## AP State Syllabus SSC 10th Class Maths Solutions 11th Lesson Trigonometry Exercise 11.2

### 10th Class Maths 11th Lesson Trigonometry Ex 11.2 Textbook Questions and Answers

Question 1.

Evaluate the following.

i) sin 45° + cos 45°

Answer:

sin 45° + cos 45°

= \(\frac{1}{\sqrt{2}}\) + \(\frac{1}{\sqrt{2}}\)

= \(\frac{1+1}{\sqrt{2}}\)

= \(\frac{2}{\sqrt{2}}\)

= \(\frac{\sqrt{2} \times \sqrt{2}}{\sqrt{2}}\)

= √2

ii)

Answer:

iii)

Answer:

iv) 2 tan^{2} 45° + cos^{2} 30° – sin^{2} 60°

Answer:

2 tan^{2} 45° + cos^{2} 30° – sin^{2} 60°

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

= \(\frac{2}{1}\) + \(\frac{3}{4}\) – \(\frac{3}{4}\)

= \(\frac{8+3-3}{4}\)

= \(\frac{8}{4}\)

= 2

v)

Answer:

Question 2.

Choose the right option and justify your choice.

i) \(\frac{2 \tan 30^{\circ}}{1+\tan ^{2} 45^{\circ}}\)

a) sin 60°

b) cos 60°

c) tan 30°

d) sin 30°

Answer:

ii) \(\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}\)

a) tan 90°

b) 1

c) sin 45°

d) 0

Answer:

\(\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}\) = \(\frac{1-(1)^{2}}{1+(1)^{2}}\)

= \(\frac{0}{1+1}\) = \(\frac{0}{2}\) = 0

iii) \(\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}\)

a) cos 60°

b) sin 60°

c) tan 60°

d) sin 30°

Answer:

Question 3.

Evaluate sin 60° cos 30° + sin 30° cos 60°. What is the value of sin (60° + 30°). What can you conclude?

Answer:

Take sin 60°.cos 30° + sin 30°.cos 60°

= \(\frac{\sqrt{3}}{2}\) . \(\frac{\sqrt{3}}{2}\) + \(\frac{1}{2}\) . \(\frac{1}{2}\)

= \(\frac{(\sqrt{3})^{2}}{4}\) + \(\frac{1}{4}\)

= \(\frac{3}{4}\) + \(\frac{1}{4}\)

= \(\frac{3+1}{4}\)

= \(\frac{4}{4}\) = 1 …… (1)

Now take sin (60° + 30°)

= sin 90° = 1 …….. (2)

From equations (1) and (2), I conclude that

sin (60°+30°) = sin 60° . cos 30° + sin 30° . cos 60°.

i.e., sin (A + B) = sin A . cos B + cos A . sin B

Question 4.

Is it right to say cos (60° + 30°) = cos 60° cos 30° – sin 60° sin 30° ?

Answer:

L.H.S. = cos (60° + 30°)

cos 90° = 0

R.H.S. = cos 60° . cos 30° – sin 60° . sin 30°.

= \(\frac{1}{2}\) . \(\frac{\sqrt{3}}{2}\) – \(\frac{\sqrt{3}}{2}\) . \(\frac{1}{2}\)

= \(\frac{\sqrt{3}}{4}\) – \(\frac{\sqrt{3}}{4}\) = 0

∴ L.H.S = R.H.S

Yes, it is right to say

cos (60°+30°) = cos 60° . cos 30° – sin 60° . sin 30°.

i.e., cos (A + B) = cos A . cos B – sin A . sin B

Question 5.

In right angle triangle △PQR, right angle is at Q and PQ = 6 cms, ∠RPQ = 60°. Determine the lengths of QR and PR.

Answer:

Given that △PQR is a right angled triangle, right angle is at Q and PQ = 6 cm, ∠RPQ = 60°.

tan 60° = \(\frac{\text { Opposite side to } \angle P}{\text { Adjacent side to } \angle P}\)

√3 = \(\frac{RQ}{6}\)

which gives RQ = 6√3 cm ……. (1)

To find the length of the side RQ, we consider

∴ The length of QR is 6√3 and RP is 12 cm.

Question 6.

In △XYZ, right angle is at Y, YZ = x, and XY = 2x then determine ∠YXZ and ∠YZX.

Answer:

Note: In the problem take

YX = x, and XZ = 2x.

Given that △XYZ is a right angled triangle and right angle at Y, and YX = x and XZ = 2x.

By Pythagoras theorem

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

(2x)^{2} = (x)^{2} + YZ^{2}

4x^{2} = x^{2} + YZ^{2}

YZ^{2} = 4x^{2} – x^{2} = 3x^{2}

YZ = \(\sqrt{3 x^{2}}\) = √3x

Now, from the △XYZ

tan X = \(\frac{XZ}{XY}\) = \(\frac{\sqrt{3} x}{x}\)

tan X = √3 = tan 60°

∴ Angle YXZ is 60°.

tan Z = \(\frac{XY}{YZ}\) = \(\frac{x}{\sqrt{3} x}\)

tan Z = \(\frac{1}{\sqrt{3}}\) = tan 30°

∴ Angle YZX is 30°.

Hence ∠YXZ and ∠YZX are 60° and 30°.

Question 7.

Is it right to say that

sin (A + B) = sin A + sin B? Justify your answer.

Answer:

Let A = 30° and B = 60°

L.H.S = sin (A + B)

= sin (30° + 60°) = sin 90° = 1

R.H.S = sin 30° + sin 60°

= \(\frac{1}{2}\) + \(\frac{\sqrt{3}}{2}\)

= \(\frac{\sqrt{3}+1}{2}\)

Hence L.H.S ≠ R.H.S

So, it is not right to say that sin (A + B) = sin A + sin B