Frank Solutions for Chapter 26 Trigonometrical Ratios of Standard Angles Class 9 Mathematics ICSE
Exercise 27.1
1. Without using tables, evaluate the following:
(i) sin 60° sin 30° + cos 30° cos 60°
(ii) sec 30° cosec 60° + cos 60° sin 30°
(iii) sec 45° sin 45° – sin 30° sec 60°
(iv) sin2 30° sin245° + sin2 60° sin2 90°
(v) tan2 30° + tan2 60° + tan2 45°
(vi) sin2 30° cos2 45° + 4 tan2 30° + sin2 90° + cos2 0°
(vii) cosec2 45° sec2 30° – sin2 30° – 4 cot2 45° + sec2 60°
(viii) cosec3 30° cos 60° tan3 45° sin2 90° sec2 45° cot 30°
(ix) (sin 90° + sin 45° + sin 30°) (sin 90° – cos 45° + cos 60°)
(x) 4(sin4 30° + cos4 60°) – 3 (cos2 45° – sin2 90°)
Answer
(i) sin 60° sin 30° + cos 30° cos 60°
sin 60° = (√3/2)
sin 30° = (1/2)
cos 30° = (√3/2)
cos 60° = (1/2)
On substituting, we get,
sin 60° sin 30° + cos 30° cos 60° = (√3/2) ×(1/2) + (√3/2)×(1/2)
= (√3/4) + (√3/4)
We get,
= (√3/2)
(ii) sec 30° cosec 60° + cos 60° sin 30°
cos 30° = (√3/2) ⇒ sec 30° = (2/√3)
sin 60° = (√3/2) ⇒ cosec 60° = (2/√3)
cos 60° = (1/2)
sin 30° = (1/2)
On substituting, we get,
sec 30° cosec 60° + cos 60° sin 30° = (2/√3)× (2/√3) + (1/2)×(1/2)
= (4/3) + (1/4)
= (16 + 3)/12
We get,
= 19/12
(iii) sec 45° sin 45° – sin 30° sec 60°
cos 45° = (1/√2) ⇒ sec 45° = √2
sin 45° = (1/√2)
sin 30° = (1/2)
cos 60° = (1/2) ⇒ sec 60° = 2
sec 45° sin 45° – sin 30° sec 60°
On substituting, we get,
= (√2)×(1/√2) – (1/2)×2
= 1 – 1
We get,
= 0
(iv) sin2 30° sin2 45° + sin2 60° sin2 90°
sin 30° = (1/2)
sin 45° = (1/√2)
sin 60° = (√3/2)
sin 90° = 1
On substituting, we get,
sin2 30° sin2 45° + sin2 60° sin2 90° = (1/2)2 (1/√2)2 + (√3/2)2 (1)2
= (1/4)×(1/2) + (3/4) (1)
= (1/8) + (3/4)
= (1 + 6)/8
We get,
= 7/8
(v) tan2 30° + tan2 60° + tan2 45°
tan 30° = (1/√3)
tan 60° = √3
tan 45° = 1
On substituting, we get,
tan2 30° + tan2 60° + tan2 45° = (1 /√3)2 + (√3)2 + 1
= (1/3) + 3 + 1
= (1 + 9 + 3)/3
We get,
= 13/3
(vi) sin2 30° cos2 45° + 4 tan2 30° + sin2 90° + cos2 0°
sin 30° = (1/2)
cos 45° = (1/√2)
tan 30° = (1/√3)
sin 90° = 1
cos 0° = 1
On substituting, we get,
sin2 30° cos2 45° + 4 tan2 30° + sin2 90° + cos2 0°
= (1/2)2 (1/√2)2 + 4 (1 / √3)2 + 1 + 1
= (1/4) (1/2) + (4/3) + 2
= (1/8) + (4/3) + 2
= (3 + 32 + 48)/24
We get,
= 83/24
(vii) cosec2 45° sec2 30° – sin2 30° – 4 cot2 45° + sec2 60°
sin 45° = (1/√2)
cosec 45° = (√2/1)
sin 30° = cos 60° = (1/2)
sec 60° = 2
cos 30° = (√3/2)
sec 30° = (2/√3)
tan 45° = 1
cot 45° = 1
On substituting, we get,
cosec2 45° sec2 30° – sin2 30° – 4 cot2 45° + sec2 60° = (√2/1)2 (2/√3)2 – (1/2)2 – 4 (1)2 + (2)2
= 2×(4/3) – (1/4) – 4 + 4
= (8/3) – (1/4)
= (32 – 3)/12
We get,
= 29/12
(viii) cosec3 30° cos 60° tan3 45° sin2 90° sec2 45° cot 30°
