RS Aggarwal Solutions Class 10 T Ratios Of Some Particular Angles

Evaluate each of the following:

Question.1: sin 60o cos 30o + cos 60o sin 30o

Solution:

On substituting the values of various T-ratios, we get:

sin 60o cos 30o + cos 60o sin 30o

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

Question.2: cos 60o cos 30o – sin 60o sin 30o

Solution:

On substituting the value of various T-ratios, we get:

cos 60o cos 30o – sin 60o sin 30o

= \(\frac{1}{2}\times \frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}\times \frac{1}{2}=\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}=0\)

Question.3: cos 45o cos 30o + sin 45o sin 30o

Solution:

On substituting the value of various T-ratios, we get:

cos 45o cos 30o + sin 45o sin 30o

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

Question.4: \(\frac{sin30^{\circ}}{cos45^{\circ}}+\frac{cot45^{\circ}}{sec60^{\circ}}-\frac{sin60^{\circ}}{tan45^{\circ}}+\frac{cos30^{\circ}}{sin90^{\circ}}\)

Solution:

On substituting the values of various T-ratios, we get:

\(\frac{sin30^{\circ}}{cos45^{\circ}}+\frac{cot45^{\circ}}{sec60^{\circ}}-\frac{sin60^{\circ}}{tan45^{\circ}}+\frac{cos30^{\circ}}{sin90^{\circ}}\)

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

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

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

Question.5: \(\frac{5cos^{2}60^{\circ}+4sec^{2}30^{\circ}-tan^{2}45^{\circ}}{sin^{2}30^{\circ}+cos^{2}30^{\circ}}\)

Solution:

On substituting the value of various T-ratios, we get:

\(\frac{5cos^{2}60^{\circ}+4sec^{2}30^{\circ}-tan^{2}45^{\circ}}{sin^{2}30^{\circ}+cos^{2}30^{\circ}}\)

= \(\frac{5\times \left ( \frac{1}{2} \right )^{2}+4\times \left ( \frac{2}{\sqrt{3}} \right )^{2}-(1)^{2}}{\left ( \frac{1}{2} \right )^{2}+\left ( \frac{\sqrt{3}}{2} \right )^{2}}\)

= \(\frac{5\times \frac{1}{4}+4\times \frac{4}{3}-1}{\frac{1}{4}+\frac{3}{4}}\)

= \(\frac{\frac{5}{4}+\frac{16}{3}-1}{\frac{1}{4}+\frac{3}{4}}\)

= \(\frac{\frac{15+64-12}{12}}{\frac{4}{4}}\) = \(\frac{67}{12}\)

Question.6: 2cos2 60o + 3sin2 45o – 3sin2 30o + 2cos2 90o

Solution:

On substituting the value of various T-ratios, we get:

2cos2 60o + 3sin2 45o – 3sin2 30o + 2cos2 90o

= \(2\times \left ( \frac{1}{2} \right )^{2}+3\times \left ( \frac{1}{\sqrt{2}} \right )^{2}-3\times \left ( \frac{1}{2} \right )^{2}+2(0)^{2}\)

= \(\frac{1}{2}+\frac{3}{2}-\frac{3}{4}=\frac{2+6-3}{4}=\frac{5}{4}\)

Question.7: cot2 30o – 2cos2 30o\(\frac{3}{4}\)sec2 45o + \(\frac{1}{4}\)cosec2 30o

Solution:

On substituting the value of various T-ratios, we get:

cot2 30o – 2cos2 30o\(\frac{3}{4}\)sec2 45o + \(\frac{1}{4}\)cosec2 30o

= \((\sqrt{3})^{2}-2\times \left ( \frac{\sqrt{3}}{2} \right )^{2}-\frac{3}{4}\times \left ( \frac{\sqrt{2}}{1} \right )^{2}+\frac{1}{4}\times (2)^{2}\)

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

= \(3-\frac{3}{2}-\frac{3}{2}+1\) = \(\frac{6-3-3+2}{2}=\frac{2}{2}=1\)

Question.8: (sin2 30o + 4cot2 45o – sec2 60o) (cosec2 45o sec2 30o)

Solution:

On substituting the value of various T-ratios, we get:

(sin2 30o + 4cot2 45o – sec2 60o)(cosec2 45o sec2 30o)

= \(\left [ \left ( \frac{1}{2} \right )^{2}+4\times (1)^{2}-(2)^{2} \right ]\left [ (\sqrt{2})^{2}\times \left ( \frac{2}{\sqrt{3}} \right )^{2} \right ]\)

