Question

अधोलिखितानां पदानां प्रकृति प्रत्यय विभागं कृत्वा लिखत्-

1. नत्वा
2. पठितुम्​

Answer 1

Answer:

Explanation:

We know that :-

Sin 60° = √3/2

Cos 30° = √3/2

Sin 30° = 1/2

cos 30° = 1/2

tan 45° = 1

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Trigonometric table :-

$\begin{gathered}\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3} }{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }& 1 & \sqrt{3} & \rm Not \: De fined \\ \\ \rm cosec A & \rm Not \: De fined & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm Not \: De fined \\ \\ \rm cot A & \rm Not \: De fined & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0 \end{array}\end{gathered}$

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​ $\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|cc}\sf Class\: interval&\sf Frequency\: (f)\\\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad}{}\\\sf 40 - 50&\sf 5\\\\\sf 50 - 60 &\sf x\\\\\sf 60-70 &\sf 15\\\\\sf 70 - 80&\sf 12\\\\\sf 80-90&\sf 7\\\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad}{}\\\sf & \sf \sum f = 67& \end{array}}\end{gathered}\end{gathered}$

Classinterval

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40−50

50−60

60−70

70−80

80−90

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∑f=67

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