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Maths M.C.Q (1 Marks)

Introduction of Trigonometry · CBSE STD 10 (CBSE - English Medium). 56 questions.

Questions · Page 2 of 2

M.C.Q (1 Marks)

MCQ 511 Mark
$\cos 60^{\circ} \cos 30^{\circ}-\sin 60^{\circ} \sec 30^{\circ}=$ ?
  • ✓
    $\cos 90^{\circ}$
  • B
    $\sin 90^{\circ}$
  • C
    $\sin 45^{\circ}$
  • D
    $\cos 45^{\circ}$
Answer
Correct option: A.
$\cos 90^{\circ}$
$\cos 60^{\circ} \cos 30^{\circ}-\sin 60^{\circ} \sin 30^{\circ}$
$=\frac{1}{2} \times \frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2} \times \frac{1}{2}$
$=\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}=0$
$=\cos 90^{\circ}$
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MCQ 521 Mark
Value of $\tan 30^{\circ} \operatorname{cosec} 60^{\circ}+\tan 60^{\circ} \sec 30^{\circ}=?$
  • A
    $\frac{3}{8}$
  • ✓
    $\frac{8}{3}$
  • C
    $\frac{2}{5}$
  • D
    None of these
Answer
Correct option: B.
$\frac{8}{3}$
$\tan 30^{\circ} \operatorname{cosec} 60^{\circ}+\tan 60^{\circ} \sec 30^{\circ} .$
$=\frac{1}{\sqrt{3}} \times \frac{2}{\sqrt{3}}+\sqrt{3} \times \frac{2}{\sqrt{3}}$
$=\frac{2}{3}+2$
$=\frac{2+6}{3}$
$=\frac{8}{3}$
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MCQ 531 Mark
Value of $\sin 60 \cdot \cos 30-\cos 60^{\circ} \sin 30^{\circ}= ?$
  • ✓
    $\frac{1}{2}$
  • B
    $\frac{1}{3}$
  • C
    $1$
  • D
    $0$
Answer
Correct option: A.
$\frac{1}{2}$
$\sin 60^{\circ} \cos 30^{\circ}-\cos 60^{\circ} \sin 30^{\circ} .$
$=\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}-\frac{1}{2} \times \frac{1}{2}$
$=\frac{3}{4}-\frac{1}{4}$
$=\frac{3-1}{4}$
$=\frac{2}{4}$
$=\frac{1}{2}$
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MCQ 541 Mark
If $\tan \theta=\frac{15}{8}$, then $\cot \theta= ?$
  • ✓
    $\frac{8}{15}$
  • B
    $\frac{17}{15}$
  • C
    $\frac{15}{17}$
  • D
    $\frac{8}{17}$
Answer
Correct option: A.
$\frac{8}{15}$
$\because \cot \theta=\frac{1}{\tan \theta}$
$=\frac{1}{\frac{15}{8}}$
$=\frac{8}{15}$
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MCQ 551 Mark
If $\sin A =\frac{9}{41}$, then $\operatorname{cosec} A =$ ?
  • A
    $\frac{9}{41}$
  • ✓
    $\frac{41}{9}$
  • C
    $\frac{41}{42}$
  • D
    None of these
Answer
Correct option: B.
$\frac{41}{9}$
(b)
$\because \operatorname{cosec} A =\frac{1}{\sin A}=\frac{1}{\frac{9}{41}}=\frac{41}{9}$
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MCQ 561 Mark
If $15 \cot A=8$, then $\sin A =$ ?
  • A
    $\frac{8}{15}$
  • B
    $\frac{8}{17}$
  • C
    $\frac{17}{8}$
  • ✓
    $\frac{15}{17}$
Answer
Correct option: D.
$\frac{15}{17}$
Image

$\therefore \tan A=\frac{15}{8}=\frac{BC}{AB}$
In $\triangle ABC$, by Pythagoras Theorem
$(AC)^2=(AB)^2+(BC)^2$
$=(8)^2+(15)^2$
$=64+225$
$=289$
$\Rightarrow AC=\sqrt{289}=17$
$\therefore \sin A=\frac{BC}{AC}=\frac{15}{17}$
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M.C.Q (1 Marks) - Page 2 - Maths STD 10 Questions - Vidyadip