= \(\left ( \frac{1}{4}+4-4 \right )\left ( 2\times \frac{4}{3} \right )\)

= \(\frac{1}{4}\times \frac{8}{3}=\frac{2}{3}\)

Question.9: \(\frac{4}{cot^{2}30^{\circ}}+\frac{1}{sin^{2}30^{\circ}}-2cos^{2}45^{\circ}-sin^{2}0^{\circ}\)

Solution:

On substituting the value of various T-ratios, we get:

\(\frac{4}{cot^{2}30^{\circ}}+\frac{1}{sin^{2}30^{\circ}}-2cos^{2}45^{\circ}-sin^{2}0^{\circ}\)

= \(\frac{4}{(\sqrt{3})^{2}}+\frac{1}{\left ( \frac{1}{2} \right )^{2}}-2\times \left ( \frac{1}{\sqrt{2}} \right )^{2}-0\)

= \(\frac{4}{3}+\frac{4}{1}-\frac{2}{2}-0\)

= \(\frac{8+24-6-0}{6}\)

= \(\frac{26}{6}=\frac{13}{3}\)

Question.10: Show that:

(i) \(\frac{1-sin60^{\circ}}{cos60^{\circ}}=\frac{tan60^{\circ}-1}{tan60^{\circ}+1}\)

(ii) \(\frac{cos30^{\circ}+sin60^{\circ}}{1+sin30^{\circ}+cos60^{\circ}}=cos30^{\circ}\)

Solution:

(i) L.H.S. = \(\frac{1-sin60^{\circ}}{cos60^{\circ}}\) = \(\frac{1-\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\frac{2-\sqrt{3}}{1}\)

R.H.S. = \(\frac{tan60^{\circ}-1}{tan60^{\circ}+1}\)

= \(\frac{\sqrt{3}-1}{\sqrt{3}+1}\times \frac{\sqrt{3}-1}{\sqrt{3}-1}\)

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

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

= \(\frac{4-2\sqrt{3}}{2}\) = \(\frac{2(2-\sqrt{3})}{2}\) = \((2-\sqrt{3})\)

L.H.S. = R.H.S.

Hence, \(\frac{1-sin60^{\circ}}{cos60^{\circ}}=\frac{tan60^{\circ}-1}{tan60^{\circ}+1}\)

(ii) L.H.S. = \(\frac{cos30^{\circ}+sin60^{\circ}}{1+sin30^{\circ}+cos60^{\circ}}=\frac{\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}}{1+\frac{1}{2}+\frac{1}{2}}=\frac{\sqrt{3}}{2}\)

R.H.S. = cos 30o = \(\frac{\sqrt{3}}{2}\)

L.H.S. = R.H.S.

Hence, \(\frac{cos30^{\circ}+sin60^{\circ}}{1+sin30^{\circ}+cos60^{\circ}}=cos30^{\circ}\)

Question.11: Verify each of the following:

(i) sin 60o cos 30o – cos 60o sin 30o = sin 30o

(ii) cos 60o cos 30o + sin 60o sin 30o = cos 30o

(iii) 2sin 30o cos 30o = sin 60o

(iv) 2sin 45o cos 45o = sin 90o

Solution:

(i) L.H.S. = sin 60o cos 30o – cos 60o sin 30o

= \(\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}-\frac{1}{2}\times \frac{1}{2}=\frac{3}{4}-\frac{1}{4}=\frac{2}{4}=\frac{1}{2}\)

R.H.S. = sin 30o = \(\frac{1}{2}\)

R.H.S. = L.H.S.

Hence, sin 60o cos 30o – cos 60o sin 30o = sin 30o

(ii) L.H.S. = cos 60o cos 30o + sin 60o sin 30o

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

R.H.S. = cos 30o = \(\frac{\sqrt{3}}{2}\)

R.H.S. = L.H.S.

Hence, cos 60o cos 30o + sin 60o sin 30o = cos 30o

(iii) L.H.S. = 2sin 30o cos 30o

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

R.H.S. = sin 60o = \(\frac{\sqrt{3}}{2}\)

R.H.S. = L.H.S.

Hence, 2sin 30o cos 30o = sin 60o

(iv) L.H.S. = 2sin 45o cos 45o

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

R.H.S. = sin 90o = 1

R.H.S. = L.H.S.

