Students can practice 10th Class Maths Study Material Telangana Chapter 11 Trigonometry InText Questions to get the best methods of solving problems.

## TS 10th Class Maths Solutions Chapter 11 Trigonometry InText Questions

Do This

Identify “Hypotenuse”, “Opposite side” and “Adjacent side” for the given angles in the given triangles. (AS_{3}) (Page No. 271)

Question 1.

For angle R

Solution:

In the ∆PQR

Opposite side = PQ

Adjacent side = QR

Hypotenuse side = PR

Question 2.

i) For angle X

ii) For angle Y

Solution:

i) In the ∆XYZ

For angle X

Opposite side = YZ

Adjacent side = XZ

Hypotenuse = XY

ii) For angle Y

Opposite side = XZ

Adjacent side = YZ

Hypotenuse = XY

Question 3.

Find (i) sin C (ii) cos C (iii) tan C in the given triangle.

Solution:

By Pythagoras theorem

AC^{2} = AB^{2} + BC^{2}

(13)^{2} = AB^{2} + (5)^{2}

AB^{2} = 169 – 25

AB^{2} = 144

∴ AB = \(\sqrt{144}\) = 12

Question 4.

In a triangle XYZ, ZY is right angle. XZ = 17 cm and YZ = 15 cm, then find

(i) sin X (ii) cos Z (iii) tan X. (AS_{1}) (Page No. 274)

Solution:

Given ∆XYZ, ∠Y is right angle.

By Pythagoras theorem

XZ^{2} = YZ^{2} + XY^{2}

17^{2} = 15^{2} + XY^{2}

XY^{2} = 17^{2} – 15^{2} = 289 – 225 = 64

XY = \(\sqrt{64}\) = 8

Question 5.

In a triangle PQR with right angle at Q, the value of ∠P is x, PQ = 7 cm and QR = 24 cm then find sin x and cos x. (AS_{1}) (Page No. 274)

Solution:

Given right angled triangle is PQR with right angle at Q. The value of ∠P is x.

By Pythagoras theorem

PR^{2} = PQ^{2} + QR^{2}

= 7^{2} + 24^{2}

PR^{2} = 49 + 576

PR^{2} = 625

PR = \(\sqrt{625}\) = 25

sin x = \(\frac{\mathrm{QR}}{\mathrm{PR}}\) = \(\frac{24}{25}\)

cos x = \(\frac{\mathrm{PQ}}{\mathrm{PR}}\) = \(\frac{7}{25}\)

Try This

Question 1.

Write lengths of “Hypotenuse”, “Opposite side” and “Adjacent side” for the given angles in the given triangles. (AS_{1}) (Page No. 271)

i) For angle C

ii) For angle A

Solution:

In ∆ ABC, ∠B = 90°

∴ AC^{2} = AB^{2} + BC^{2} (By Pythagoras theorem)

⇒ AB^{2} = AC^{2} – BC^{2}

Substitute BC = 4 cm and AC = 5 cm in eq. (1), we get

AB^{2} = 5^{2} – 4^{2} = 25 – 16 = 9

∴ AB = \(\sqrt{9}\) = 3

i) For angle C :

Length of hypotenuse = AC = 5 cm

Length of opposite side = AB = 3 cm

Length of adjacent side = BC = 4 cm

ii) For angle A :

Length of hypotenuse AC = 5 cm

Length of opposite side = BC = 4 cm

Length of adjacent side = AB = 3 cm

Question 2.

In a right angle triangle ABC, right angle is at C, BC + CA = 23 cm and BC – CA = 7 cm, then find sin A and tan B. (AS_{1}) (Page No. 274)

Solution:

∆ABC, ∠C = 90°

Given that,

⇒ BC = \(\frac{30}{2}\) = 15

BC + CA = 23

15 + CA = 23

CA = 23 – 15 = 8

AB^{2} = BC^{2} + CA^{2} (By Pythagoras theorem)

= 15^{2} + 8^{2} = 225 + 64 = 289

∴ AB = \(\sqrt{289}\) = 17

sin A = \(\frac{\mathrm{BC}}{\mathrm{AB}}\) = \(\frac{15}{17}\) ; tan B = \(\frac{\mathrm{AC}}{\mathrm{BC}}\) = \(\frac{8}{15}\)

Question 3.

What will be the ratio of sides for sec A and cot A ? (AS_{3}) (Page No. 275)

Solution:

sec A = \(\frac{1}{\cos A}\) = \(\frac{\text { Hypotenuse } }{\text { Side opposite to angle A }}\)

cot A = \(\frac{1}{\tan A}\) = \(\frac{\text { Side adjacent to angle A} }{\text { Side opposite to angle A }}\)

Think – Discuss

Question 1.