sin 30° = (1/2)
cosec 30° = 2
cos 60° = (1/2)
sec 60° = 2
cos 45° = (1/√2)
sec 45° = √2
tan 45° = 1
sin 90° = 1
tan 30° = (1/√3)
cot 30° = √3
On substituting, we get,
cosec3 30° cos 60° tan3 45° sin2 90° sec2 45° cot 30°
= (2)3 (1/2) (1)3 (1)2 (√2)2 (√3)
= 8 x (1/2)×2 ×√ 3
We get,
= 8√3
(ix) (sin 90° + sin 45° + sin 30°) (sin 90° – cos 45° + cos 60°)
sin 30° = (1/2)
sin 45° = (1/√2)
sin 90° = 1
cos 45° = (1/√2)
cos 60° = (1/2)
(sin 90° + sin 45° + sin 30°) (sin 90° – cos 45° + cos 60°)
On substituting, we get,
= {(1) + (1/√2) + (1/2)} {(1) – (1/√2) + (1/2)}
= {(3/2) + (1/√2)} {(3/2) – (1/√2)}
= (3/2)2 – (1/√2)2
= (9/4) – (1/2)
We get,
= (9 – 2)/4
= 7/4
(x) 4 (sin4 30° + cos4 60°) – 3 (cos2 45° – sin2 90°)
sin 30° = (1/2)
sin 90° = 1
cos 45° = (1/√2)
cos 60° = (1/2)
On substituting, we get,
4 (sin4 30° + cos4 60°) – 3 (cos2 45° – sin2 90°)
= 4 {(1/2)4 + (1/2)4} – 3 {(1/√2)2 – (1)2}
= 4 {(1/16) + (1/16)} – 3 {(1/2) – 1}
On calculating further, we get,
= 4×(2/16) + 3×(1/2)
= (1/2) + (3/2)
= 4/2
= 2
2. Without using tables, find the value of the following:
(i) (sin 30° – sin 90° + 2 cos 0°)/tan 30° tan 60°
(ii) (sin 30°/sin 45°) + (tan 45°/sec 60°) – (sin 60°/cot 45°) – (cos 30°/sin 90°)
(iii) (tan 45°/cosec 30°) + (sec 60°/cot 45°) – (5 sin 90°/2 cos 0°)
(iv) (tan2 60° + 4 cos2 45° + 3 sec2 30° + 5 cos 90°)/(cosec 30° + sec 60° – cot2 30°)
(v) (4/cot2 30°) + (1/sin2 60°) – cos2 45°
Answer
(i) (sin 30° – sin 90° + 2 cos 0°)/tan 30° tan 60°
= {(1/2) – (1) + 2×1}/(1/√3)×√3
On further calculation, we get,
= {(1/2) – 1 + 2}/1
= (1/2) – 1 + 2
= (1/2) + 1
We get,
= 3/2
(ii) (sin 30°/sin 45°) + (tan 45°/sec 60°) – (sin 60°/cot 45°) – (cos 30°/sin 90°)
= {(1/2)/(1/√2)} + (1/2) – {(√3/2)/1} – {(√3/2)/1}
On calculating further, we get,
= (√2/2) + (1/2) – (√3/2) – (√3/2)
= (√2 + 1 – 2√3)/2
(iii) (tan 45°/cosec 30°) + (sec 60°/cot 45°) – (5 sin 90°/2 cos 0°)
= (1/2) + (2/1) – (5×1)/(2×1)
On further calculation, we get,
= (1/2) + (2/1) – (5/2)
= (1 + 4 – 5)/2
We get,
= 0
(iv) (tan2 60° + 4 cos2 45° + 3 sec2 30° + 5 cos 90°)/(cosec 30° + sec 60° – cot2 30°)
= {(√3)2 + 4×(1/√2)2 + 3×(2/√3)2 + 5×0}/(2) + (2) – (√3)2
On further calculation, we get,
= {3 + 4×(1/2) + 3×(4/3) + 0}/(2 + 2 – 3)
= (3 + 2 + 4)/(4 – 3)
We get,
= 9
(v) (4/cot2 30°) + (1/sin2 60°) – cos2 45°
= {4/(√3)2} + {1/(√3/2)2} – (1/√2)2
On further calculation, we get,
= (4/3) + {1/(3/4)} – (1/2)
= (4/3) + (4/3) – (1/2)
= (8 + 8 – 3)/6
We get,
= 13/6
3. Prove that:
(a) sin 60° . cos 30° – sin 60°. sin 30° = (1/2)
(b) cos 60°. Cos 30° – sin 60°. sin 30° = 0
(c) sec2 45° – tan2 45° = 1
(d) {(cot 30° + 1)/(cot 30° – 1)}2 = (sec 30° + 1)/(sec 30° – 1)
Answer
(a) Consider L.H.S.