Hence, 2sin 45o cos 45o = sin 90o

Question.12: If A = 45o, verify that:

(i) sin 2A = 2sin A cos A

(ii) cos 2A = 2cos2 A – 1 = 1 – 2sin2 A

Solution:

A = 45o \(\Rightarrow\) 2A = 90o

(i) sin 2A = sin 90o = 1

\(Therefore,\) 2sin A cos A = 2sin 45o cos 45o = \(2\times \frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}}=1\)

Therefore, sin 2A = 2sin A cos A

(ii) cos 2A = cos 90o = 0

2cos2 A – 1 = 2cos2 45o – 1 = \(2\left ( \frac{1}{\sqrt{2}} \right )^{2}-1=1-1=0\)

1 – 2sin2 A = 1 – 2sin2 45o = \(1-2\times \left ( \frac{1}{\sqrt{2}} \right )^{2}=1-1=0\)

\(Therefore, \) cos 2A = 2cos2 A – 1 = 1 – 2sin2 A

Question.13: If A = 30o, verify that:

(i) \(sin2A=\frac{2tanA}{1+tan^{2}A}\)

(ii) \(cos2A=\frac{1-tan^{2}A}{1+tan^{2}A}\)

(iii) \(tan2A=\frac{2tanA}{1-tan^{2}A}\)

Solution:

A = 30o \(\Rightarrow\) 2A = 60o

(i) sin 2A = sin 60o = \(\frac{\sqrt{3}}{2}\)

Also \(\frac{2tanA}{1+tan^{2}A}=\frac{2tan30^{\circ}}{1+tan^{2}30^{\circ}}=\frac{2\times \frac{1}{\sqrt{3}}}{1+\left ( \frac{1}{\sqrt{3}} \right )^{2}}=\frac{\frac{2}{\sqrt{3}}}{1+\frac{1}{3}}\)

= \(\frac{\frac{2}{\sqrt{3}}}{\frac{4}{3}}=\frac{2}{\sqrt{3}}\times \frac{3}{4}=\frac{\sqrt{3}}{2}\)

Hence, \(sin2A=\frac{2tanA}{1+tan^{2}A}\)

(ii) cos 2A = cos 60o = \(\frac{1}{2}\)

Also, \(\frac{1-tan^{2}A}{1+tan^{2}A}=\frac{1-tan^{2}30^{\circ}}{1+tan^{2}30^{\circ}}=\frac{1-\left ( \frac{1}{\sqrt{3}} \right )^{2}}{1+\left ( \frac{1}{\sqrt{3}} \right )^{2}}\)

= \(\frac{\left ( 1-\frac{1}{3} \right )}{\left ( 1+\frac{1}{3} \right )}=\frac{\left ( \frac{2}{3} \right )}{\left ( \frac{4}{3} \right )}=\left ( \frac{2}{3}\times \frac{3}{4} \right )=\frac{1}{2}\)

Hence, \(cos2A=\frac{1-tan^{2}A}{1+tan^{2}A}\)

(iii) tan 2A = tan 60o = \(\sqrt{3}\)

Also, \(\frac{2tanA}{1-tan^{2}A}=\frac{2tan30^{\circ}}{1-tan^{2}30^{\circ}}=\frac{2\times \frac{1}{\sqrt{3}}}{1-\left ( \frac{1}{\sqrt{3}} \right )^{2}}=\frac{2\times \frac{1}{\sqrt{3}}}{1-\frac{1}{3}}\)

= \(\frac{\left ( \frac{2}{\sqrt{3}} \right )}{\left ( \frac{2}{3} \right )}=\left ( \frac{2}{\sqrt{3}}\times \frac{3}{2} \right )=\sqrt{3}\)

Hence, \(tan2A=\frac{2tanA}{1-tan^{2}A}\)

Question.14: If A = 60o and B = 30o, verify that:

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

(ii) cos (A + B) = cos A cos B – sin A sin B

Solution:

A = 60o and B = 30o, (A + B) = (60o + 30o) = 90o

(i) L.H.S. = sin (A + B) = sin 90o = 1

R.H.S. = sin A cos B + cos A sin B

= sin 60o cos 30o + cos 60o sin 30o

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

Hence, sin (A + B) = sin A cos B + cos A sin B

(ii) L.H.S. = cos (A + B) = cos 90o = 0

R.H.S. = cos A cos B – sin A sin B

= cos 60o cos 30o – sin 60o sin 30o

= \(\frac{1}{2}\times \frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}\times \frac{1}{2}=\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}=0\)

Hence, cos (A + B) = cos A cos B – sin A sin B

Question.15: If A = 60o and B = 30o, verify that:

