CBSE Class X

Question 3:

If tan 2A = cot (A− 18°), where 2A is an acute angle, find the value of A.

Answer:

Given that,

tan 2A = cot (A− 18°)

cot (90° − 2A) = cot (A −18°)

90° − 2A = A− 18°

108° = 3A

A = 36°

Question 2:

Show that

(I) tan 48° tan 23° tan 42° tan 67° = 1

(II)cos 38° cos 52° − sin 38° sin 52° = 0

Answer:

(I) tan 48° tan 23° tan 42° tan 67°

= tan (90° − 42°) tan (90° − 67°) tan 42° tan 67°

= cot 42° cot 67° tan 42° tan 67°

= (cot 42° tan 42°) (cot 67° tan 67°)

= (1) (1)

= 1

(II) cos 38° cos 52° − sin 38° sin 52°

= cos (90° − 52°) cos (90°−38°) − sin 38° sin 52°

= sin 52° sin 38° − sin 38° sin 52°

= 0

Question 1:

Evaluate

Answer:

(III)cos 48° − sin 42° = cos (90°− 42°) − sin 42°

= sin 42° − sin 42°

= 0

(IV) cosec 31° − sec 59° = cosec (90° − 59°) − sec 59°

= sec 59° − sec 59°

= 0

Question 4:

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

(i) sin (A + B) = sin A + sin B

(ii) The value of sinθ increases as θ increases

(iii) The value of cos θ increases as θ increases

(iv) sinθ = cos θ for all values of θ

(v) cot A is not defined for A = 0°

Answer:

(i) sin (A + B) = sin A + sin B

Let A = 30° and B = 60°

sin (A + B) = sin (30° + 60°)

= sin 90°

= 1

sin A + sin B = sin 30° + sin 60°

Clearly, sin (A + B) ≠ sin A + sin B

Hence, the given statement is false.

(ii) The value of sin θ increases as θ increases in the interval of 0° < θ < 90° as

sin 0° = 0

sin 30º = 1/2 = 0.5

sin 45º = 1/√2 = 0.707

sin 60° = √3/2 = 0.866

sin 90° = 1

Hence, the given statement is true.

(iii) cos 0° = 1

cos 30° = √3/2 = 0.866

cos 45° = 1/√2 = 0.707

cos 60° = 1/2 = 0.5

cos90° = 0

It can be observed that the value of cos θ does not increase in the interval of 0° < θ < 90°.

Hence, the given statement is false.

(iv) sin θ = cos θ for all values of θ.

This is true when θ = 45°

As sin 45° = 1/√2

cos 45° = 1/√2

It is not true for all other values of θ.

As sin 30° = 1/2 and cos 30° = √3/2,

Hence, the given statement is false.

(v) cot A is not defined for A = 0°

As cot A = cos A/sin A,

cot 0° = cos 0°/sin 0° = 1/0 = undefined

Hence, the given statement is true.

Question 3:

If tan (A + B) = √3 and tan (A – B) = 1/√3 ; 0° < A + B ≤ 90°; A > B, find A and B.

Answer:

tan (A + B) = √3

⇒ tan (A + B) = tan 60

⇒ A + B = 60 …..(1)

tan (A – B) = 1/√3

⇒ tan (A – B) = tan 30

⇒ A – B = 30 ….. (2)

On adding both equations, we obtain

2A = 90

⇒ A = 45

From equation (1), we obtain

45 + B = 60

B = 15

Therefore, ∠A = 45° and ∠B = 15°

Question 2:

Choose the correct option and justify your choice.

(A). sin60°       (B). cos60°      (C). tan60°         (D). sin30°

(A). tan90°      (B). 1                  (C). sin45°           (D). 0

(iii) sin2A = 2sinA is true when A =

(A). 0°              (B). 30°              (C). 45°                 (D). 60°

(A). cos60°      (B). sin60°        (C). tan60°          (D). sin30°

Answer:

Out of the given alternatives, only sin 60º = √3/2

Hence, (A) is correct.

Hence, (D) is correct.

(iii)Out of the given alternatives, only A = 0° is correct.

As sin 2A = sin 0° = 0

2 sinA = 2sin 0° = 2(0) = 0

Hence, (A) is correct.

Out of the given alternatives, only tan 60° = √3

Hence, (C) is correct.

Question 1:

Evaluate the following

(i) sin60° cos30° + sin30° cos 60°

(ii) 2tan²45° + cos²30° − sin²60°

Answer:

(i) sin60° cos30° + sin30° cos 60°

 

Question 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 cosecant of angle A.

(iv) cot A is the product of cot and A

(v) sin θ = 4/3 , for some angle θ

Answer:

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

But 12/5 > 1

∴tan A > 1

So, tan A < 1 is not always true.

Hence, the given statement is false.

(ii) sec A = 12/5

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

Applying Pythagoras theorem in ΔABC, 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.

Hence, 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,

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.

Question 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.

Answer:

Given that, PR + QR = 25

PQ = 5

Let PR be x.