Discuss between your friends that

i) sin x = \(\frac{4}{3}\) does exist for some value of angle x ?

ii) The value of sin A and cos A is always less than 1. Why ?

iii) tan A is product of tan and A. (AS_{2}) (Page No. 274)

Solution:

i) The value of sin0 always lies between 0 and 1. Here, sin x = \(\frac{4}{3}\) which is greater than 1. So, it does not exist.

ii) Draw a circle of radius 1 unit with centre at the origin. Let P(a, b) be any point on the circle with angle AOP = θ

sin θ = \(\frac{\mathrm{AP}}{\mathrm{OP}}=\frac{\mathrm{b}}{1}\) = y – co-ordinate

cos θ = \(\frac{\mathrm{OA}}{\mathrm{OP}}=\frac{\mathrm{a}}{1}\) = x – co-ordinate

One complete rotation of point ‘P’ in a circular an angle 360° at the centre.

Quarter rotation substends an angle of AOB equals to 90°.

Half rotation substends an angle of AOC equal to 180°.

Three quarter substends an angle of AOD equals to 270°.

The co-ordinates of the points A, B, C and D respectively (1, 0), (0, 1) (-1, 0) and (0, -1).

According to the co-ordinates

cos 0° = 1 sin 0° = 0

cos 90° = 0 sin 90° = 1

cos 180° = – 1 sin 180° = 0

cos 270° = 0 sin 270° = -1

cos 360° = 1 sin 360° = 0

Hence, the value of “sine” and “cosine” is always less than 1.

iii) The symbol tan A is used as an abbreviation for “the tan of the angle A”, tan A is not the product of “tan” and A “tan” separated from “A” has no meaning.

Question 2.

Is \(\frac{\sin \mathrm{A}}{\cos \mathrm{A}}\) equal to tan A ? (Page No. 275)

Solution:

Question 3.

Is \(\frac{\cos \mathrm{A}}{\sin \mathrm{A}}\) equal to cot A ? (Page No. 275)

Solution:

Do This

Question 1.

Find cosec 60°, sec 60° and cot 60°. (AS_{1}) (Page No. 279)

Solution:

Consider an equilateral triangle ABC. Since each angle is 60° in an equilateral triangle, we have ∠A = ∠B = ∠C = 60° and the sides of equilateral triangle is AB = BC = CA = 2a units

Draw the perpendicular line AD from vertex A to BC as shown in the given figure.

Perpendicular AD acts as “angle bisector of angle A” and “bisector of the side BC” in the equilateral triangle ABC.

Therefore, ∠BAD = ∠CAD = 30°.

Since point D divides the side BC into equal halves.

BD = \(\frac{1}{2}\) BC = \(\frac{2 \mathrm{a}}{2}\) = a units

Consider right angle triangle ABD in the above given figure.

We have AB = 2a, and BD = a

Then AD^{2} = AB^{2} – BD^{2} (By Pythagoras theorem)

= (2a)^{2} – (a)^{2} = 3a^{2}

Therefore, AD = a\(\sqrt{3}\)

From, definitions of trigonometric ratios.

sin 60° = \(\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{a} \sqrt{3}}{2 \mathrm{a}}=\frac{\sqrt{3}}{2}\)

cos 60° = \(\frac{\mathrm{BD}}{\mathrm{AB}}=\frac{\mathrm{a}}{2 \mathrm{a}}=\frac{1}{2}\)

So, similarly tan 60° = \(\sqrt{3}\)

Try This

Question 1.

Find sin 30°, cos 30°, tan 30°, cosec 30°, sec 30° and cot 30° by using the ratio concepts. (AS_{1}) (Page No. 279)

Solution:

Let ABC be an equilateral triangle.

∴ ∠A = ∠B = ∠C = 60°

Let AB BC = CA ‘a’ units

Draw AD ⊥ BC, AD bisects BC.

∴ BD = a/2

In ∆ABD, ∠ADB = 90°; ∠B = 60°

∴ ∠BAD = 180° – (90° + 60°)

= 180° – 150° = 30°

By Pythagoras theorem,

AB^{2} = AD^{2} + BD^{2}

⇒ AD^{2} = AB^{2} – BD^{2}

Question 2.

Find the values for tan 90°, cosec 90°, sec 90° and cot 90°. (AS_{1}) (Page No. 281)

Solution:

Let us see what happens when angle made by AC with ray AB increases. When angle A is increased, height of point C increases and the foot of the perpendicular shifts from B to X and then to Y and so on. So, when the angle becomes 90°, base (adjacent side of the angle) would become zero; the height of C from AB ray increases and it would be equal to AC.

AB = 0 and BC = AC

Think – Discuss

Question 1.