sin 60°. cos 30° – cos 60°. sin 30°
= (√3/2) x (√3/2) – (1/2) x (1/2)
On simplification, we get,
= (3/4) – (1/4)
= (3 – 1)/4
= 2/4
We get,
= 1/2
= R.H. S.
Hence, proved
(b) Consider L.H.S.
cos 60°. cos 30° – sin 60°. sin 30°
= (1/2)×(√3/2) – (√3/2)×(1/2)
On calculating further, we get,
= (√3/4) – (√3/4)
We get,
= 0
= R.H.S.
Hence, proved
(c) Consider L.H.S.
sec2 45° – tan2 45°
= (√2)2 – (1)2
= 2 – 1
= 1
= R.H.S.
Hence, proved
(d) Consider L.H.S.
{(cot 30° + 1)/(cot 30° – 1)}2
= {(√3) + 1/(√3) – 1}2
= {(√3 + 1)/(√3 – 1) x (√3 + 1)/(√3 + 1)}2
On further calculation, we get,
= {(√3)2 + (1)2 + 2√3}/{(√3)2 + (1)2 – 2√3}
= (3 + 1 + 2√3)/(3 + 1 – 2√3)
= (4 + 2√3)/(4 – 2√3)
Taking 2 as common, we get,
= 2 (2 + √3)/2 (2 – √3)
= (2 + √3)/(2 – √3)
= {(2/√3) + 1}/{(2/√3) – 1}
= (sec 30° + 1)/(sec 30° – 1)
= R.H.S.
Hence, proved
4. Find the value of ‘A’, if
(a) 2 cos A = 1
(b) 2 sin 2A = 1
(c) cosec 3A = (2/√3)
(d) 2 cos 3A = 1
(e) √3 cot A = 1
(f) cot 3A = 1
Answer
(a) 2 cos A = 1
⇒ cos A = (1/2)
⇒ cos A = cos 60°
⇒ A = 60°
Therefore, the value of ‘A’ is 60°
(b) 2 sin 2A = 1
⇒ sin 2A = (1/2)
⇒ sin 2A = sin 30°
⇒ 2A = 30°
We get,
A = 15°
Therefore, the value of ‘A’ is 15°
(c) cosec 3A = (2/√3)
⇒ cosec 3A = cosec 60°
⇒ 3A = 60°
We get,
A = 20°
Therefore, the value of ‘A’ is 20°
(d) 2 cos 3A = 1
⇒ cos 3A = (1/2)
⇒ cos 3A = cos 60°
⇒ 3A = 60°
We get,
A = 20°
Therefore, the value of ‘A’ is 20°
(e) √3 cot A = 1
⇒ cot A = (1/√3)
⇒ cot A = cot 60°
⇒ A = 60°
Therefore, the value of ‘A’ is 60°
(f) cot 3A = 1
⇒ cot 3A = cot 45°
⇒ 3A = 45°
We get,
A = 15°
Therefore, the value of ‘A’ is 15°
5. Find the value of ‘A’, if
(a) (1 – cosec A) (2 – sec A) = 0
(b) (2 – cosec 2A) cos 3A = 0
Answer
(a) (1 – cosec A) (2 – sec A) = 0
Here,
1 – cosec A = 0 and 2 – sec A = 0
On calculating further, we get,
cosec A = 1 and sec A = 2
⇒ cosec A = cosec 90° and sec A = sec 60°
⇒ A = 90° and A = 60°
(b) (2 – cosec 2A) cos 3A = 0
Here,
2 – cosec 2A = 0 and cos 3A = 0
On further calculation, we get,
⇒ cosec 2A = 2 and cos 3A = 0
⇒ cosec 2A = cosec 30° and cos 3A = cos 90°
We get,
2A = 30° and 3A = 90°
⇒ A = 15° and A = 30°
6. Find the value of ‘x’ in each of the following:
(a)
(a)
From the figure,
We have,
