Introduction of Trigonometry Important Questions for class 10 Chapter 8 (ex 8.1,ex8.2)
Trigonometric Ratio त्रिकोणमितीय अनुपात Sin A Cos A & TanA ratioSin A=Sin 600
CosA = Cos 60
Tan A = Tan 60
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।
Pythagoras Theorem (ex 8.3)
एक नजर ExampleEx 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।
tan A = BC/AB =4/3
4K/3K
Pythagoras Theorem
AC2 = AB2 + BC2
AC2 = (4K)2 + (3K)2
AC2 =
4X4XKXK + 3X3XKXK
= 16K2 + 9K2
AC2=25K2
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-1Let 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
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, in which AB = 29 units, BC = 21 units and ∠ ABC = θ Determine 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) cos2 θ + sin2 θ,
(ii) cos2 θ – sin2 θ.
Step-1 Step-2 Step-3 Step-4EX 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।
AC=?
As per Pythogoras Theorem
AB2 + BC2 = AC2
Since AB = BC = 1
(1)2 + (1)2 = 2 = AC2
AC2 = 2
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।
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
5 Questions & Hints
Question-1
If tan θ + cot θ = 5, find the value of tan2θ + cot2θ. (2012)
Hint-
Question 2.
If sec 2A = cosec (A – 27°) where 2A is an acute angle, find the measure of ∠A. (2012, 2017D)
Hint
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)
23
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)
39
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-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
Next 5 Questions & Hint
Question 7.
If cosec θ = 5/4, find the value of cot θ.
(2014)
Question 8
If θ = 450, then what is the value of
2 sec2450+ 3 Cosec2450
(2014)
Question 10.
Evaluate: sin219° + sin271°.
(2015)
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