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

60 questions · 60 still being checked

EXERCISE 8.4 11–18 (part 7 of 7)

  1. Exercise 11

    If asin⁡θ+bcos⁡θ=c\displaystyle a \sin \theta+b \cos \theta=c, then prove that acos⁡θ−bsin⁡θ=a2+b2−c2\displaystyle a \cos \theta-b \sin \theta=\sqrt{a^2+b^2-c^2}.

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    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    \[a\sin\theta+b\cos\theta=c \] \[(a\sin\theta+b\cos\theta)^2+(a\cos\theta-b\sin\theta)^2=a^2(\sin^2\theta+\cos^2\theta)+b^2(\cos^2\theta+\sin^2\theta)=a^2+b^2 \] \[c^2+(a\cos\theta-b\sin\theta)^2=a^2+b^2 \] \[(a\cos\theta-b\sin\theta)^2=a^2+b^2-c^2 \] Taking the positive root: \[a\cos\theta-b\sin\theta=\sqrt{a^2+b^2-c^2} \] Answer: \(\displaystyle a\cos\theta-b\sin\theta=\sqrt{a^2+b^2-c^2}\)

  2. Exercise 12

    Prove that 1+sec⁡θ−tan⁡θ1+sec⁡θ+tan⁡θ=1−sin⁡θcos⁡θ\displaystyle \frac{1+\sec \theta-\tan \theta}{1+\sec \theta+\tan \theta}=\frac{1-\sin \theta}{\cos \theta}

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    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    \[\sec^2\theta-\tan^2\theta=1 \quad\Rightarrow\quad \sec\theta-\tan\theta=\frac{1}{\sec\theta+\tan\theta} \] \[1+\sec\theta-\tan\theta=1+\frac{1}{\sec\theta+\tan\theta}=\frac{(\sec\theta+\tan\theta)+1}{\sec\theta+\tan\theta} \] \[\frac{1+\sec\theta-\tan\theta}{1+\sec\theta+\tan\theta}=\frac{1}{\sec\theta+\tan\theta} \] \[\frac{1}{\sec\theta+\tan\theta}=\frac{\cos\theta}{1+\sin\theta}=\frac{\cos\theta(1-\sin\theta)}{1-\sin^2\theta}=\frac{1-\sin\theta}{\cos\theta} \] Answer: \(\displaystyle \dfrac{1+\sec\theta-\tan\theta}{1+\sec\theta+\tan\theta}=\dfrac{1-\sin\theta}{\cos\theta}\)

  3. Exercise 13

    The angle of elevation of the top of a tower 30\displaystyle 30 m high from the foot of another tower in the same plane is 60\displaystyle 60° and the angle of elevation of the top of the second tower from the foot of the first tower is 30\displaystyle 30°. Find the distance between the two towers and also the height of the other tower.

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    NCERT’s answer
    \(\displaystyle 10 \sqrt{3} \mathrm{~m} ; 10 \mathrm{~m}\)
    Let \(\displaystyle AB=30\) m be the first tower, \(\displaystyle CD\) the second, \(\displaystyle BD=x\); \(\displaystyle \angle ADB=60^\circ,\ \angle CBD=30^\circ\). NCERT_Solution_Class10_Maths_Exemplar_Ch8_Ex8-4_Q13 \[\tan 60^\circ = \frac{AB}{BD} \] \[x = \frac{30}{\sqrt3} = 10\sqrt3 \] \[\tan 30^\circ = \frac{CD}{BD} \] \[CD = 10\sqrt3 \times \frac{1}{\sqrt3} = 10 \] Answer: distance between the towers \(\displaystyle 10\sqrt3\approx17.3\) m; height of the other tower \(\displaystyle 10\) m.

  4. Exercise 14

    From the top of a tower h m\displaystyle h \mathrm{~m} high, the angles of depression of two objects, which are in line with the foot of the tower are α\displaystyle \alpha and β(β>α)\displaystyle \beta(\beta>\alpha). Find the distance between the two objects.

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    NCERT’s answer
    \(\displaystyle h(\cot \alpha-\cot \beta)\)
    Let \(\displaystyle AB=h\) be the tower, \(\displaystyle P\) the nearer object (depression \(\displaystyle \beta\)), \(\displaystyle Q\) the farther (depression \(\displaystyle \alpha\)), both in line with foot \(\displaystyle B\). NCERT_Solution_Class10_Maths_Exemplar_Ch8_Ex8-4_Q14 \[\angle APB=\beta,\quad \angle AQB=\alpha \quad \text{(alternate angles)} \] \[\tan\beta = \frac{h}{BP} \implies BP = h\cot\beta \] \[\tan\alpha = \frac{h}{BQ} \implies BQ = h\cot\alpha \] \[PQ = BQ - BP = h(\cot\alpha-\cot\beta) \] Answer: \(\displaystyle PQ=h(\cot\alpha-\cot\beta)\) m.

  5. Exercise 15

    A ladder rests against a vertical wall at an inclination α\displaystyle \alpha to the horizontal. Its foot is pulled away from the wall through a distance p\displaystyle p so that its upper end slides a distance q\displaystyle q down the wall and then the ladder makes an angle β\displaystyle \beta to the horizontal. Show that pq=cos⁡β−cos⁡αsin⁡α−sin⁡β\displaystyle \frac{p}{q}=\frac{\cos \beta-\cos \alpha}{\sin \alpha-\sin \beta}.

