Introduction to Trigonometry-Exercise 8.1-Class 10

1. In ∆ABC right angled at B, AB = 24 cm, BC = 7 m. Determine

(i) sin A, cos A

(ii) sin C, cos C 

Solution:

(i)

Applying Pythagoras theorem for ∆ABC,

we obtain AC2 = AB2 + BC2

= (24 cm)2 + (7 cm)2

= (576 + 49) cm2

= 625 cm2

∴ AC = √(625)cm = 25 cm

3

We know that,

sinA = Opposite side/hypotenuse

 

sinA = BC/AC  = 7/25 

cosA =  Adjacent side/hypotenuse

 

cosA = AB/AC  = 24/25 

 

(ii)

4.png

sinC = Opposite side/hypotenuse

24/25

cosC =  Adjacent side/hypotenuse

7/25


2. In the given figure find tan P − cot R 

5.png

Solution:

Applying Pythagoras theorem for ∆PQR,

we obtain PR2 = PQ2 + QR2

(13 cm)2 = (12 cm)2 + QR2

169 cm2 = 144 cm2 + QR2

25 cm2 = QR2

QR = 5 cm

6.png

tanP = opposite side/Adjacent side

tanP = 5/12

cotR =  adjacent side/Opposite side

cotR =  5/12

tan P − cot R = 5/125/12 = 0


3. If sin A = ¾ , calculate cos A and tan A. 

Solution:

Let ∆ABC be a right-angled triangle, right-angled at point B.

7.png

Given that, Let BC be 3k.

sinA = 3/4

cosA =  3/4

Therefore, AC will be 4k, where k is a positive integer.

Applying Pythagoras theorem in ∆ABC,

we obtain AC2 = AB2 + BC2

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

16k2 − 9k2 = AB2

7k2 = AB2

AB = √7k

8.png


 

 

4. Given 15 cot A = 8. Find sin A and sec A 

Solution:

Consider a right-angled triangle, right-angled at B.

9

cotR =  adjacent side/Opposite side

cotR =  AB/BC 

It is given that,

cotA = 8/15

Let AB be 8k.

Therefore, BC will be 15k, where k is a positive integer.

Applying Pythagoras theorem in ∆ABC,

we obtain AC2 = AB2 + BC2

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

= 64k2 + 225k2

= 289k2

AC = 17k

SinA = (opposite side)/(Hypotenuse side) = BC/AC = 15k/17k = 15/17

secA = hypotenuse/(adjacent side) = AC/AB = 15k/8k = 15/8


5. Given sec θ = 13/12 , calculate all other trigonometric ratios

Solution:

Consider a right-angle triangle ∆ABC, right-angled at point B.

10

secθ = hypotenuse/(adjacent side) = AC/AB = 13/12

If AC is 13k, AB will be 12k, where k is a positive integer.

Applying Pythagoras theorem in ∆ABC, we obtain

(AC)2 = (AB)2 + (BC)2

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

169k2 = 144k2 + BC2

25k2 = BC2

BC = 5k

sinθ = opposie side to θ)/hypotenuse = BC/AC = 5k/13k = 5/13

cosθ = (adjacent side to θ)/hypotenuse = AB/AC = 12k/13k = 12/13

tanθ = opposite side tp θ/adjacent side to θ = BC/AB = 5k/12k = 5/12

cotθ = adjacent side to θ/opposite side to θ = AB/BC = 12k/5k = 12/5

cosecθ = hypotenuse/adjacent side to θ = AC/BC = 13k/12k = 13/12


6. If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B. 

Solution:

Let us consider a triangle ABC in which CD ⊥ AB.

11.png

It is given that ,

cos A = cos B

AD/AC = BD/BC ————(1)

We have to prove ∠A = ∠B. To prove this, let us extend AC to P such that BC = CP.

12.png

From equation (1), we obtain

AD/BD = AC/BC

⇒AD/BD = AC/CP [By construction we have BC = CP]————(2)

By using the converse of B.P.T, CD||BP

⇒ ∠ACD = ∠CPB (Corresponding angles)———–(3)

∠BCD = ∠CBP (Alternate interior angles)———-(4)

By construction, we have BC = CP.

∴ ∠CBP = ∠CPB (Angle opposite to equal sides of a triangle)——(5)

From equations (3), (4), and (5), we obtain

∠ACD = ∠BCD————- (6)

In ∆CAD and ∆CBD, ∠ACD = ∠BCD [Using equation (6)]

∠CDA = ∠CDB [Both 90°]

Therefore, the remaining angles should be equal.

∴ ∠CAD = ∠CBD

⇒∠A = ∠B


7. If cot θ = 7/8 , evaluate

(i) [(1+sinθ)(1-sinθ)]/[(1+cosθ)(1-cosθ)]

(ii) cot²θ

Solution:

Let us consider a right triangle ABC, right-angled at point B.

13.png

cotθ = adjacent side to θ/opposite side to θ  = BC/AB = 7/8

If BC is 7k, then AB will be 8k, where k is a positive integer.

