Rs Aggarwal 2015 Solutions for Class 10 Math Chapter 6 T Ratios Of Some Particular Angles are provided here with simple step-by-step explanations. These solutions for T Ratios Of Some Particular Angles are extremely popular among Class 10 students for Math T Ratios Of Some Particular Angles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2015 Book of Class 10 Math Chapter 6 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2015 Solutions. All Rs Aggarwal 2015 Solutions for class Class 10 Math are prepared by experts and are 100% accurate.

Page No 288:

Question 1:

sin 60° cos 30° + cos 60° sin 30°

Answer:

On substituting the values of various T-ratios, we get:
 sin 60o cos 30o + cos 60o sin 30o 
 =32×32+12×12=34+14=44=1

Page No 288:

Question 2:

cos 60° cos 30° − sin 60° sin 30°

Answer:

On substituting the values of various T-ratios, we get:
 cos 60o cos 30o − sin 60o sin 30o 
 =12×32-32×12=34-34=0

Page No 288:

Question 3:

cos 45° cos 30° + sin 45° sin 30°

Answer:

On substituting the values of various T-ratios, we get:
 cos 45o cos 30o + sin 45o  sin 30o 
 = 12×32 + 12×12 = 322 + 122 =3 +122

Page No 288:

Question 4:

2 cos2 60° + 3 sin2 45° − 3 sin2 30° + 2 cos2 90°

Answer:

On substituting the values of various T-ratios, we get:
 2 cos2 60o + 3 sin2 45o − 3 sin2 30o + 2 cos2 90o
 =2×122+3×122-3×122 + 2×02=2×14+3×12-3×14+0= 12+32-34=2+6-34=54

Page No 288:

Question 5:

cot230° − 2cos230° − 34 sec245° + 14 cosec230°

Answer:

On substituting the values of various T-ratios, we get:
  cot2 30o − 2 cos2 30o − 34 sec2 45o + 14 cosec2 30o
 = 32-2×322-34×22+14×22= 3-2×34-34×2+14×4= 3-32-32+1 = 4-32+32 = 4-3= 1

Page No 288:

Question 6:

(sin230° + 4cot245° − sec260°)(cosec245° sec230°)

Answer:

On substituting the values of various T-ratios, we get:
  (sin2 30o + 4 cot2 45o − sec2 60o )(cosec2 45o sec2 30o)
 =122+4×12-22 22 232=14+4-4 2×43 =14×83=23

Page No 288:

Question 7:

4cot230°+1sin230°-2cos245°-sin20°

Answer:

On substituting the values of various T-ratios, we get:
 
 4cot2 30o+1sin2 30o-2 cos2 45o-sin2 0o=432+1122-2×122-02=43+114-2×12-0=43+4-1 =43+3=4+93=133 

Page No 288:

Question 8:

tan260°+4cos245°+3cosec260°+2cos290°2cosec30°+3sec60°-73cot230°

Answer:

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

   tan260o+4cos245o+3 cosec260o+2 cos290o2 cosec30o+3 sec60o-73cot230o=32+4×122+3×232+2×022×2+3×2-73×32=3+4×12+3×43+04+6-73×3=3+2+44+6-7=93=3   

Page No 288:

Question 9:

sin30°cos45°+cot45°sec60°-sin60°tan45°-cos30°sin90°

Answer:

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

  sin 30ocos 45o+cot 45osec 60o-sin 60otan 45o-cos 30osin 90o= 1212+12-321-321= 22+12-32-32= 2+1-232

Page No 288:

Question 10:

Show that:
(i) 1-sin60°cos60°=tan60°-1tan60°+1
(ii) cos30°+sin60°1+sin30°+cos60°=cos30°

Answer:

(i)
 LHS=1-sin 60ocos 60o=1-3212=2-3212=2-32×2=2-3RHS= tan 60o-1tan 60o+1=3-13+1=3-13+1×3 -13 -1=3-1232-12=3+1-233-1=4-232=2-3

Hence, LHS = RHS

 1-sin 60ocos 60o=tan 60o-1tan 60o+1

(ii)
 LHS = cos 30o+sin 60o1+sin 30o+cos 60o=32+32 1+12+12 =3+322+1+12=32Also, RHS = cos 30o=32

Hence, LHS = RHS

   cos 30o+sin 60o1+sin 30o+cos 60o=cos 30o1sin60°cos60°

Page No 288:

Question 11:

Verify each of the following:
(i) sin 60° cos 30° − cos 60° sin 30° = sin 30°
(ii) cos 60° cos 30° + sin 60° sin 30° = cos 30°
(iii) 2 sin 30° cos 30° = sin 60°
(iv) 2 sin 45° cos 45° = sin 90°

Answer:

(i) sin 60o cos 30o − cos 60o sin 30o
 =32×32-12×12=34-14=24=12Also, sin 30o=12
∴ sin 60o cos 30o − cos 60o sin 30o = sin 30o

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

=12×32+32×12=34+34=32Also, cos 30o =32
∴ cos 60o cos 30o + sin 60o sin 30o = cos 30o

(iii) 2 sin 30o cos 30o
=2×12×32=32Also, sin 60o =32
∴ 2 sin 30o cos 30o = sin 60o

(iv) 2 sin 45o cos 45o
=2×12×12=1
Also, sin 90o = 1
∴ 2 sin 45o cos 45o = sin 90o



Page No 289:

Question 12:

If A = 45°, verify that:
(i) sin 2A = 2 sin A cos A
(ii) cos 2A = 2 cos2 A − 1 = 1 − 2 sin2 A

Answer:

A = 45o
⇒ 2A = 2 × 45o = 90o

(i) sin 2A = sin 90o = 1
2 sin A cos A = 2 sin 45o cos 45o 2×12×12 = 1
∴ sin 2A = 2 sin A cos A

(ii) cos 2A = cos 90o = 0
2 cos2 A − 1 = 2 cos2 45o − 1 =  2×122 - 1 = 2×12 -1 = 1-1 = 0
Now, 1 − 2 sin2 A1-2×122 = 1 - 2×12  =1 - 1 = 0
∴ cos 2A 2 cos2 A − 1 = 1 − 2 sin2 A

Page No 289:

Question 13:

If A = 30°, verify that:
(i) sin2A=2tanA1+tan2A
(ii) cos2A=1-tan2A1+tan2A
(iii) tan2A=2tanA1-tan2A

Answer:

A = 30o
⇒ 2A = 2 × 30o = 60o

​(i) sin 2A = sin 60o = 32
2 tan A1+tan2 A=2 tan 30o1+tan2 30o=2×131+132 =231 + 13=2343=23×34=32
sin 2A=2tan A1+tan2A

(ii) cos 2A = cos 60o = 12

1-tan2 A1+tan2 A=1-tan2 30o1+tan2 30o=1-1321+132 =1 - 131 + 13=2343=23×34=12
cos 2A=1-tan2 A1+tan2 A

(iii) tan 2A = tan 60o = 3
2 tan A1-tan2 A=2 tan 30o1-tan2 30o=2×131-132 =231-13=2323=23×32=3
tan 2A=2 tan A1-tan2 A=2tanA1+tan2A

Page No 289:

Question 14:

Using the formula, cos A=1+cos2A2 , find the value of cos 30°, it being given that cos 60° = 12.

Answer:

A = 30o
⇒ 2A = 2 × 30o = 60o

By substituting the value of the given T-ratio, we get:
cos A=1+cos 2A2 cos 30o=1+ cos 60o2=1+122=322  =34=32
∴ cos 30o = 32

Page No 289:

Question 15:

Using the formula, sin A=1-cos2A2 , find the value of sin 30°, it being given that cos 60° = 12.

Answer:

A = 30o
⇒ 2A = 2 × 30o = 60o

By substituting the value of the given T-ratio, we get:

sin A=1-cos 2A2sin 30o=1-cos 60o2=1-122=122  =14=12
∴ sin 30o = 12

Page No 289:

Question 16:

Using the formula, tan 2 A=2tanA1-tan2A , find the value of tan 60°, it being given that tan 30° = 13.

Answer:

A = 30o
⇒ 2A = 2 × 30o = 60o

By substituting the value of the given T-ratio, we get:
tan 2A = 2 tan A1-tan2 A tan 60o=2 tan 30o1-tan2 30o=2×131-132=231-13=2323=23×32=3
∴ tan 60o = 3

Page No 289:

Question 18:

If A = 60° and B = 30°, 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

Answer:

A = 60o and B = 30o
Now, A + B = 60o + 30o​ = 90o
Also, A − B = 60o − 30o = 30o

(i) sin (A + B) = sin 90o = 1
sin A cos B + cos A sin B = sin 60o cos 30o + cos 60o sin 30o
32×32+12×12=34 + 14=1
∴ sin (A + B) = sin A cos B + cos A sin B


(ii) cos (A + B) = cos 90o = 0
cos A cos B − sin A sin B =cos 60o cos 30o − sin 60o sin 30o  =12×32 - 32×12 = 34 - 34 = 0
∴​ cos (A + B) = cos A cos B − sin A sin B
 

Page No 289:

Question 19:

If A and B are acute angles such that tan A = 13, tan B12 and tan (A + B) = tan A + tan B1-tan A tan B , show that A + B = 45°.