(i) sin (A – B) = sin A cos B – cos A sin B

(ii) cos (A – B) = cos A cos B + sin A sin B

(iii) tan (A – B) = \(\frac{tanA-tanB}{1+tanA.tanB}\)

Solution:

A = 60o and B = 30o, (A – B) = (60o – 30o) = 30o

(i) L.H.S. = sin (A – B) = sin 30o = \(\frac{1}{2}\)

R.H.S. = sin A cos B – cos A sin B

= sin 60o cos 30o – cos 60o sin 30o

= \(\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}-\frac{1}{2}\times \frac{1}{2}\)

= \(\frac{3}{4}-\frac{1}{4}=\frac{2}{4}=\frac{1}{2}\)

Hence, sin (A – B) = sin A cos B – cos A sin B

(ii) L.H.S. = cos (A – B) = cos 30o = \(\frac{\sqrt{3}}{2}\)

R.H.S. = cos A cos B + sin A sin B

= cos 60o cos 30o + sin 60o sin 30o

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

Hence, cos (A – B) = cos A cos B + sin A sin B

(iii) L.H.S. = tan (A – B) = tan 30o = \(\frac{1}{\sqrt{3}}\)

R.H.S. = \(\frac{tanA-tanB}{1+tanA.tanB}\)

= \(\frac{tan60^{\circ}-tan30^{\circ}}{1+tan60^{\circ}.tan30^{\circ}}\)

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

Hence, tan (A – B) = \(\frac{tanA-tanB}{1+tanA.tanB}\)

Question.16: If A and B are acute angles such that tan A = \(\frac{1}{3}\), tan B = \(\frac{1}{2}\) and tan (A + B) = \(\frac{tanA-tanB}{1-tanA.tanB}\), show that A + B = 45o.

Solution:

tan (A + B) = \(\frac{tanA+tanB}{1-tanA.tanB}\)

tan (A + B) = \(\frac{\left ( \frac{1}{3}+\frac{1}{2} \right )}{1-\frac{1}{3}\times \frac{1}{2}}\) [\(Since, tanA=\frac{1}{3},tanB=\frac{1}{2}\)]

= \(\frac{\frac{5}{6}}{\frac{5}{6}}=\frac{5}{6}\times \frac{6}{5}=1\)

tan (A + B) = 1 \(\Rightarrow\) tan (A + B) = tan 45o

Hence, (A + B) = 45o

Question.17: Using the formula, tan 2A = \(\frac{2tanA}{1-tan^{2}A}\), find the value of tan 60o, it being given that tan 30o = \(\frac{1}{\sqrt{3}}\).

Solution:

Given: tan 2A = \(\frac{2tanA}{1-tan^{2}A}\) and tan 30o = \(\frac{1}{\sqrt{3}}\)

Let A = 30o \(\Rightarrow\) 2A = 60o

\(\Rightarrow\) tan 60o = tan 2A = \(\frac{2tanA}{1-tan^{2}A}\)

= \(\frac{2\times \frac{1}{\sqrt{3}}}{1-\left ( \frac{1}{\sqrt{3}} \right )^{2}}\)

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

= \(\frac{2}{\sqrt{3}}\times \frac{3}{2}=\sqrt{3}\)

Hence, tan 60o = \(\sqrt{3}\)

Question.18: Using the formula, cos A = \(\sqrt{\frac{1+cos2A}{2}}\), find the value of cos 30o, it being given that cos 60o = \(\frac{1}{2}\).

Solution:

Given: cos A = \(\sqrt{\frac{1+cos2A}{2}}\) and cos 60o = \(\frac{1}{2}\)

Let, A = 30o \(\Rightarrow\) 2A = 60o

\(\Rightarrow\) cos 30o = cos A = \(\sqrt{\frac{1+cos2A}{2}}\)

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

= \(\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}\)

Hence, cos 30o = \(\frac{\sqrt{3}}{2}\)

Question.19: Using the formula, sin A = \(\sqrt{\frac{1-cos2A}{2}}\), find the value of sin 30o, it being given that cos 60o = \(\frac{1}{2}\).

