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Maths Case study (4 Marks)

Introduction of Trigonometry · Jharkhand Board (JAC) STD 10 (Jharkhand - English Medium). 5 questions.

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Case study (4 Marks)

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5 questions · self-marked practice — reveal the answer and mark yourself.

Question 14 Marks

Ritu's daughter is feeling so hungry and so thought to eat something. She looked into the fridge and found some bread pieces. She decided to make a sandwich. She cut the piece of bread diagonally and found that it forms a right angled triangle, with sides 4cm, $4\sqrt{3}\text{ cm}$ and 8cm.

On the basis of above information, answer the following questions.
  1. The value of $\angle\text{M}=$
  1. The value of $\angle\text{K}=$
  2. Find the value of $\tan\text{M}$.
    Or
    $\sec^2\text{M}-1=$
Answer
1. 30º2. 60º
3.$\frac{1}{\sqrt{3}}$
Or
$\tan^2\text{M}$
 
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Question 24 Marks

Aanya and her father go to meet her friend Juhi for a party. When they reached to Juhi's place, Aanya saw the roof of the house, which is triangular in shape. If she imagined the dimensions of the roof as given in the figure, then answer the following questions.
  1. If D is the mid point of AC, then BD =
  2. Measure of $\angle\text{A}=$
  3. Find the value of $\sin\text{A}+\cos\text{A}$.
    Or
    Find the value of $\tan^2\text{C}+\tan^2\text{A}$.
Answer
1. 6 m2. 45º
3. $\sqrt{2}$
Or
2
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Question 34 Marks

Anita, a student of class 10th, has to made a project on 'Introduction to Trigonometry'. She decides to make a bird house which is triangular in shape. She uses cardboard to make the bird house as shown in the figure. Considering the front side of bird house as right angled triangle PQR, right angled at R, answer the following questions.
  1. If $\angle\text{PQR}=\theta$, then $\cos\theta=$
  2. The value of $\sec\theta=$
  3. The value of $\frac{\tan\theta}{1+\tan^2\theta}=$
    Or
    The value of $\cot^2\theta-\text{cosec}^2\theta=$
Answer
1. $\cos\theta=\frac{\text{QR}}{\text{PQ}}=\frac{12}{13}$2. $\sec\theta=\frac{1}{\cos\theta}=\frac{13}{12}$
3. $\tan\theta=\frac{\text{PR}}{\text{RQ}}=\frac{5}{12}$
Or
$\therefore\frac{\tan\theta}{1+\tan^2\theta}=\frac{\frac{5}{12}}{1+\frac{25}{144}}=\frac{\frac{5}{12}}{\frac{169}{144}}=\frac{60}{169}$
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Question 44 Marks

Two aeroplanes leave an airport, one after the other. After moving on runway, one flies due North and other flies due South. The speed of two aeroplanes is 400km/ hr and 500km/ hr respectively. Considering PQ as runway and A and B are any two points in the path followed by two planes, then answer the following questions.
  1. Find $\tan\theta$ if $\angle\text{APQ}=\theta.$
  2. Find $\cot\text{B}$.
  3. Find $\tan\text{A}$.
    Or
    Find $\sec\text{A}$.
Answer
1. ln $\triangle\text{APQ},\tan\theta=\frac{\text{AQ}}{\text{PQ}}=\frac{1.2}{1.6}=\frac{3}{4}$2. In $\triangle\text{PBQ},\cot\text{B}=\frac{\text{QB}}{\text{PQ}}=\frac{3}{1.6}=\frac{15}{8}$
3. In $\triangle\text{APQ},\tan\text{A}=\frac{\text{PQ}}{\text{AQ}}=\frac{1.6}{1.2}=\frac{4}{3}$
Or
We have, $\tan^2\text{A}+ 1 = \sec^2\text{A}$
$\Rightarrow\sqrt{\Big(\frac{4}{3}\Big)^2+1}$
$=\sqrt{\frac{16}{9}+1}=\sqrt{\frac{25}{9}}=\frac{5}{3}$
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Question 54 Marks

Three friends - Anshu, Vijay and Vishal are playing hide and seek in a park. Anshu and Vijay hide in the shrubs and Vishal have to find both of them. If the positions of three friends are at A, B and C respectively as shown in the figure and forms a right angled triangle such that AB = 9 m, BC $=\sqrt{3}\text{ m}$ and $\angle\text{B}=90^\circ$, then answer the following questions.
  1. The measure of $\angle\text{A}$ is:
  2. The length of AC is:
  3. $\cos 2\text{A} =$
    Or
    $\text{Sin}\Big(\frac{\text{C}}{2}\Big)=$
Answer
1. We have, AB = 9 m, BC $=\sqrt{3}\text{ m}$ in $\triangle\text{ABC}$, we have
$\tan\text{A}=\frac{\text{BC}}{\text{AB}}=\frac{3\sqrt{3}}{9}=\frac{1}{\sqrt{3}}$
$\Rightarrow\tan\text{A} = \tan30^\circ \Rightarrow\angle\text{A}=30^\circ$2. Since, $\sin\text{A}=\frac{\text{BC}}{\text{AC}}\Rightarrow\text{sin}30^\circ=\frac{\text{BC}}{\text{AC}}$
$\Rightarrow\frac{1}{2}=\frac{3\sqrt{3}}{\text{AC}}\Rightarrow\text{AC}=6\sqrt{3}\text{ m}$
3.$\because\angle\text{A}=30^\circ$
$\therefore\cos2\text{A}=\cos(2 \times 30^\circ)=\cos60^\circ=\frac{1}{2}$
Or
$\because\angle\text{C}=60^\circ$
$\therefore\text{sin}\Big(\frac{\text{C}}{2}\Big)=\text{sin}\Big(\frac{60^\circ}{2}\Big)=\text{sin}30^\circ=\frac{1}{2}$
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