SolveItClass 10 · NCERT

NCERT Solutions · Class 10 Mathematics Introduction to Trigonometry

19 questions · 19 still being checked

EXERCISE 8.1 1–11 (part 1 of 3)

  1. Exercise 1

    In △ABC\displaystyle \triangle \mathrm{ABC}, right-angled at B,AB=24 cm,BC=7 cm\displaystyle \mathrm{B}, \mathrm{AB}=24 \mathrm{~cm}, \mathrm{BC}=7 \mathrm{~cm}. Determine :
    (i)
    sin⁡A,cos⁡A\displaystyle \sin \mathrm{A}, \cos \mathrm{A}
    (ii)
    sin⁡C,cos⁡C\displaystyle \sin \mathrm{C}, \cos \mathrm{C}
    NCERT’s answer
    (i)
    $\displaystyle \sin \mathrm{A}=\frac{7}{25}, \cos \mathrm{~A}=\frac{24}{25}$ (ii) $\displaystyle \sin \mathrm{C}=\frac{24}{25}, \cos \mathrm{C}=\frac{7}{25}$

    Working being prepared

  2. Exercise 2

    In Fig. 8.13\displaystyle 8.13, find tan⁡P−cot⁡R\displaystyle \tan \mathrm{P}-\cot \mathrm{R}.NCERT_Question_Class10_Maths_Ch8_Ex8-1_Q2
    NCERT’s answer
    $\displaystyle 0$

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  3. Exercise 3

    If sin⁡A=34\displaystyle \sin \mathrm{A}=\frac{3}{4}, calculate cos⁡A\displaystyle \cos \mathrm{A} and tan⁡A\displaystyle \tan \mathrm{A}.
    NCERT’s answer
    $\displaystyle \cos \mathrm{A}=\frac{\sqrt{7}}{4}, \tan \mathrm{~A}=\frac{3}{\sqrt{7}}$

    Working being prepared

  4. Exercise 4

    Given 15cot⁡ A=8\displaystyle 15 \cot \mathrm{~A}=8, find sin⁡A\displaystyle \sin \mathrm{A} and sec⁡A\displaystyle \sec \mathrm{A}.
    NCERT’s answer
    $\displaystyle \sin \mathrm{A}=\frac{15}{17}, \sec \mathrm{~A}=\frac{17}{8}$

    Working being prepared

  5. Exercise 5

    Given sec⁡θ=1312\displaystyle \sec \theta=\frac{13}{12}, calculate all other trigonometric ratios.
    NCERT’s answer
    $\displaystyle \sin \theta=\frac{5}{13}, \cos \theta=\frac{12}{13}, \tan \theta=\frac{5}{12}, \cot \theta=\frac{12}{5}, \operatorname{cosec} \theta=\frac{13}{5}$

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  6. Exercise 6

    If ∠A\displaystyle \angle \mathrm{A} and ∠B\displaystyle \angle \mathrm{B} are acute angles such that cos⁡A=cos⁡B\displaystyle \cos \mathrm{A}=\cos \mathrm{B}, then show that ∠A=∠B\displaystyle \angle \mathrm{A}=\angle \mathrm{B}.

    Solution being prepared

  7. Exercise 7

    If cot⁡θ=78\displaystyle \cot \theta=\frac{7}{8}, evaluate :
    (i)
    (1+sin⁡θ)(1−sin⁡θ)(1+cos⁡θ)(1−cos⁡θ)\displaystyle \frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)},
    (ii)
    cot⁡2θ\displaystyle \cot ^{2} \theta
    NCERT’s answer
    (i)
    $\displaystyle \frac{49}{64}$ (ii) $\displaystyle \frac{49}{64}$

    Working being prepared

  8. Exercise 8

    If 3cot⁡ A=4\displaystyle 3 \cot \mathrm{~A}=4, check whether 1−tan⁡2 A1+tan⁡2 A=cos⁡2 A−sin⁡2 A\displaystyle \frac{1-\tan ^{2} \mathrm{~A}}{1+\tan ^{2} \mathrm{~A}}=\cos ^{2} \mathrm{~A}-\sin ^{2} \mathrm{~A} or not.
    NCERT’s answer
    Yes

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  9. Exercise 9

    In triangle ABC, right-angled at B, if tan⁡A=13\displaystyle \tan \mathrm{A}=\frac{1}{\sqrt{3}}, find the value of:
    (i)
    sin⁡Acos⁡C+cos⁡Asin⁡C\displaystyle \sin \mathrm{A} \cos \mathrm{C}+\cos \mathrm{A} \sin \mathrm{C}
    (ii)
    cos⁡Acos⁡C−sin⁡Asin⁡C\displaystyle \cos \mathrm{A} \cos \mathrm{C}-\sin \mathrm{A} \sin \mathrm{C}
    NCERT’s answer
    (i)
    $\displaystyle 1$ (ii) $\displaystyle 0$

    Working being prepared

  10. Exercise 10

    In △PQR\displaystyle \triangle \mathrm{PQR}, right- angled at Q,PR+QR=25 cm\displaystyle \mathrm{Q}, \mathrm{PR}+\mathrm{QR}=25 \mathrm{~cm} and PQ=5 cm\displaystyle \mathrm{PQ}=5 \mathrm{~cm}. Determine the values of sin P, cos P and tan P.
    NCERT’s answer
    $\displaystyle \sin \mathrm{P}=\frac{12}{13}, \cos \mathrm{P}=\frac{5}{13}, \tan \mathrm{P}=\frac{12}{5}$

    Working being prepared

  11. Exercise 11

    State whether the following are true or false. Justify your answer.
    (i)
    The value of tan A is always less than 1.
    (ii)
    sec⁡A=125\displaystyle \sec \mathrm{A}=\frac{12}{5} for some value of angle A .
    (iii)
    cos⁡A\displaystyle \cos \mathrm{A} is the abbreviation used for the cosecant of angle A .
    (iv)
    cot⁡A\displaystyle \cot \mathrm{A} is the product of cot⁡\displaystyle \cot and A .
    (v)
    sin⁡θ=43\displaystyle \sin \theta=\frac{4}{3} for some angle θ\displaystyle \theta.
    NCERT’s answer
    (i)
    False (ii) True (iii) False (iv) False (v) False

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