Solution:

Given: sin A = \(\sqrt{\frac{1-cos2A}{2}}\) and cos 60o = \(\frac{1}{2}\)

Let, A = 30o \(\Rightarrow\) 2A = 60o

\(\Rightarrow\) sin 30o = sin A = \(\sqrt{\frac{1-cos2A}{2}}\)

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

= \(\sqrt{\frac{1}{4}}=\frac{1}{2}\)

Hence, sin 30o = \(\frac{1}{2}\)

Question.20: In the adjoining figure, \(\Delta\)ABC is a right-angled triangle in which \(\angle\)A = 30o and AC = 20 cm. Find

(i) BC,

(ii) AB

https://lh3.googleusercontent.com/7Swuh8zb2nHAeZfePuqx57zWPLx3pXcU5ieIrQ0nLJ088rYIKzr53GtPracVQCy8jsSmiUzabrYLpKrcydAHHnAsS40YpHLARiM1PyRb9CYFHjxPj5vUqHyLAYzxco0BdLnH5W04ZcfZlraS0w

Solution:

From right angled \(\Delta\)ABC,

We have \(\frac{BC}{AC}\) = sin 30o

\(\Rightarrow\) \(\frac{BC}{20}=\frac{1}{2}\Rightarrow BC=10cm\)

By Pythagoras theorem,

(AB)2 = (AC)2 – (BC)2

\(\Rightarrow\) AB = \(\sqrt{(AC)^{2}-(BC)^{2}}\)

\(\Rightarrow\) AB = \(\sqrt{(20)^{2}-(10)^{2}}\)

\(\Rightarrow\) AB = \(\sqrt{300}=10\sqrt{3}cm\)

Hence, BC = 10 cm and AB = \(10\sqrt{3}cm\)

Question.21: In the adjoining figure, \(\Delta\)ABC is a right-angled triangle at B and \(\angle\)A = 30o. If BC = 6 cm, find

(i) AB,

(ii) AC

https://lh3.googleusercontent.com/qr1h_GFMVLmx-kd3b9K7XfgqURkiIJTbOAQzuZ6G5Zor4FG966CD1soSWebD8TCK-7jIHZsLuAAgvcdYy8XhlqhmFaS6F06Sg7GVjkjyntmIWAzstfaV-xlCIXAekNy154Fxi2lZVz7X8ZZOVQ

Solution:

We have \(\frac{BC}{AC}\) = sin 30o

\(\Rightarrow\) \(\frac{6}{AC}=\frac{1}{2}\)

\(\Rightarrow\) AC = 12 cm

By Pythagoras theorem,

(AB)2 = (AC)2 – (BC)2

\(\Rightarrow\) AB = \(\sqrt{(AC)^{2}-(BC)^{2}}\)

\(\Rightarrow\) AB = \(\sqrt{(12)^{2}-(6)^{2}}\)

\(\Rightarrow\) AB = \(\sqrt{144-36} \)

\(\Rightarrow\) AB = \(\sqrt{108}=6\sqrt{3}cm\)

Hence, AB = \(6\sqrt{3}cm\) and AC = 12 cm

Question.22: In the adjoining figure, \(\Delta\)ABC is a right-angled triangle at B and \(\angle\)A = 45o. If AC = \(3\sqrt{2}\) cm, find

(i) BC,

(ii) AB.

https://lh3.googleusercontent.com/H0zbVRuxTe85b1mP-Usf20tCisM4DbAkBQqpby127BWiMeikrINSAdFte9ouNj_HsKi31qRlXtqEZdQ3pCzRxwbP8bULfm2ocqkkyd-93pOSIR0YQtAfnQUQcupZcOFPMtQlfpy2MrAsmYRoIQ

Solution:

(i) \(\frac{BC}{AC}\) = sin 45o

\(\Rightarrow\) \(\frac{BC}{3\sqrt{2}}=\frac{1}{\sqrt{2}}\)

\(\Rightarrow\) BC = 3

(ii) By Pythagoras theorem:

(AB)2 = (AC)2 – (BC)2 = \(\sqrt{(3\sqrt{2})^{2}-(3)^{2}}\)

\(\Rightarrow\) AB = \(\sqrt{18-9}=\sqrt{9}=3cm\)

Hence, (i) BC = 3 cm and (ii) AB = 3 cm

Question.23: If sin (A + B) = 1 and cos (A – B) = 1, 0o \(\leq\) (A + B) \(\leq\) 90o and A > B then find A and B.

Solution:

sin (A + B) = 1 \(\Rightarrow\) sin (A + B) = sin 90o

\(\Rightarrow\) A + B = 90o – – – – – – – (1)

cos (A – B) = 1 \(\Rightarrow\) cos (A – B) = cos 0o

\(\Rightarrow\) A – B = 0o – – – – – – – – – (2)

Adding (1) and (2), we get 2A = 90o \(\Rightarrow\) A = 45o

Putting A = 45o in (1) we get

45o + B = 90o \(\Rightarrow\) B = 45o

Hence, A = 45o and B = 45o

Question.24: If sin (A – B) = \(\frac{1}{2}\) and cos (A + B) = \(\frac{1}{2}\), 0o < (A + B) < 90o and A > B then find A and B.