Therefore, QR = 25 − x

 

Applying Pythagoras theorem in ΔPQR, 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

Question 9:

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

(ii) cos A cos C − sin A sin C

Answer:

 

tan A = 1/√3

BC/AB = 1/√3

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

In ΔABC,

AC² = AB² + BC²

= (√3 k)² + (k)²

= 3k² + k² = 4k²

∴ AC = 2k

 

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

 

Question 8:

If 3 cot A = 4, Check whether

Answer:

It is given that 3cot A = 4

Or, cot A = 4/3

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

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

In ΔABC,

(AC)² = (AB)² + (BC)²

= (4k)² + (3k)²

= 16k² + 9k²

= 25k²

AC = 5k

Question 7:

If cot θ = 7/8, evaluate

Answer:

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

 

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

Applying Pythagoras theorem in ΔABC, we obtain

AC² = AB² + BC²

= (8k)² + (7k)²

= 64k² + 49k²

= 113k²

AC = √113 k

Question 6:

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

Answer:

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

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.

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)

And, ∠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 ΔCAD and ΔCBD,

∠ACD = ∠BCD [Using equation (6)]

∠CDA = ∠CDB [Both 90°]

Therefore, the remaining angles should be equal.

∴∠CAD = ∠CBD

⇒ ∠A = ∠B

Alternatively,

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

It is given that,

cos A = cos B

⇒ AD = k BD … (1)

And, AC = k BC … (2)

Using Pythagoras theorem for triangles CAD and CBD, we obtain

CD² = AC² − AD² … (3)

And, CD² = BC² − BD² … (4)

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

AC² − AD² = BC² − BD²

⇒ (k BC)² − (k BD)² = BC² − BD²

⇒ k² (BC² − BD²) = BC² − BD²

⇒ k² = 1

⇒ k = 1

Putting this value in equation (2), we obtain

AC = BC

⇒ ∠A = ∠B(Angles opposite to equal sides of a triangle)

Question 5:

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

Answer:

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

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

Applying Pythagoras theorem in ΔABC, we obtain

(AC)² = (AB)² + (BC)²

(13k)² = (12k)² + (BC)²

169k² = 144k² + BC²

25k² = BC²

BC = 5k

 

Question 4:

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

Answer:

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

Let AB be 8k.Therefore, BC will be 15k, where k is a positive integer.

Applying Pythagoras theorem in ΔABC, we obtain

AC² = AB² + BC²

= (8k)² + (15k)²

= 64k² + 225k²

= 289k²

AC = 17k

Question 3:

If sin A = 3/4, calculate cos A and tan A.

Answer:

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

Given that,

sin A= 3/4

BC/AC = 3/4

Let BC be 3k. Therefore, AC will be 4k, where k is a positive integer.

Applying Pythagoras theorem in ΔABC, we obtain

AC² = AB² + BC²

(4k)² = AB² + (3k)²

16k ² − 9k ² = AB²

7k ² = AB²

AB = √7 k

 

Question 2:

In the given figure find tan P − cot R

Answer:

Applying Pythagoras theorem for ΔPQR, we obtain

PR² = PQ² + QR²

(13 cm)² = (12 cm)² + QR²

169 cm² = 144 cm² + QR²

25 cm² = QR²

QR = 5 cm

 

Question 1:

In Δ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine :

(i) sin A, cos A     (ii) sin C, cos C

Answer:

Applying Pythagoras theorem for ΔABC, we obtain

AC² = AB² + BC²

= (24 cm)² + (7 cm)²

= (576 + 49) cm²

= 625 cm²

∴ AC = √625 cm = 25 cm

Question 8:

ABCD is a rectangle formed by the points A (− 1, − 1), B (− 1, 4), C (5, 4) and D (5, − 1). P, Q, R and S are the mid-points of AB, BC, CD, and DA respectively. Is the quadrilateral PQRS is a square? a rectangle? or a rhombus? Justify your answer.

Answer:

It can be observed that all sides of the given quadrilateral are of the same measure.

However, the diagonals are of different lengths. Therefore, PQRS is a rhombus. 

Question 7:

Let A (4, 2), B (6, 5) and C (1, 4) be the vertices of ΔABC.

(i) The median from A meets BC at D. Find the coordinates of point D.

(ii) Find the coordinates of the point P on AD such that AP: PD = 2:1

(iii) Find the coordinates of point Q and R on medians BE and CF respectively such that BQ: QE = 2:1 and CR: RF = 2:1.

(iv) What do you observe?

(v) If A(x₁, y₁), B(x₂, y₂), and C(x₃, y₃) are the vertices of ΔABC, find the coordinates of the centroid of the triangle.

Answer:

(i) Median AD of the triangle will divide the side BC in two equal parts.

Therefore, D is the mid-point of side BC.

 

(ii) Point P divides the side AD in a ratio 2:1.

(iii) Median BE of the triangle will divide the side AC in two equal parts.

Therefore, E is the mid-point of side AC.

Point Q divides the side BE in a ratio 2:1.

Median CF of the triangle will divide the side AB in two equal parts. Therefore, F is the mid-point of side AB.

(iv) It can be observed that the coordinates of point P, Q, R are the same.
Therefore, all these are representing the same point on the plane i.e., the centroid of the triangle.

(v) Consider a triangle, ΔABC, having its vertices as A(x₁, y₁), B(x₂, y₂), and C(x₃, y₃).

Median AD of the triangle will divide the side BC in two equal parts. Therefore, D is the mid-point of side BC.