Answer:

Given:
tan A = 13 and tan B= 12tan (A+B) = tan A+tan B1-tan A tan BOn substituting these values in RHS of the expression, we get:tan A+tan B1-tan A tan B=13+121-13×12=561-16=5656=1 tan (A+B)=1=tan 45o       [ tan 45o =1]A+B=45o

tan A = 13 and tan B = 12By substituting the values, we get;tan (A+ B) = tan A + tan B1- tan A tan B = 13 +121-13×12 = 561-16 = 5656 = 1tan (A + B) = 1 = tan 45oHence, A + B = 4513

Page No 289:

Question 20:

In the adjoining figure, ∆ABC is a right-angled triangle in which ∠B = 90°, ∠A = 30° and AC = 20 cm.
Find (i) BC, (ii) AB.
Figure

Answer:

From the given right-angled triangle, we have:
BCAC = sin 30oBC20 = 12 BC = 202 = 10 cm Also, ABAC = cos 30oAB20 = 32 AB = 20×32 = 103  cm BC = 10 cm and AB = 103  cm

Page No 289:

Question 17:

If A = B = 45°, verify that:
(i) sin (A + B) = sin A cos B + cos A sin B
(ii) sin (A B) = sin A cos B − cos A sin B
(iii) cos (A + B) = cos A cos B − sin A sin B
(iv) cos (AB) = cos A cos B + sin A sin B

Answer:

A = B = 45o
Now, A + B = 45o + 45o​ = 90o
Also, AB = 45o − 45o = 0o

(i) sin (A + B) = sin 90o = 1
sin A cos B + cos A sin B = sin 45o cos 45o + cos 45o sin 45o
= 12×12 + 12×12 = 12 + 12 = 1
∴ sin (A + B) = sin A cos B + cos A sin B

(ii) sin (AB) = sin 0o = 0
sin A cos B − cos A sin B = sin 45o cos 45o − cos 45o sin 45o
= 12×12 - 12×12 = 12 - 12 = 0
∴ sin (AB) = sin A cos B − cos A sin B

(iii) cos (A + B) = cos 90o = 0
cos A cos B − sin A sin B = cos 45o cos 45o − sin 45o sin 45o
= 12×12 - 12×12 = 12 - 12 = 0
∴​ cos (A + B) = cos A cos B − sin A sin B

(iv) cos (AB) = cos 0o = 1
cos A cos B + sin A sin B = cos 45o cos 45o + sin 45o sin 45o
= 12×12 + 12×12 = 12 + 12 = 1
∴​ cos (AB) = cos A cos B + sin A sin B



Page No 290:

Question 21:

In the adjoining figure, ∆ ABC is a right-angled at B and ∠A = 30°. If BC = 6 cm,
Find (i) AB, (ii) AC.
Figure

Answer:

From the given right-angled triangle, we have:

BCAB=tan 30o6AB=13  AB=63 cmAlso, BCAC=sin 30o6AC=12  AC=2×6=12 cmAB = 63 cm and AC = 12 cm

Page No 290:

Question 22:

In the adjoining figure, ∆ABC is a right-angled at B and ∠A = 45°. If AC32 cm,
find (i) BC, (ii) AB.
Figure

Answer:

From right-angled ∆ABC, we have:

    BCAC=sin 45oBC32=12  BC=3 cm  Also, ABAC=cos 45oAB32=12  AB=3 cm BC=3 cm and AB=3 cm

Page No 290:

Question 23:

If sin (A + B) = 1 and cos (A − B) = 1, 0° ≤ (A + B) ≤ 90° and A > B, then find A and B.