Solution:

sin (A – B) = \(\frac{1}{2}\) \(\Rightarrow\) sin (A – B) = sin 30o

\(\Rightarrow\) A – B = 30o – – – – – – – (1)

cos (A + B) = \(\frac{1}{2}\) \(\Rightarrow\) cos (A + B) = cos 60o

\(\Rightarrow\) A + B = 60o – – – – – – – – – (2)

Solving (1) and (2), we get 2A = 90o \(\Rightarrow\) A = 45o

Putting A = 45o in (1) we get

45o – B = 30o \(\Rightarrow\) B = 15o

Hence, A = 45o and B = 15o

Question.25: If tan (A – B) = \(\frac{1}{\sqrt{3}}\) and tan (A + B) = \(\sqrt{3}\), 0o < (A + B) < 90o and A > B then find A and B.

Solution:

tan (A – B) = \(\frac{1}{\sqrt{3}}\) \(\Rightarrow\) tan (A – B) = tan 30o

\(\Rightarrow\) A – B = 30o – – – – – – – (1)

tan (A + B) = \(\sqrt{3}\) \(\Rightarrow\) tan (A + B) = tan 60o

\(\Rightarrow\) A + B = 60o – – – – – – – – (2) [tan 60o = \(\sqrt{3}\)]

Solving (1) and (2), we get 2A = 90o \(\Rightarrow\) A = 45o

Putting A = 45o in (1) we get

45o – B = 30o \(\Rightarrow\) B = 15o

Hence, A = 45o and B = 15o

Question.26: If 3x = cosec \(\theta\) and \(\frac{3}{x}\) = cot \(\theta\), find the value of \(3\left ( x^{2}-\frac{1}{x^{2}} \right )\).

Solution:

\(3(x^{2} – \frac{1}{x^{2}})\)

\(= \frac{9}{3}(x^{2} – \frac{1}{x^{2}})\)

\(= \frac{1}{3}(9x^{2} – \frac{9}{x^{2}})\)

\(= \frac{1}{3}\left [ (3x)^{2} – \left ( \frac{3}{x} \right )^{2} \right ]\)

\(= \frac{1}{3}\left [ (cosec \; \theta)^{2} – \left ( \cot \theta \right )^{2} \right ]\)

\(= \frac{1}{3}\left [ (cosec ^{2}\; \theta – \cot ^{2} \theta ) \right ]\)

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

Question.27: If sin (A + B) = sin A cos B + cos A sin B and cos (A – B) = cos A cos B + sin A sin B, find the values of

(i) sin 75o and

(ii) cos 15o

Solution:

Let A = \(45^{\circ}\) and B = \(30^{\circ}\)

(i) As, \(\sin (A + B) = \sin A \cos B + \cos A \sin B\)

\(\Rightarrow \sin (45^{\circ} + 30 ^{\circ}) = \sin 45^{\circ} \cos 30 ^{\circ} + \cos 45^{\circ} \sin 30 ^{\circ}\)

\(\Rightarrow \sin (75 ^{\circ}) = \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \times \frac{1}{2}\)

\(\Rightarrow \sin (75 ^{\circ}) = \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2 \sqrt{2}}\)

\(Therefore, \sin (75 ^{\circ}) = \frac{\sqrt{3} + 1}{2\sqrt{2}}\)

(ii) As, \(\cos (A – B) = \cos A \cos B + \sin A \sin B\)

\(\Rightarrow \cos (45^{\circ} – 30 ^{\circ}) = \cos 45^{\circ} \cos 30 ^{\circ} + \sin 45^{\circ} \sin 30 ^{\circ}\)

\(\Rightarrow \cos (15 ^{\circ}) = \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \times \frac{1}{2}\)

\(\Rightarrow \cos (15 ^{\circ}) = \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2 \sqrt{2}}\)

\(Therefore, \cos (75 ^{\circ}) = \frac{\sqrt{3} + 1}{2\sqrt{2}}\)


Practise This Question

A wheel of radius 4π cm touches a flat surface, and the corresponding point of contact is A. Now the wheel rolls on the flat surface till the point A lies vertically above the corresponding point of contact in this position as shown. What is the distance covered by the centre of the wheel?