Discuss between your friend about the following conditions :

What can you say about cosec 0° = \(\frac{1}{\sin 0}\) ? Is it not defined ? Why ? (AS_{2}) (Page No. 280)

Solution:

sin 0° = 0

cosec 0° = \(\frac{1}{\sin 0}\) = \(\frac{1}{0}\) = not defined

Reason : Division by ‘0’ is not allowed, hence \(\frac{1}{0}\) is indeterminate.

Question 2.

Is it defined cot 0° = \(\frac{1}{\tan 0}\) ? Why ? (Page No. 281)

Solution:

tan 0° = 0

cot 0° = \(\frac{1}{\tan 0}\) = \(\frac{1}{0}\) undefined

Reason : Division by ‘0’ is not allowed, hence \(\frac{1}{\tan 0}\) is indeterminate.

Question 3.

sec 0° = 1. Why ? (Page No. 281)

Solution:

sec 0° = \(\frac{1}{\cos 0^{\circ}}\) [∵ cos 0° = 1]

= \(\frac{1}{1}\) = 1

Question 4.

What can you say about the values of sin A and cos A, as the value of angle A increases from 0° to 90° ? (AS_{3}) (Page No. 282)

i) If A > B, then sin A > sin B. Is it true ?

ii) If A > B, then cos A > cos B. Is it true ? Discuss.

Solution:

i) Given statement

“If A > B, then sin A > sin B”.

Yes, this statement is true.

Because, it is clear from the table below that the sin A increases as A increases.

ii) Given statement

“If A > B, then cos A > cos B”.

No, this statement is not true.

Because it is clear from the table below that cos A decreases as A increases.

Think – Discuss

Question 1.

For which value of acute angle

i) \(\frac{\cos \theta}{1-\sin \theta}+\frac{\cos \theta}{1+\sin \theta}\) = 4 is true ? For which value of 0° ≤ θ ≤ 90°, above equa-tion is not defined ? (AS_{1}, AS_{2}) (Page No. 285)

Solution:

Given \(\frac{\cos \theta}{1-\sin \theta}+\frac{\cos \theta}{1+\sin \theta}\) = 4

cos θ = \(\frac{1}{2}\)

cos θ = cos 60°

∴ θ = 60°

Given statement is true for the acute angle i.e., θ = 60°

Question 2.

Check and discuss the above relations AB

[sin (90° – x) = \(\frac{\mathrm{A B}}{\mathrm{A C}}\) = cos x and BC

cos (90° – x) = \(\frac{\mathrm{B C}}{\mathrm{A C}}\) = sin x,

tan (90° – x) = \(\frac{\mathrm{A B}}{\mathrm{A C}}\) = cot x and

cot (90° – x) = \(\frac{\mathrm{B C}}{\mathrm{A B}}\) = tan x,

cosec (90° – x) = \(\frac{\mathrm{A C}}{\mathrm{A B}}\) = sec x and

sec (90° – x) = \(\frac{\mathrm{A C}}{\mathrm{B C}}\) = cosec x] in the case of angles between 0° and 90°, whether they hold for these angles or not ? So,

(i) sin (90° – A) = cos A

(ii) cos (90° – A) = sin A

(iii) tan (90° – A) = cot A and

(iv) cot (90° – A) = tan A

(vi) cosec (90° – A) = cosec A

(v) sec (90° – A) = cosec A (AS_{2}) (Page No. 286)

Solution:

Let A = 30°

i) sin (90° – A ) = cos A

⇒ sin (90° – 30°) = cos 30°

⇒ sin 60° = cos 30° = \(\frac{\sqrt{3}}{2}\)

ii) cos (90° – A ) = sin A

⇒ cos (90° – 30°) = sin 30°

⇒ cos 60° – sin 30° = \(\frac{1}{2}\)

iii) tan (90° – A ) = cot A

⇒ tan (90° – 30°) = cot 30°

⇒ tan 60° = cot 30° = \(\sqrt{3}\)

iv) cot (90° – A ) = tan A

⇒ cot (90° – 30°) = tan 30°

⇒ cot 60° = tan 30° = \(\frac{1}{\sqrt{3}}\)

v) sec (90° – A ) = cosec A

⇒ sec (90° – 30°) = cosec 30°

⇒ sec 60° = cosec 30° = 2

vi) cosec (90° – A ) = sec A

⇒ cosec (90° – 30°) = sec 30°

⇒ cosec 60° = sec 30° = \(\frac{2}{\sqrt{3}}\)

So, the above relations hold for all the angles between 0° and 90°.

