AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 11 Trigonometry Ex 11.1 Textbook Questions and Answers.

## AP State Syllabus SSC 10th Class Maths Solutions 11th Lesson Trigonometry Exercise 11.1

### 10th Class Maths 11th Lesson Trigonometry Ex 11.1 Textbook Questions and Answers

Question 1.

In right angle triangle ABC, 8 cm, 15 cm and 17 cm are the lengths of AB, BC and CA respectively. Then, find out sin A, cos A and tan A.

Answer:

Given that

△ABC is a right angle triangle and Lengths of AB, BC and CA are 8 cm, 15 cm and 17 cm respectively.

Among the given lengths CA is longest.

Hence CA is the hypotenuse in △ABC and its opposite vertex having right angle.

i.e., ∠B = 90°.

With reference to ∠A, we have opposite side = BC = 15 cm

adjacent side = AB = 8 cm

and hypotenuse = AC = 17

sin A = \(\frac{\text { Opposite side of } \angle \mathrm{A}}{\text { Hypotenuse }}\) = \(\frac{BC}{AC}\) = \(\frac{15}{17}\)

cos A = \(\frac{\text { Adjacent side of } \angle \mathrm{A}}{\text { Hypotenuse }}\) = \(\frac{AB}{AC}\) = \(\frac{8}{17}\)

tan A = \(\frac{\text { Opposite side of } \angle \mathrm{A}}{\text { Adjacent side of } \angle \mathrm{A}}\) = \(\frac{BC}{AB}\) = \(\frac{15}{8}\)

∴ sin A = \(\frac{15}{17}\);

cos A = \(\frac{8}{17}\)

tan A = \(\frac{15}{8}\)

Question 2.

The sides of a right angle triangle PQR are PQ = 7 cm, QR = 25 cm and ∠P = 90° respectively. Then find, tan Q – tan R.

Answer:

Given that △PQR is a right angled triangle and PQ = 7 cm, QR = 25 cm.

By Pythagoras theorem QR^{2} = PQ^{2} + PR^{2}

(25)^{2} = (7)^{2} + PR^{2}

PR^{2} = (25)^{2} – (7)^{2} = 625 – 49 = 576

PR = √576 = 24 cm

tan Q = \(\frac{PR}{PQ}\) = \(\frac{24}{7}\);

tan R = \(\frac{PQ}{PR}\) = \(\frac{7}{24}\)

∴ tan Q – tan R = \(\frac{24}{7}\) – \(\frac{7}{24}\)

= \(\frac{(24)^{2}-(7)^{2}}{168}\)

= \(\frac{576-49}{168}\)

= \(\frac{527}{168}\)

Question 3.

In a right angle triangle ABC with right angle at B, in which a = 24 units, b = 25 units and ∠BAC = θ. Then, find cos θ and tan θ.

Answer:

Given that ABC is a right angle triangle with right angle at B, and BC = a = 24 units, CA = b = 25 units and ∠BAC = θ.

By Pythagoras theorem

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

(25)^{2} = AB^{2} + (24)^{2}

AB^{2} = 25^{2} – 24^{2} = 625 – 576

AB^{2} = 49

AB = √49 = 1

With reference to ∠BAC = θ, we have

Opposite side to θ = BC = 24 units.

Adjacent side to θ = AB = 7 units.

Hypotenuse = AC = 25 units.

Now

cos θ = \(\frac{\text { Adjacent side of } \theta}{\text { Hypotenuse }}\) = \(\frac{AB}{AC}\) = \(\frac{7}{25}\)

tan θ = \(\frac{\text { Opposite side of } \theta}{\text { Adjacent side of } \theta}\) = \(\frac{BC}{AB}\) = \(\frac{24}{7}\)

Hence cos θ = \(\frac{7}{25}\) and tan θ = \(\frac{24}{7}\)

Question 4.

If cos A = \(\frac{12}{13}\), then find sin A and tan A.

Answer:

From the identity

sin^{2} A + cos^{2} A = 1

⇒ sin^{2} A = 1 – cos^{2} A

= 1 – \(\left(\frac{12}{13}\right)^{2}\)

= 1 – \(\frac{144}{169}\)

= \(\frac{169-144}{169}\)

= \(\frac{25}{169}\)

∴ sin A = \(\sqrt{\frac{25}{169}}\) = \(\frac{5}{13}\)

∴ sin A = \(\frac{5}{13}\); tan A = \(\frac{5}{12}\)

Question 5.

If 3 tan A = 4, then find sin A and cos A.

Answer:

Given 3 tan A = 4

⇒ tan A = \(\frac{4}{3}\)

From the identify sec^{2} A – tan^{2} A = 1

⇒ 1 + tan^{2} A = sec^{2} A

If cos A = \(\frac{3}{5}\) then from

sin^{2} A + cos^{2} A = 1

We can write sin^{2}A = 1 – cos^{2}A

= 1 – \(\left(\frac{3}{5}\right)^{2}\)

= 1 – \(\frac{9}{25}\)

⇒ sin^{2} A = \(\frac{16}{25}\)

⇒ sin A = \(\frac{4}{5}\)

∴ sin A = \(\frac{4}{5}\); cos A = \(\frac{3}{5}\)

Question 6.

In △ABC and △XYZ, if ∠A and ∠X are acute angles such that cos A = cos X then show that ∠A = ∠X.

Answer:

In the given triangle, cos A = cos X

⇒ \(\frac{AC}{AX}\) = \(\frac{XC}{AX}\)

⇒ AC = XC

⇒ ∠A = ∠X (∵ Angles opposite to equal sides are also equal)

Question 7.

Given cot θ = \(\frac{7}{8}\), then evaluate

Answer:

cot^{2} θ = (cot θ)^{2}

= \(\left(\frac{7}{8}\right)^{2}\) = \(\frac{49}{64}\) …… (1)

= sec θ + tan θ

So cot θ = \(\frac{7}{8}\)

⇒ tan θ = \(\frac{8}{7}\)

⇒ tan^{2} θ = \(\left(\frac{8}{7}\right)^{2}\) = \(\frac{64}{49}\)

From sec^{2} θ – tan^{2} θ = 1

⇒ 1 + tan^{2} θ = sec^{2} θ

Question 8.

In a right angle triangle ABC, right angle is at B, if tan A = √3, then find the value of

i) sin A cos C + cos A sin C

ii) cos A cos C – sin A sin C

Answer:

Given, tan A = \(\frac{\sqrt{3}}{1}\)

Hence \(\frac{\text { Opposite side }}{\text { Adjacent side }}=\frac{\sqrt{3}}{1}\)

Let opposite side = √3k and adjacent side = 1k

In right angled △ABC,

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

(By Pythagoras theorem)

⇒ AC^{2} = (1k)^{2} + (√3k)^{2}

⇒ AC^{2} = 1k^{2} + 3k^{2}

⇒ AC^{2} = 4k^{2}

∴ AC = \(\sqrt{4 k^{2}}\) = 2k