Ncert Solutions For Class 10 Maths Ex 8.1

Ncert Solutions For Class 10 Maths Chapter 8 Ex 8.1


Q1) In ABC , 90 at B, AB=24cm, BC = 7cm.

Determine:

(i)sin(A), cos(A)

(ii) sin(C), cos(C)

Ans.) In ABC , B=90

By Applying Pythagoras theorem, we get

AC2=AB2+BC2

(24)2+72 =(576+49)

AC2 = 625cm2

à AC = 25cm

(i) sin(A) = BC/AC = 7/25

Cos(A) = AB/AC = 24/25

(ii) sin(C) = AB/AC =24/25

cos(C) = BC/AC = 7/25

 

Q2) In the given figure find tan(P) – cot(R)

Ans.) PR = 13cm,PQ = 12cm and QR = 5cm

According to Pythagorean theorem,

132=QR2+122 169=QR2+144 QR2=169144=25 QR=25=5

tan(P) = oppositesideadjacentside=QRPQ=512

cot(P) = adjacentsideoppositeside = PQQR = 512

tan(P) – cot(R) = 512512=0

Therefore ,tan(P) – cot(R) = 0

 

Q3) If sin(A) = 3/4, calculate cos(A) and tan(A)

Ans.) Let ABC , be a right-angled triangle, right-angled at B.

We know that sin(A) = BC/AC = 3/4

Let BC be 3k and AC will be 4k where k is a positive real number.

By Pythagoras theorem we get,

AC2=AB2+BC2

 

(4k)2=AB2+(3k)2

 

16k29k2=AB2

 

AB2=7k2

 

AB=7k

 

cos(A) = AB/AC = 7k/4k=7/4

tan(A) = BC/AB =3k/7=3/7

 

Q4) In question given below 15cot(A) = 8 ,find sin A and sec A.

Ans.)  Let ABC be a right angled triangle, right-angled at B.

We know that cot(A) = AB/BC = 8/15

Given

Let AB side be 8k and BC side 15k

Where k is positive real number

By Pythagoras theorem we get,

AC2=AB2+BC2

 

AC2=(8k)2+(15k)2

 

AC2=64k2+225k2

 

AC2=289k2

AC = 17k

sin(A) = BC/AC = 15k/17k = 15/7

sec(A) =AC/AB =17k/8k = 17/8

 

Q5) Given sec Ѳ =13/12, calculate all other trigonometric ratios.

Ans.) Let  ABC be right-angled triangle, right-angled at B.

We know that sec Ѳ =OP/OM =13/12(Given)

Let side OP be 13k and side OM will be 12k where k is positive real number.

By Pythagoras theorem we get,

OP2=OM2+MP2

 

(13k)2=(12k)2+MP2

 

169(k)2144(k)2=MP2

 

MP2=25k2

MP = 5

Now,

sin Ѳ = MP/OP = 5k/13k =5/13

cos Ѳ = OM/OP = 12k/13k = 12/13

tan Ѳ = MP/OM = 5k/12k = 5/12

cot Ѳ = OM/MP = 12k/5k = 12/5

cosec Ѳ = OP/MP = 13k/5k = 13/5

 

Q6) If A and B are acute angles such that

 cos(A) = cos(B), then show A =B .

Ans.) Let  ABC in which CDAB .

A/q,

cos(A) = cos(B)

à AD/AC = BD/BC

à AD/BD = AC/BC

Let  AD/BD =AC/BC =k

AD =kBD …. (i)

AC=kBC  …. (ii)

By applying Pythagoras theorem in CAD and CBD we get,

CD2=AC2AD2 ….(iv)

From the equations (iii) and (iv) we get,

AC2AD2=BC2BD2 AC2AD2=BC2BD2 k2(BC2BD2)=BC2BD2 k2=1

Putting this value in equation (ii) , we obtain

AC = BC

A=B (Angles opposite to equal side are equal-isosceles triangle)

 

Q7) If  cot Ѳ = 7/8, evaluate :

(i) (1+sin Ѳ)(1-sin Ѳ) / (1+cos Ѳ)(1-cos Ѳ)

(ii) cot2Θ

Ans.) Let ABC in which  B=90

and C=Θ

A/q,

cot Ѳ =BC/AB = 7/8

Let BC = 7k and AB = 8k, where k is a positive real number

According to Pythagoras theorem in ABC we get.