Applying Pythagoras theorem in ∆ABC, we obtain

AC2 = AB2 + BC2

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

= 64k2 + 49k2

= 113k2

AC = √113k

sinθ = opposie side to θ)/hypotenuse = AB/AC = 8k/√113k = 8/√113

cosθ = (adjacent side to θ)/hypotenuse = BC/AC = 7k/√113k = 7/√113

(i) [(1+sinθ)(1-sinθ)]/[(1+cosθ)(1-cosθ)] = (1-sin²θ)/(1-cos²θ)

= [1 – (8/√113)²]/[1 – (7/√113)²]

= [1 – (64/113)]/[1 – (49/113)]

= [(49/113)]/[(64/113)]

= 49/64

(ii)cot²θ = cotθ² = (7/8)² = 49/64


8. If 3 cot A = 4, Check whether [(1-tan²A)/(1+tan²A)] = cos²A – sin²A or not

Solution:

It is given that 3cot A = 4 Or, cot A = 4/3

Consider a right triangle ABC, right-angled at point B.

14.png

cotA = adjacent side to A/opposite side to A = AB/BC = 4/3

If AB is 4k, then BC will be 3k, where k is a positive integer.

In ∆ABC, (AC)2 = (AB)2 + (BC)2

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

= 16k2 + 9k2

= 25k2

AC = 5k

sinA = opposie side to A)/hypotenuse = BC/AC = 3k/5k = 3/5

cosA = (adjacent side to A)/hypotenuse = AB/AC = 4k/5k = 4/5

tanθ = opposite side tp θ/adjacent side to θ = BC/AB = 3k/4k = 3/4

To check [(1-tan²A)/(1+tan²A)]  = cos²A – sin²A

LHS, [(1-tan²A)/(1+tan²A)] = [1-(3/4)²/(1+3/4)²] 

= [1-(9/16)]/[1+(9/16)]

= [(7/16)]/[(25/16)]

= 7/25,LHS

RHS, cos²A – sin²A

= (3/5)² – (4/5)²

= (9/25) – (16/25)

= [9-16]/25

= 7/25,RHS

Thus, LHS = RHS

Hence, [(1-tan²A)/(1+tan²A)]  = cos²A – sin²A


9. In ∆ABC, right angled at B. If tanA = 1/√3 , find the value of

(i) sin A cos C + cos A sin C

(ii) cos A cos C − sin A sin C  

Solution:

15

Given tanA = 1/√3

i.e., BC/AB = 1/√3

If BC is k, then AB will be , where k is a positive integer.

In ∆ABC, AC2 = AB2 + BC2

AC2 = (√3k)2 + k2

AC2 = 3k2 + k2 = 4k2

AC = √(4k2) = 2k

sinA= opposie side to A)/hypotenuse = BC/AC = k/2k = 1/2

cosA = (adjacent side to A)/hypotenuse = AB/AC = √3k/2k = √3/2

sinC= opposie side to C/hypotenuse = AB/AC = √3k/2k = √3/2

cosC = (adjacent side to C)/hypotenuse = BC/AC = k/2k = 1/2

To find, (i) sin A cos C + cos A sin C

=   1/2 x 1/2 + √3/2 x √3/2

= 1/4 + 3/4

4/4

=  1

to find (ii) cos A cos C − sin A sin C  

=   √3/2 x 1/2 – 1/2 x √3/2

=   √3/4 – √3/4

= 0


10. In ∆PQR, right angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P. 

Solution:

Given that, PR + QR = 25

PQ = 5

Let PR be x.

Therefore, QR = 25 − x

16.png

Applying Pythagoras theorem in ∆PQR, we obtain

PR2 = PQ2 + QR2

x2 = (5)2 + (25 − x)2

x2 = 25 + 625 + x2 − 50x

50x = 650

x = 13 Therefore, PR = 13 cm , QR = (25 − 13) cm = 12 cm

sinP = opposie side to P)/hypotenuse = RQ/PR = 12/13

cosP = (adjacent side to P)/hypotenuse = PQ/PR = 5/13

tanP = opposite side tp P/adjacent side to P = RQ/PQ = 12/5


11. State whether the following are true or false. Justify your answer.

(i) The value of tan A is always less than 1.

(ii) sec A = 12/5 for some value of angle A.

(iii) cos A is the abbreviation used for the cosec of angle A.

(iv) cot A is the product of cot and A (v) sin θ = 4/3 , for some angle θ 

Solution:

(i) Consider a ∆ABC, right-angled at B.

17.png

tanA = opposite side tp A/adjacent side to A = 12/5

But 12/5 > 1

Therefore, tanA > 1

So, tan A < 1 is not always true. Hence, the given statement is false

(ii) secA = 12/5

17

sec A = AC/AB = 12/5

Let AC be 12k, AB will be 5k, where k is a positive integer.

Applying Pythagoras theorem in ∆ABC, we obtain

AC2 = AB2 + BC2

(12k)2 = (5k)2 + BC2

144k2 = 25k2 + BC2

BC2 = 119k2

BC = 10.9k

It can be observed that for given two sides AC = 12k and AB = 5k,

BC should be such that,

AC − AB < BC < AC + AB

12k − 5k < BC < 12k + 5k

7k < BC < 17 k

However, BC = 10.9k.

Clearly, such a triangle is possible and hence, such value of sec A is possible. Therefore, the given statement is true.

(iii) Abbreviation used for cosecant of angle A is cosec A. And cos A is the abbreviation used for cosine of angle A. Hence, the given statement is false.

(iv) cot A is not the product of cot and A. It is the cotangent of ∠A. Hence, the given statement is false.

(v) sin θ = 4/3

We know that in a right-angled triangle,

sinθ = opposie side to θ)/hypotenuse

In a right-angled triangle, hypotenuse is always greater than the remaining two sides. Therefore, such value of sin θ is not possible. Hence, the given statement is false.


 

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