Important Questions For Class 10 Maths Chapter 8| Introduction Of Trigonometry

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Introduction of Trigonometry Important Questions for class 10 Chapter 8 (ex 8.1,ex8.2)

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Trigonometric Ratio त्रिकोणमितीय अनुपात

Sin A Cos A & TanA ratio

Sin A=Sin 60^{0
CosA = Cos 60}^{Tan A = Tan 60}

Reverse (उल्टा) Sin A, Cos A & TanA

The Value of the trignometric ratio of an angle do not vary with the lengths of the sides of the triangle; if the angles remains the same.
यदि कोण समान बना रहता हो, तो एक कोण के त्रिकोणमितीय अनुपातों के मानोंं में त्रिभुज की भुजाओं की लम्बाइयों के साथ कोई परिवर्तन नहीं होता है।

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Pythagoras Theorem (ex 8.3)

एक नजर

Example

Ex 1. Given tan A =4/3, find the other trigonometric ratios of the angle A
यदि tan A = 4/3 हो तो कोण A के अन्य त्रिकोणमितीय अनुपात ज्ञात कीजिए।

Step-1

tan A = BC/AB =4/3
4K/3K

Step- 2

Pythagoras Theorem
AC² = AB² + BC²
AC² = (4K)² + (3K)²
AC² =
4X4XKXK + 3X3XKXK
= 16K² + 9K²
AC²=25K²

Step-3

AC = 5

EX2 If ∠ B and ∠ Q are acute angles such that sin B = sin Q, then prove that ∠ B = ∠ Q.

यदि B और ∠Q हैं न्यून कोण जैसे sin B = sin Q, तो सिद्ध कीजिए कि ∠ B = ∠ Q.

Step-1

Let us consider two right angle triangles PQR and ABC where
sin Q = sin B
आइए हम दो समकोणों पर विचार करें त्रिभुज PQR और ABC जहाँ
sin Q = sin B

Step-2

Step-3

Step-4

So Triangle ABC and triangliePQR are similar therefore
अत: त्रिभुज ABC और त्रिभुजPQR समरूप हैं इसलिए
∠ B = ∠ Q

EX 3 Consider ∆ ACB, right-angled at C, inwhich AB = 29 units, BC = 21 units and ∠ ABC = θDetermine the values of
(ACB पर विचार करें, जो C पर समकोण है, जिसमें AB = 29 इकाई, BC = 21 इकाई और ABC = के मान निर्धारित करें।)

(i) cos²θ + sin²θ,

(ii) cos²θ – sin²θ.

Step-1

a²– b² = (a+b)(a–b)

Step-2

Step-3

Step-4

EX 4 In a right triangle ABC, right-angled at B, if tan A = 1, then verify that 2 sin A cos A = 1.
एक समकोण त्रिभुज ABC में, B पर समकोण है, यदि tan A = 1 है, तो सत्यापित करें कि 2 sin A cos A = 1 है।

Step-1

Step-2

Step-3

AC=?
As per Pythogoras Theorem
AB² + BC² = AC²
Since AB = BC = 1
(1)² + (1)² = 2 = AC²
AC² = 2

Step-4

Ex 5 In ∆ OPQ, right-angled at P, OP = 7 cm and OQ – PQ = 1 cm (see Fig.). Determine the values of sin Q and cos Q
OPQ में, P पर समकोण, OP = 7 सेमी और OQ – PQ = 1 सेमी (देखिए आकृति )। sin Q और cos Q के मान ज्ञात कीजिए।

Step-1

Step-2

OQ² = OP² + PQ²
(1+PQ)² = OP² + PQ²
(a+b)² = a² + 2ab +b²
1² +2XPQ + PQ² = OP² + PQ²
1 + 2PQ = OP²
1 + 2PQ = 7²
2PQ = 49-1=48
PQ = 48/2 =24
So OQ = 1+PQ = 1 + 24
OQ = 25

Step-3

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5 Questions & Hints

Question-1

If tan θ + cot θ = 5, find the value of tan²θ + cot²θ. (2012)
Hint-

Trigonometry Basic Click करें

Question 2.
If sec 2A = cosec (A – 27°) where 2A is an acute angle, find the measure of ∠A. (2012, 2017D)

Hint

Question 3

Hint

Question 4.
If sin θ – cos θ = 0, find the value of sin⁴θ + cos⁴θ.
(2012, 2017D)

Question 5.
If sec θ + tan θ = 7, then evaluate
sec θ – tan θ.
(2017OD)

Flash Cards for 1 to 5 Questions

Question 1.
If tan θ + cot θ = 5, find the value of tan²θ + cot²θ. (2012)

Question Explanation:

If tan θ + cot θ = 5, find the value of tan²θ + cot²θ. (2012)
Solution:
tan θ + cot θ = 5 … [Given
tan²θ + cot²θ + 2 tan θ cot θ = 25 … [Squaring both sides
tan²θ + cot²θ + 2 = 25
∴ tan²θ + cot²θ = 23

Question 2.
If sec 2A = cosec (A – 27°) where 2A is an acute angle, find the measure of ∠A. (2012, 2017)

Question Explanation:

Question 2.
If sec 2A = cosec (A – 27°) where 2A is an acute angle, find the measure of ∠A. (2012, 2017)

Hint

Solution:
sec 2A = cosec (A – 27°)
cosec(90° – 2A) = cosec(A – 27°) …[∵ sec θ = cosec (90° – θ)
90° – 2A = A – 27°
90° + 27° = 2A + A
⇒ 3A = 117°
∴ ∠A =117/3= 39°

Question-4
If sin θ – cos θ = 0, find the value of sin⁴θ + cos⁴θ. (2012, 2017)

1/2

Question Explanation:

If sin θ – cos θ = 0, find the value of sin⁴θ + cos⁴θ. (2012, 2017)

Question 3

Question Explanation:

Hint

Solution:

Question-5
If sec θ + tan θ = 7, then evaluate
sec θ – tan θ.
(2017OD)

1/7

Question Explanation:

Question-5
If sec θ + tan θ = 7, then evaluate
sec θ – tan θ.
(2017OD)

Solution:
We know that,
sec²θ – tan²θ = 1
(sec θ + tan θ) (sec θ – tan θ) = 1
(7) (sec θ – tan θ) = 1 …[sec θ + tan θ = 7; (Given)
∴ sec θ – tan θ =1/7

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