Exercise 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=169−144=25

QR=25−−√=5

tan(P) = oppositesideadjacentside=QRPQ=512

cot(P) = adjacentsideoppositeside = PQQR = 512

tan(P) – cot(R) = 512−512=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

16k2−9k2=AB2

AB2=7k2

AB=7–√k

cos(A) = AB/AC = 7–√k/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)2−144(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 CD⊥AB .

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=AC2−AD2 ….(iv)

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

AC2−AD2=BC2−BD2

AC2−AD2=BC2−BD2

k2(BC2−BD2)=BC2−BD2

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=113−−−√k

sin Ѳ = AB/AC = 8k/113−−−√k=8/113−−−√

and cos Ѳ = BC/AC = 7k/113−−−√k=7/113−−−√

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

= 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 (1−tan2A)/(1+tan2A)=cos2A−sin2A 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. = (1−tan2A)(1+tan2A)=1−(3/4)2/1+(3/4)2=(1−9/16)/(1+9/16)=(16−9)/(16+9)=7/25

R.H.S. =cos2A−sin2A=(4/5)2−(3/4)2=(16/25)−(9/25)=7/25

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

Hence, (1−tan2A)/(1+tan2A)=cos2A−sin2A

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=(3k−−√2))+K2

EG2=3k2+K2

EG2=4k2

EG=2K

sinE = FG/EG = 1/2

cosE = EF/EG =  3√2  ,
sin G = EF/EG = 3√2 cosE = FG/EG = 1/2
(i) sin EcosG + cosE sin G = (1/2\ast1/2) + (3√2∗3√2)= 1/4+3/4 = 4/4 = 1
(ii) cosEcosG – sin E sin C = (3√2∗12)−(3√2∗12)= (3√4)−(3√4)= 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+(25−x)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.

Exercise 8.3

1) Calculate:

                (i) sin18∘cos72∘

                (ii) tan26∘cot64∘

                (iii) cos 48° – sin 42°

                (iv) cosec 31° – sec 59°