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    Ladder \(\displaystyle AB\), length \(\displaystyle L\), leans on the wall at \(\displaystyle A\) and slides to \(\displaystyle CD\), \(\displaystyle CD=L\); \(\displaystyle O\) is the foot of the wall. NCERT_Solution_Class10_Maths_Exemplar_Ch8_Ex8-4_Q15 \[\triangle AOB:\quad OB=L\cos\alpha,\quad OA=L\sin\alpha \] \[\triangle COD:\quad OD=L\cos\beta,\quad OC=L\sin\beta \] \[p = BD = OD-OB = L(\cos\beta-\cos\alpha) \] \[q = AC = OA-OC = L(\sin\alpha-\sin\beta) \] \[\frac{p}{q} = \frac{L(\cos\beta-\cos\alpha)}{L(\sin\alpha-\sin\beta)} = \frac{\cos\beta-\cos\alpha}{\sin\alpha-\sin\beta} \] Answer: \(\displaystyle \dfrac{p}{q}=\dfrac{\cos\beta-\cos\alpha}{\sin\alpha-\sin\beta}\)

  6. Exercise 16

    The angle of elevation of the top of a vertical tower from a point on the ground is 60\displaystyle 60°. From another point 10\displaystyle 10 m vertically above the first, its angle of elevation is 45\displaystyle 45°. Find the height of the tower.

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    NCERT’s answer
    \(\displaystyle 5(\sqrt{3}+3) \mathrm{m}\)
    Let \(\displaystyle AB=h\) be the tower, \(\displaystyle C\) the ground point, \(\displaystyle D\) the point \(\displaystyle 10\) m above it; \(\displaystyle DM\perp AB\), so \(\displaystyle DM=BC\), \(\displaystyle AM=h-10\). NCERT_Solution_Class10_Maths_Exemplar_Ch8_Ex8-4_Q16 \[\tan 60^\circ = \frac{h}{BC} \implies BC=\frac{h}{\sqrt3} \] \[\tan 45^\circ = \frac{h-10}{BC} \implies h-10=\frac{h}{\sqrt3} \] \[h\left(1-\frac{1}{\sqrt3}\right)=10 \implies h=\frac{10\sqrt3}{\sqrt3-1} \] \[h=\frac{10\sqrt3(\sqrt3+1)}{(\sqrt3-1)(\sqrt3+1)}=\frac{10(3+\sqrt3)}{2}=5(3+\sqrt3) \] Answer: \(\displaystyle h=5(3+\sqrt3)\) m \(\displaystyle \approx23.66\) m.

  7. Exercise 17

    A window of a house is h\displaystyle h metres above the ground. From the window, the angles of elevation and depression of the top and the bottom of another house situated on the opposite side of the lane are found to be α\displaystyle \alpha and β\displaystyle \beta, respectively. Prove that the height of the other house is h(1+tan⁡αcot⁡β)\displaystyle h(1+\tan \alpha \cot \beta) metres.

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    Let \(\displaystyle W\) be the window, \(\displaystyle OW=h\); \(\displaystyle F,\,T\) the foot and top of the other house, \(\displaystyle FT=H\); \(\displaystyle WM\) horizontal, \(\displaystyle WM=OF=d\). NCERT_Solution_Class10_Maths_Exemplar_Ch8_Ex8-4_Q17 \[\tan\beta=\frac{MF}{WM}=\frac{h}{d} \implies d=h\cot\beta \] \[\tan\alpha=\frac{MT}{WM}=\frac{H-h}{d} \] \[H-h = d\tan\alpha = h\cot\beta\tan\alpha \] \[H = h(1+\tan\alpha\cot\beta) \] Answer: \(\displaystyle H=h(1+\tan\alpha\cot\beta)\) m.

  8. Exercise 18

    The lower window of a house is at a height of 2\displaystyle 2 m above the ground and its upper window is 4\displaystyle 4 m vertically above the lower window. At certain instant the angles of elevation of a balloon from these windows are observed to be 60\displaystyle 60° and 30\displaystyle 30°, respectively. Find the height of the balloon above the ground.

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    NCERT’s answer
    $\displaystyle 8$ m
    Let \(\displaystyle P,\,Q\) be the lower and upper windows, \(\displaystyle OP=2\), \(\displaystyle OQ=6\); \(\displaystyle B\) the balloon, \(\displaystyle BG=H\); \(\displaystyle PM,\,QN\) horizontal, \(\displaystyle PM=QN=d\). Then \(\displaystyle BM=H-2\), \(\displaystyle BN=H-6\). NCERT_Solution_Class10_Maths_Exemplar_Ch8_Ex8-4_Q18 \[\tan 60^\circ = \frac{BM}{PM}=\frac{H-2}{d} \implies d=\frac{H-2}{\sqrt3} \] \[\tan 30^\circ = \frac{BN}{QN}=\frac{H-6}{d} \implies d=\sqrt3(H-6) \] \[\frac{H-2}{\sqrt3} = \sqrt3(H-6) \implies H-2=3(H-6) \] \[2H=16 \implies H=8 \] Answer: the balloon is \(\displaystyle 8\) m above the ground.