Answer:

Here, sin (A + B) = 1
⇒ sin (A+ B) = sin 90o                     [∵ sin 90o = 1]
A + B = 90o​                       ...(i)

Also, cos (AB) = 1
⇒​ cos (A − B) = cos 0o                    [∵​ cos 0o = 1]
A B = 0o                        ...(ii)

Solving (i) and (ii), we get:
A = 45o and B = 45o

Page No 290:

Question 24:

If sin (AB) = 12 and cos (A + B) = 12, 0° < (A + B) < 90° and A > B, then find A and B.

Answer:

Here, sin (A − B) = 12 
⇒ sin (A B) = sin 30o                [∵ sin 30o = 12]
A − B = 30o​                                  ...(i)

Also, cos (A + B) = 12
⇒​ cos (A + B) =  cos 60o              [∵​ cos 60o = 12]
A + B = 60o                                 ...(ii)

Solving (i) and (ii), we get:
A = 45o and B = 15o

Page No 290:

Question 25:

If tan (A − B) = 13 and tan (A + B) = 3, 0° < (A + B) < 90° and A > B, then find A and B.

Answer:

Here, tan (A − B) = 13 
⇒ tan (A B) = tan 30o       [∵ tan 30o = 13]
A − B = 30o                     ...(i)

Also, tan (A + B) = 3 
⇒​ tan (A + B) =  tan 60o        [∵ tan 60o = 3]
A + B = 60o                           ...(ii)  

Solving (i) and (ii), we get:
A = 45o and B = 15o

Page No 290:

Question 26:

Find the value of sin 30° geometrically.

Answer:

Consider an equilateral ∆ABC with each side equal to 2a.
Then, each angle of ​∆ABC = 60o
From A, draw ADBC.
Figure
Then, it is clear that BD = DC = a
Also, ∠ADB =  90o
Now, in  ​∆ADB, we have:
ADB + ∠DBA + ∠BAD = 180o           (Angle sum property)
⇒ 90o + 60o + ∠BAD = 180o
⇒ ∠BAD = 180o − 150o
⇒ ∠BAD = 30o
∴ sin 30oBDAB = a2a = 12

Page No 290:

Question 27:

Find the value of sin 60° geometrically.

Answer:

Consider an equilateral ∆ABC with each side equal to 2a.
Then, each angle of ​∆ABC = 60o
From A, draw ADBC.
Figure
Then, it is clear that BD = DC = a
Also, ∠ADB =  90o
Now, in  ​∆ ADB, we have:
ADB + ∠DBA + ∠BAD = 180o       (angle sum property)
⇒ 90o + 60o + ∠ BAD = 180o
⇒ ∠BAD = 180o − 150o = 30o
∴ ∠BAD = 30o

In ∆ADB, by Pythagoras theorem, we have:
AB2=AD2+BD2AD=AB2-BD2=2a2-a2=4a2 - a2=3a
∴ sin 60oADAB = 3a2a = 32



Page No 291:

Question 1:

(sec260°− 1) = ?
(a) 2
(b) 3
(c) 4
(d) 0

Answer:

(b) 3

sec260o − 1 = (2)2 − 1 = 4 − 1 = 3

Page No 291:

Question 2:

If x tan 45° cos 60° = sin 60° cot 60°, then x = ?
(a) 1
(b) 12
(c) 12
(d) 3

Answer:

(a) 1

     x tan 45o cos 60o = sin 60o cot 60ox 1 12=3213x 12=12x=1

Page No 291:

Question 3:

If (tan245° − cos230°) = x sin 45° cos 45°, then x = ?
(a) 2
(b) −2
(c) 12
(d) -12

Answer:

(c) 12

tan2 45o - cos2 30o = x sin 45o cos 45ox=tan2 45o-cos2 30o sin 45o cos 45o=12-32212×12=1-3412=1412=14×2=12

Page No 291:

Question 4:

If tan x = 3 cot x, then x = ?
(a) 45°
(b) 60°
(c) 30°
(d) 15°

Answer:

(b) 60o

     tan x=3 cot xtan xcot x=3 tan2 x=3               cot x=1tan x tan x=3=tan 60o x= 60o

Page No 291:

Question 5:

If 2 sin 2θ = 3, then θ = ?
(a) 30°
(b) 45°
(c) 60°
(d) 90°

Answer:

(a) 30o

    2 sin 2θ=3sin 2θ=32=sin 60osin 2θ=sin 60o2θ=60oθ=30o

Page No 291:

Question 6:

If 2cos 3θ = 1, then θ = ?
(a) 10°
(b) 15°
(c) 20°
(d) 30°

Answer:

(c) 20o

     2 cos 3θ=1 cos 3θ=12cos 3θ=cos 60o                    cos 60o=12 3θ=60oθ=60o3=20o

Page No 291:

Question 7:

If 3 tan 2θ − 3 = 0, then θ = ?
(a) 15°
(b) 30°
(c) 45°
(d) 60°

Answer:

(b) 30o

3tan 2θ -3=03tan 2θ =3tan 2θ=33tan 2θ=3         [tan 60o =3 ]tan 2θ=tan 60o2θ=60oθ=30o 

Page No 291:

Question 8:

If sin (A + B) = 1 and cos (A − B) = 32, where A > B, then B = ?
(a) 30°
(b) 45°
(c) 60°
(d) 90°

Answer:

(a) 30o

sin (A+B)=1=sin 90o                       [sin 90o=1] A+B=90o               ...(i)cos (A-B)=32 = cos 30o             cos 30o=32 A-B=30o               ...(ii)Solving (i) and (ii) we get: A = 60o and B = 30o B = 30o



Page No 292:

Question 9:

If tan (A + B) = 3 and tan (AB) = 13, where A ≥ B and (A + B) is acute, then A = ?
(a) 15°
(b) 30°
(c) 45°
(d) 60°

Answer:

(c) 45o

     tan (A +B)=3 =tan 60o              tan 60o = 3 A+B=60o                       ...(1)     tan (A- B)=13=tan 30o              tan 30o = 13  A-B=30o                       ...(2)     Solving (1) and (2), we get: A=45o and B=15o A = 45o

Page No 292:

Question 10:

In ∆ ABC, ∠B = 90°, AB = 3 cm and AC = 6 cm.
Then, ∠A = ?
(a) 30°
(b) 60°
(c) 45°
(d) None of these
Figure

Answer:

(b) 60o

In right-angled ∆ ABC, we have:
cos A=ABAC=36=12=cos 60o A=60o

Page No 292:

Question 11:

In ∆ABC, ∠B = 90°, ∠A = 30° and AB = 9 cm.
Then, BC = ?
(a) 3 cm
(b) 23 cm
(c) 33 cm
(d) 6 cm
Figure

Answer:

(c ) 33 cm

In right-angled ∆ ABC, we have:
BCAB=tan 30o=13 BC=AB3=93=33 cm BC=33 cm

Page No 292:

Question 12:

The value of (cos 60° cos 30° − sin 60° sin 30°) is
(a) 0
(b) 32
(c) 12
(d) 1

Answer:

(a) 0

cos 60o cos 30o - sin 60o sin 30o=12×32-32×12=34-34=0

Page No 292:

Question 13:

(sin 60° cos 30° − cos 60° sin 30°) = ?
(a) 0
(b) 1
(c) 12
(d) 32

Answer:

(c)12

sin 60o cos 30o - cos 60o sin 30o=32×32-12×12=34-14=24=12

Page No 292:

Question 14:

(tan 30° cosec 60° + tan 60° sec 30°) = ?
(a) 213
(b) 223
(c) 313
(d) 123

Answer:

(b ) 223

tan 30o cosec 60o + tan 60o sec 30o = 13×23 + 3×23 = 23 + 2 = 223

Page No 292:

Question 15:

(cos 0° + sin 30° + sin 45°) (sin 90° + cos 60° − cos 45°) = ?
(a) 56
(b) 58
(c) 35
(d) 74

Answer:

(d) 74

cos 0o+sin 30o+sin 45osin 90o+cos 60o-cos 45o= 1+12+121+12-12=32+1232-12=322-122=94 -12=9-24=74

Page No 292:

Question 16:

(sin230° + 4cot245° − sec260°) = ?
(a) 0
(b) 14
(c) 4
(d) 1

Answer:

(b) 14

sin2 30o+4 cot2 45o-sec2 60o=122+4×12-22=14+4-4=14

Page No 292:

Question 17:

(3cos260° + 2cot230° − 5sin245°) = ?
(a) 136
(b) 174
(c) 1
(d) 4

Answer:

(b) 174

3 cos2 60o+2 cot2 30o-5 sin2 45o = 3×122+2×32-5×122 = 34+6-52=3+24-104=174

Page No 292:

Question 18:

cos230°cos245°+4sec260°+12cos290°-2tan260°=?
(a) 738
(b) 758
(c) 818
(d) 838

Answer:

(d) 838

cos2 30o cos2 45o+4 sec2 60o+12 cos2 90o-2 tan2 60o= 322×122+4×22+12×02-2×32= 34×12+16-6=38+10=3+808=838



Page No 293:

Question 19:

sin230°cos245°+4tan230°+12sin290°+18cot260°=?
(a) 2
(b) 3
(c) 4
(d) 52

Answer:

(a) 2


sin2 30o cos2 45o+4 tan2 30o+12 sin2 90o+18cot2 60o = 122×122+4×132+12×12+18×132= 14×12+43+12+124 = 18+43+12+124=3+32+12+124=4824=2

Page No 293:

Question 20:

43cot230°+3sin260°-2cosec260°-34tan230°=?
(a) 3
(b) 83
(c) 103
(d) 94

Answer:

(c) 103

43cot2 30o +3 sin2 60o-2 cosec2 60o-34tan2 30o = 43×32+3×322-2×232-34×132= 4+94- 83-14 = 48+27-32-312=75-3512=4012=103

Page No 293:

Question 21:

tan260°+4sin245°+3sec230°+5cos290°cosec30°+sec60°-cot230°=?
(a) 97
(b) 9
(c) 79
(d) 6

Answer:

(b) 9


   tan2 60o+4 sin2 45o+3 sec2 30o+5 cos2 90o cosec 30o+sec 60o-cot2 30o= 32+4×122+3×232+5×022+2-32= 3+2+4 4-3=9   

Page No 293:

Question 22:

14cot430°-cosec460°+32sec245°-tan230°-5cos260°=?
(a) 359

(b) 458

(c) 5518

(d) 4936

Answer:

(c) 5518

14cot4 30o-cosec4 60o+32sec2 45o-tan2 30o-5 cos2 60o=1434-234+3222-132-5122=149-169+322-13-54=1481-169+326-13-54=6536+156-54=65+90-4536=11036=5518

Page No 293:

Question 23:

4sin430°+cos460°-23sin260°-cos245°+12tan260°=?
(a) 83
(b) 116
(c) 136
(d) 138

Answer:

(b) 116

4sin4 30o+cos4 60o-23sin2 60o-cos2 45o+12tan2 60o=4124+124-23322-122+12×32=4116+116-2334-12+32=4×18-23×14+32=12-16+32=3-1+96=116

Page No 293:

Question 24:

23cos430°-sin445°-3sin260°-sec245°+14cot230°=?
(a) 5312
(b) 7324
(c) 11324
(d) 8312

Answer:

(c) 11324


23cos4 30o-sin4 45o-3sin2 60o-sec2 45o+14cot2 30o=23324-124-3322-22+14×32=23916-14-334-2+34=23×516-3×-54+34=524+154+34=5+90+1824=11324

Page No 293:

Question 25:

If sin α = 12 and cos β = 12, then (α + β) = ?
(a) 0°
(b) 30°
(c) 60°
(d) 90°

Answer:

(d) 90o

sinα = 12 = sin 30o α = 30ocos β = 12 =cos 60o β =60o  α + β = 30o + 60o = 90o

Page No 293:

Question 26:

If (sin θ − cos θ) = 0, then (sin4 θ + cos4 θ) = ?
(a) 1
(b) 12
(c) 14
(d) 34

Answer:

(b) 12

sin θ-cos θ=0 sin θ=cos θsin θcos θ=1 tan θ=1 θ = 45osin4 θ + cos4 θ =sin4 45o + cos4 45o =124+124=14+14=12

Page No 293:

Question 27:

If ∆ABC is right-angled at C, then cos (A + B) = ?
(a) 0
(b) 12
(c) 1
(d) 32

Answer:

(a) 0

A + ∠B + ∠C = 180o   (by angle sum property)
It is given that ∠C = 90o.
∴ ∠A + ∠B = 90o
cos (A + B) = cos (90o) = 0

Page No 293:

Question 28:

A pole 6 m high casts a shadow 23 m long on the ground. The sun's altitude is
(a) 30°
(b) 45°
(c) 60°
(d) 90°

Answer:

(c) 60o

Let AB be the pole and AC be its shadow.
 Then, AB = 6 m and AC = 23 mLet ACB = θoThen, tan θ = ABAC = 6 23 = 3 = tan 60o θ = 60o



View NCERT Solutions for all chapters of Class 10