Telangana SCERT Solution Class X (10) Maths Chapter 11 Trigonometry Exercise 11.1
Exercise 11.1
1> In right angled triangle ABC, 8cm, 15cm and 17cm are lengths of ABC, BC and CA respectively. Then find SinA, CosA and tanA
Solution:
In Right triangle ABC
AB = 8cm
BC = 15 cm and
CA = 17cm
SinA = Opposite side/hypotenuse
SinA = 15/17
CosA = Opposite side/hypotenuse
= AB/CA
CosA = 8/17
tanA = Opposite side/Adjecent Side
= BC/AB
tanA = 15/8
2> The sides of a right angled triangle PQR are PQ = 7cm, PR = 25cm and ∠ Q = 900 respectively, then find tanP – tanR
In right Triangle= ∠Q = 900
PQ = 7cm
PR = 25cm
According to Pythagoras Theorem
PQ2 + QR2 = PR2
72 + QR2 = 252
49 + QR2 = 625
QR2 = 625 – 49
QR 2 = 576
QR = √ 576
QR = 24cm
tanP = Opposite Side/Adjecent Side
= QR/PQ
tanp = 24/7
tanR = Opposite side/adjacent Side
= PQ /QR
tanR = 7/24
Therefore, tanP – tanR = 24/7 – 7/24
= 576-49/7×24
Tanp – tanR = 527/168
3> In a right angled triangle ABC with right angle at , in which a = 24 units, b = 25 units and ∠ BAC = 0 Then find cos and tan
Solution:
In Right triangle ABC ∠ B = 900
a = BC = 24Units
b= CA = 25 Units
According to Pythagoras Theorem
AB2 + BC2 = AC2
AB2 + 242 = 252
AB2 + 576 = 625
AB2 = 625 – 576
AB2 = 49
AB = √49
AB= 7
COSθ = Adjecent side/Hypotenus
= AB/CA
COSθ = 7/25
Tanθ = Opposide side/Adjecent side
= BC/AB
Tanθ = 24/7
Q4> If cosA = 12/3, then find SinA and tanA (A<90o)
= Solution :
CoSA = 12/13
CosA = 12/13
CosA = Adj. Side/Hypotenus
= AB/AC
CosA = 12/13
Let Adjecnt side = 12m then Hypotenuse 13m. (Where, m = any positive number)
From the right triangle ABC, adjacent Side = AB = 12m, Hypotenuse = AC =13m
According to Pythagoras thorem,
AB2 + BC2 = CA2
(12m)2 + BC2 = (13)2
144m2 + BC2 = 169m2
BC2 = 169m2 – 144m2
BC2 = 25m2
BC = ∠ 25m2
BC = 5m
SinA = Opposite Side/Hypotenus
SinA = BC/AC = 5m/13m = 5/13
TanA = opposite Side/Adjecent Side
= BC/AB
= 5m/12m
= TanA = 5/12
5> If, 3tanA = 4, then find SinA and CosA
Solution:
Given 3 tanA =4
tanA = 4/3
TanA = opposide side / Adjecent Side
tanA = 4/3
Let Opposide Side = 4m then adj. side = 3m
( m = any positive number)
From right triangle ABC, Opposite Side = BC = 4m, Adjecent Side = AB = 3m
According to Pythogoras Theorem,
AB2 + BC2 = AC2
(3m)2 + (4m)2 = AC2
9m2 + 16m2 = AC2
25m2 = AC2
AC = 5m
SinA = opposite side/ Hypotenus
= BC/AC
= 4m/5m
SinA = 4/5
CoS = Adjecent Side / Hypotenus
= AB/AC
= 3m/5m
CosA = 3/5
6> In △ ABC and △ XYZ, if ∠A and ∠X are a cute angles such that CosA = CosX then show that ∠ A = ∠ X
Solution:
CosA = CosX (Therefore, Given)
We have,
CosA = AB/Ac and COSx = XY/XZ
AB/AC = XY/XZ
Let AB/Ac = XY/XZ
AB = mac , XY = mxz
AB/XY = AC/XZ
Using Pythagoras Theorem,
AB2 + BC2 = AC2
BC = √AC2 – AB2
BC/YZ = √ Ac2 – AB2 / √ XZ2 – XY2
= √ Ac2 –m2AC2 / √ XZ2 – m2XY2
= √ Ac2 – (1-m2)/ XZ2 (1-m2)
= √ AC2 / XZ2
BC/YZ = AC/XZ
From Equations (3) & (4),
AB/XY = BC/YZ = AC/XZ
7> Given CotΘ = 7/8 Then evaluate
i> (1+Sinq) (1-sinq)/(1+Cosq)(1-cosq)
ii> (1+sinq)/Cosq
Solution:
Cotθ = 7/8
Cot = Adj. Side/ Oppo. side
cotθ = 7/8
In Right triangle ABC, Let Adj side = AB = 7m then Oppo. Side = bc = 8m
Using Pythagoras theorem
AB2 + BC2 = AC2
(7m)2 + (8m2) = AC2
49m2 + 64m2 = AC2
113m2 = AC2
AC = √ 1132
AC = √ 113
Sin = oppo. side / Hypotenus
= BC / AC
SinΘ = 8m/ √ 113m
CosΘ = Adj. Side/hypotenuse
= AB/AC
= 7m/√ 113m
CosΘ = 7m/√ 113m
i> (1+ sinΘ) (1-SinΘ)/ (1+CosΘ) (1-CosΘ)
= (1+8/√113) (1-8/√113) / (1+7/√113)(1-7√113)
(a+b)(a-b) = (a2-b2)
= 1-64/113/1-49/113
= 113-64/113 /113-49/113
= 49/64
8> In a right Angled triangle ABC, right angle is at B. If tanA = √3, then find the value of
i> SinA Cosc + CosA SinC
ii> CosA Cosc – SinA SinC
Solution:
tanA = √3
tanA = Opp side / Adj side = √3/1
In Right angle Triangle ABC, Let Opp. side = BC = √3m
Using Pythagoras theorem,
AC2 = AB2 + BC2
AC2 = (√3)2 + m2
= 3m2 + m2
= 4m2
Therefore, AC = √4m2
AC = 2m
SinA = opp.side/hypotenus = BC/AC
= √33/2m
SinA = √3/2
CosA = Adj. Side/hypotenuse
= AB/AC
= m/2m
CosA = 1/2
SinC = Opp. Side/Hyptenus
Sin Θ = AB/AC
= 3/2m
SinΘ = ½
CosΘ = Adj side/hypotenus
= BC/AC
= √3m/2m
CosΘ = √3/2
SinA CosC + CosA SinC = √3/2 x √3/2 + ½ x ½
= ¾ + ¼
= 3+1/4
= 4/4
= 1
CosA cosC – SinA SinC = ½ x √2/2 – √3/2 x ½
= √3/4 – √3/4
= 0
Here is your solution of Telangana SCERT Class 10 Maths Solution Chapter 11 Trigonometry Exercise 11.1
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