Do This

i) If sin C = \(\frac{15}{17}\), then find cos C. (AS_{1}) (Page No. 290)

Solution:

Given sin C = \(\frac{15}{17}\)

cos C = \(\sqrt{1-\sin ^2 \mathrm{C}}\) (from identity – 1)

= \(\sqrt{1-\left(\frac{15}{17}\right)^2}\) = \(\sqrt{\frac{289-225}{289}}\) = \(\sqrt{\frac{64}{289}}\) = \(\frac{8}{17}\)

ii) If tan x = \(\frac{5}{12}\), then find sec x. (AS_{1}) (Page No. 290)

Solution:

Given tan x = \(\frac{5}{12}\)

We know that sec^{2}x – tan^{2}x = 1

sec^{2}x = 1 + tan^{2}x

iii) If cosec θ = \(\frac{25}{7}\), then find cot θ. (AS_{1}) (Page No. 290)

Solution:

Given cosec θ = \(\frac{25}{7}\)

We know that cosec^{2}θ – cot^{2}θ = 1

cot^{2}θ = cosec^{2}θ – 1

Try This

Question 1.

Evaluate the following and justify your answer. (AS_{4}) (Page No. 290)

i) \(\frac{\sin ^2 15+\sin ^2 75}{\cos ^2 36+\cos ^2 54}\)

ii) sin 5° cos 85° + cos 5° sin 85°

iii) sec 16° cosec 74° – cot 74° tan 16°

Solution:

i) \(\frac{\sin ^2 15+\sin ^2 75}{\cos ^2 36+\cos ^2 54}\)

We can write sin 15°= sin (90° – 75°)

= cos 75°

∴ sin^{2} 15° = cos^{2} 75°

Similarly, cos 36°= cos (90° – 54°) = sin 54°

∴ cos^{2} 36° = sin^{2} 54°

sin^{2} 15° + sin^{2} 75°= cos^{2} 75° + sin^{2} 75°

(∵ sin^{2} 15° = cos^{2} 75°)

= 1 ———– (1) (∵ cos^{2}θ + sin^{2}θ = 1)

(Here θ = 75°)

cos^{2} 36° + cos^{2} 54°= sin^{2} 54° + cos^{2} 54°

(∵ cos^{2} 36° = sin^{2} 54°)

= 1 ———– (2) (∵ sin^{2}θ + cos^{2}θ = 1)

(Here θ = 54°)

From (1) & (2), we get

\(\frac{\sin ^2 15+\sin ^2 75}{\cos ^2 36+\cos ^2 54}\) = \(\frac{1}{1}\) = 1

ii) sin 5° cos 85° + cos 5° sin 85° …………….. (1)

sin 5°= sin (90° – 85°) = cos 85°

cos 5° = cos (90° – 85°) = sin 85°

Substitute these values of sin 5° and cos 5° in (1).

We get

sin 5° cos 85° + cos 5° sin 85°

= cos 85°. cos 85° + sin 85°. sin 85°

= cos^{2} 85° + sin^{2} 85°

= 1 (∵ cos^{2} θ + sin^{2} θ = 1, Here θ = 85°)

iii) sec 16° cosec 74° – cot 74° tan 16° …………….. (1)

cosec 74° = cosec (90° – 16°) = sec 16°

[(∵ cosec (90° – θ) = sec θ) and cot(90° – θ) = tan θ]

cot 74° = cot (90° – 16°) = tan 16°

Substitute the equivalents of cosec 74° and cot 74° in (1), we get

sec 16° . cosec 74° – cot 74° . tan 16°

= sec 16° . sec 16° – tari 16° . tan 16°

= sec^{2} 16° – tan^{2} 16° = 1

(∵ sec^{2} θ – tan^{2}θ = 1) Here θ = 16°

Think – Discuss

Question 1.

Are these identities true for 0° ≤ A ≤ 90° ? If not, for which values of A they are true ?

i) sec^{2} A – tan^{2} A = 1

ii) cosec^{2} A – cot^{2} A = 1 (AS_{2}) (Page No. 290)

Solution:

i) Given identity is sec^{2} A – tan^{2} A = 1

Let A = 0°

LHS = sec^{2} 0° – tan^{2} 0°

= 1 – 0 = 1 = R.H.S.

Let A =90°

tan A and sec A are not defined.

So it is true.

∴ For all given values of ‘A’ such that 0° ≤ A ≤ 90° this trigonometric identity is true.

ii) Given identity is cosec^{2} A – cot^{2} A = 1

Let A = 0°

cosec A and cot A are not defined for A = 0°.

Therefore identity is true for A = 0°

Let A = 90°

cosec A = cosec 90° = 1

cot A = cot 90° = 0

∴ L.H.S. = 1^{2} – 0^{2} = 1 – 0 = 1 = R.H.S.

∴ This identity is true for all values of A, such that 0° ≤ A ≤ 90°