frac{2\pi}{3}
[D]\frac{3\pi}{5}

\mathbf{\frac{5\pi}{12}}
Two angles = A and B where A > B.
\therefore A + B = 135\textdegree
= \left ( \frac{135\times \pi}{180} \right ) radian
=> A + B = \left ( \frac{3\pi}{4} \right ) radian…..(1)
A - B = \frac{\pi}{12}…..(2)
On adding these equations,
2A = \frac{3\pi}{4} + \frac{\pi}{12}
= \frac{9\pi+\pi}{12} = \frac{10\pi}{12} = \frac{5\pi}{6}
\therefore A = \frac{5\pi}{12}radian
Hence option [B] is the right answer.

8.

If the sum and difference of two angles are \frac{22}{9} radian and 36\textdegree respectively, then the value of smaller angle in degree taking the value of \pi as \frac{22}{7} is :
[A]48\textdegree
[B]60\textdegree
[C]56\textdegree
[D]52\textdegree

\mathbf{52\textdegree}
\because \pi radian = 180\textdegree
\therefore \frac{22}{9}radian = \frac{180}{\pi}\times\frac{22}{9}
= \frac{180}{22}\times \frac{22\times7}{9} = 140\textdegree….(1)
According to the question,
A + B = 140\textdegree
and, A - B = 36\textdegree ……(2)
On adding
2A = 176\textdegree
=> A = \frac{176}{2} = 88\textdegree
From equation (1),
\therefore 88\textdegree+B = 140\textdegree
=> B = 140\textdegree - 88\textdegree = 52\textdegree
Hence option [D] is the right answer.

9.

If \cos x+\cos y  = 2, the value of \sin x+\sin y is :
[A]-1
[B]1
[C]0
[D]2

0
\cos x+\cos y  = 2
\because \cos x\leq 1
=> \cos x = 1; \cos y = 1
=> x = y = 0\textdegree [\cos0\textdegree = 1]
\therefore \sin x + \sin y = 0
Hence option [C] is the right answer.

10.

The minimun value of 2\sin ^{2}\theta +3\cos ^{2}\theta is :
[A]0
[B]2
[C]3
[D]1

2
2\sin ^{2}\theta +3\cos ^{2}\theta
=> 2\sin ^{2}\theta +2\cos ^{2}\theta +\cos ^{2}\theta
=> 2\left ( \sin ^{2}\theta +\cos ^{2}\theta  \right ) + \cos ^{2}\theta
=> 2 + \cos ^{2}\theta
Minimum value of \cos\theta = -1
But \cos ^{2}\theta \geq 0, where \theta  = 90\textdegree
[\cos  0\textdegree = 1, \cos  90\textdegree = 0]
Hence required minimum value = 2 + 0 = 0
Option [B] is the right answer.



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