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NCERT Exemplar · Class 10 Mathematics Introduction to Trigonometry and its Applications

60 questions · 60 still being checked

EXERCISE 8.1 11–15 (part 2 of 7)

  1. Choose the correct answer from the given four options:

    Exercise 11

    The value of the expression [sin⁡222∘+sin⁡268∘cos⁡222∘+cos⁡268∘+sin⁡263∘+cos⁡63∘sin⁡27∘]\displaystyle \left[\frac{\sin ^2 22^{\circ}+\sin ^2 68^{\circ}}{\cos ^2 22^{\circ}+\cos ^2 68^{\circ}}+\sin ^2 63^{\circ}+\cos 63^{\circ} \sin 27^{\circ}\right] is
    (A)
    3\displaystyle 3 (B) 2\displaystyle 2 (C) 1\displaystyle 1 (D) 0\displaystyle 0

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    NCERT’s answer
    (B)
    (B) \(\displaystyle 2\) \[\sin68^\circ=\cos22^\circ,\quad \cos68^\circ=\sin22^\circ \] \[\frac{\sin^222^\circ+\sin^268^\circ}{\cos^222^\circ+\cos^268^\circ}=\frac{\sin^222^\circ+\cos^222^\circ}{\cos^222^\circ+\sin^222^\circ}=1 \] \[\cos63^\circ=\sin27^\circ \] \[\sin^263^\circ+\cos63^\circ\sin27^\circ=\cos^227^\circ+\sin^227^\circ=1 \] \[1+1=2 \]

  2. Exercise 12

    If 4tan⁡θ=3\displaystyle 4 \tan \theta=3, then (4sin⁡θ−cos⁡θ4sin⁡θ+cos⁡θ)\displaystyle \left(\frac{4 \sin \theta-\cos \theta}{4 \sin \theta+\cos \theta}\right) is equal to
    (A)
    23\displaystyle \frac{2}{3}
    (B)
    13\displaystyle \frac{1}{3}
    (C)
    12\displaystyle \frac{1}{2}
    (D)
    34\displaystyle \frac{3}{4}

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    NCERT’s answer
    (C)
    (C) \(\displaystyle \dfrac12\) \[4\tan\theta=3 \implies \tan\theta=\tfrac34 \] \[\frac{4\sin\theta-\cos\theta}{4\sin\theta+\cos\theta}=\frac{4\tan\theta-1}{4\tan\theta+1} \] \[=\frac{3-1}{3+1}=\frac12 \]

  3. Exercise 13

    If sin⁡θ−cos⁡θ=0\displaystyle \sin \theta-\cos \theta=0, then the value of (sin⁡4θ+cos⁡4θ)\displaystyle \left(\sin ^4 \theta+\cos ^4 \theta\right) is
    (A)
    1\displaystyle 1 (B) 34\displaystyle \frac{3}{4}
    (C)
    12\displaystyle \frac{1}{2}
    (D)
    14\displaystyle \frac{1}{4}

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    NCERT’s answer
    (C)
    (C) \(\displaystyle \dfrac12\) \[\sin\theta-\cos\theta=0 \implies \sin\theta=\cos\theta \] \[\sin^2\theta=\cos^2\theta,\quad \sin^2\theta+\cos^2\theta=1 \implies \sin^2\theta=\cos^2\theta=\tfrac12 \] \[\sin^4\theta+\cos^4\theta=\left(\tfrac12\right)^2+\left(\tfrac12\right)^2=\tfrac12 \]

  4. Exercise 14

    sin⁡(45∘+θ)−cos⁡(45∘−θ)\displaystyle \sin \left(45^{\circ}+\theta\right)-\cos \left(45^{\circ}-\theta\right) is equal to
    (A)
    2cos⁡θ\displaystyle 2 \cos \theta
    (B)
    0\displaystyle 0 (C) 2sin⁡θ\displaystyle 2 \sin \theta

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    NCERT’s answer
    (B)
    (B) \(\displaystyle 0\) \[\cos(45^\circ-\theta)=\sin\big(90^\circ-(45^\circ-\theta)\big)=\sin(45^\circ+\theta) \] \[\sin(45^\circ+\theta)-\cos(45^\circ-\theta)=\sin(45^\circ+\theta)-\sin(45^\circ+\theta)=0 \]

  5. Exercise 15

    A pole 6\displaystyle 6 m high casts a shadow 23 m\displaystyle 2 \sqrt{3} \mathrm{~m} long on the ground, then the Sun's elevation is
    (A)
    60\displaystyle 60°
    (B)
    45\displaystyle 45°
    (C)
    30\displaystyle 30°
    (D)
    90\displaystyle 90°

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    NCERT’s answer
    (A)
    (A) \(\displaystyle 60^\circ\) \[\tan\theta=\frac{\text{pole}}{\text{shadow}}=\frac{6}{2\sqrt3}=\sqrt3 \] \[\theta=60^\circ \] NCERT_Solution_Class10_Maths_Exemplar_Ch8_Ex8-1_Q15