 

AC2=AB2+BC2

 

AC2=(8k)2+(7k)2

 

AC2=64k2+49k2

 

AC2=113k2

 

AC=113k

 

sin Ѳ = AB/AC = 8k/113k=8/113

and cos Ѳ = BC/AC = 7k/113k=7/113

 

(i) (1+sin Ѳ)(1-sinѲ)/(1+cosѲ)(1-cos Ѳ) = (1sin2Θ)/(1cos2Θ)

= 1(8/113)2/1(7/113)2

= {1-(64/113)}/{1-(49/113)} = {(113-64)/113}/{(113-49)/113} = 49/64

 

(ii) cot2Θ=(7/8)2=49/64

 

Q8) If 3cot(A) = 4/3, check whether (1tan2A)/(1+tan2A)=cos2Asin2A or not.

Ans.) Let ABC in which B=90

A/q,

cot(A) = AB/BC = 4/3

Let AB = 4k an BC =3k, where k is a positive real number.

AC2=AB2+BC2

 

AC2=(4k)2+(3k)2

 

AC2=16k2+9k2

 

AC2=25k2

 

AC=5k

 

tan(A) = BC/AB = 3/4

sin(A) = BC/AC = 3/5

cos(A) = AB/AC = 4/5

L.H.S. = (1tan2A)(1+tan2A)=1(3/4)2/1+(3/4)2=(19/16)/(1+9/16)=(169)/(16+9)=7/25

R.H.S. =cos2Asin2A=(4/5)2(3/4)2=(16/25)(9/25)=7/25

R.H.S. =L.H.S.

Hence, (1tan2A)/(1+tan2A)=cos2Asin2A

 

Q9) In triangle EFG, right-angled at F, if tan E =1/√3 find the value of:
(i) sin EcosG + cosE sin G
(ii) cosEcosG – sin E sin G

Answer

LetΔEFG in which F=90, E/q

tanE=FCEF tanE=FCEF=13

Where k is the positive real number of the problem

By Pythagoras theorem in ΔEFG we get:

EG2=EF2+FG2 EG2=(3k2))+K2 EG2=3k2+K2 EG2=4k2 EG=2K

 

sinE = FG/EG = 1/2

cosE = EF/EG =  32  ,
sin G = EF/EG = 32 cosE = FG/EG = 1/2
(i) sin EcosG + cosE sin G = (1/2\ast1/2) + (3232)= 1/4+3/4 = 4/4 = 1
(ii) cosEcosG – sin E sin C = (3212)(3212)= (34)(34)= 0

 

Q10)In Δ MNO, right-angled at N, MO + NO = 25 cm and MN = 5 cm. Determine the values of sin M, cos M and tan M.

Answer

Given that, MO + NO = 25 , MN = 5
Let MO be x.  ∴ NO = 25 – x

By Pythagoras theorem ,
MO2=MN2+NO2
X2=52+(25x)2
50x = 650
x = 13
∴ MO = 13 cm
NO = (25 – 13) cm = 12 cm

sinM = NO/MO = 12/13

cosM = MN/MO = 5/13

tanM = NO/MN = 12/5

 

Q11)  State whether the following are true or false. Justify your answer.
(i) The value of tan M is always less than 1.
(ii) secM = 12/5 for some value of angle M.
(iii) cosM is the abbreviation used for the cosecant of angle M.
(iv) cot M is the product of cot and M.
(v) sin θ = 4/3 for some angle θ.

Answer

(i) False.

In ΔMNC in which N = 90,

MN = 3, NC = 4 and MC = 5

Value of tan M = 4/3 which is greater than.

The triangle can be formed with sides equal to 3, 4 and hypotenuse = 5 as

it will follow the Pythagoras theorem.

MC2=MN2+NC2
52=32+42
25 = 9 + 16
25 = 25

(ii) True.
Let a ΔMNC in which ∠N = 90º,MC be 12k and MB be 5k, where k is a positive real number.
By Pythagoras theorem we get,
MC2=MN2+NC2
(12k)2=(5k)2+NC2
NC2+25k2=144K2
NC2=119k2

Such a triangle is possible as it will follow the Pythagoras theorem.
(iii) False.

Abbreviation used for cosecant of angle M is cosec M.cosM is the abbreviation used for cosine of angle M.

(iv) False.

cotM is not the product of cot and M. It is the cotangent of M.
(v) False.

sinΘ = Height/Hypotenuse

We know that in a right angled triangle, Hypotenuse is the longest side.

∴ sinΘwill always less than 1 and it can never be 4/3 for any value of Θ.

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