Trigonometry

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.

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.

$\left ( \frac{3\pi }{5} \right )$ 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.

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.

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.

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.

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.

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.

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.

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.

Trigonometry

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