sin 60° = BC/AC
√3/2 = 12/x
On simplification, we get,
x = (2×12)/√3
⇒ x = 24/√3
⇒ x = (8×3)/√3
We get,
x = 8√3
(b)
From the figure,
We have
tan 45° = BC/AB
⇒ 1 = 24/x
We get,
x = 24
(c)
We have,
cos x = AB/AC
⇒ cos x = 12/24
⇒ cos x = 1/2
⇒ cos x = cos 60°
We get,
x = 60°
(d)
We have,
sin x = BC/AC
⇒ sin x = (15/√2)/15
⇒ sin x = (1/√2)
⇒ sin x = sin 45°
We get,
⇒ x = 45°
7. Find the length of AD.
Given: ∠ABC = 60°, ∠DBC = 45° and BC = 24 cm.
tan 60° = AC/BC
√3 = AC/24
We get,
AC = 24√3 cm
In △DBC,
tan 45° = DC/BC
⇒ 1 = DC/24
We get,
DC = 24 cm
Now,
AC = AD + DC
AD = AC – DC
Substituting the values of AC and DC, we get,
AD = 24√3 – 24
⇒ AD = 24 (√3 – 1) cm
Therefore, the length of AD is 24(√3 – 1) cm
8. Find lengths of diagonals AC and BD. Given AB = 24 cm and ∠BAD = 60°
∴ The given figure is a rhombus
We know that,
Diagonals of a rhombus bisect each other at right angles and also bisect the angle of vertex
Let the diagonals AC and BD intersect each other at point O
Hence,
OA = OC = (1/2) AC
OB = OD = (1/2) BD
∠AOB = 90°
Given
∠BAD = 60°
⇒ ∠OAB = (1/2) ∠BAD
⇒ ∠OAB = 30°
In right-angled △AOB,
sin 30° = OB/AB
= (1/2)
Given AB = 24,
⇒OB/24 = (1/2)
OB = 24/2
We get,
OB = 12 cm
cos 30° = OA/AB
= √3/2
⇒ OA/24 = √3/2
OA = (24√3)/2
We get,
OA = 12√3 cm
Therefore,
Length of diagonal AC = 2×OA = 2×12√3 = 24√3 cm and
Length of diagonal BD = 2×OB = 2×12 = 24 cm
9. Find the length of EC.
⇒ AB = 28 cm
In right △ABE,
tan 30° = BE/AB
⇒ 1/√3 = BE/28
We get,
BE = (28/√3)
In right △ABC,
tan 60° = CB/AB
⇒ √3 = CB/28
We get,
CB = 28√3
Hence,
Length of EC = CB + BE
= 28√3 + (28/√3)
On further calculation, we get,
= (84 + 28)/√3
= (112/√3)
Hence, the length of EC is (112/√3) cm
10. In the given figure, AB and EC are parallel to each other. Sides AD and BC are 1.5 cm each and are perpendicular to AB. Given that ∠AED = 45° and ∠ACD = 30°. Find:
(a) AB
(b) AC
(c) AE
tan 30° = AD/DC
(1/√3) = 1.5/DC (given AD = 1.5 cm)
⇒ DC = 1.5√3
Here,
AB || DC and AD ⊥ EC, ABCD is a parallelogram
Therefore, opposite sides are equal
⇒ AB = DC = 1.5√3 cm
(b) In right △ADC,
sin 30° = AD/AC
⇒ (1/2) = 1.5/AC (given AD = 1.5 cm)
⇒ AC = 2×1.5
We get,
AC = 3 cm
(c) In right △ADE,
sin 45° = AD/AE
⇒ (1/√2) = 1.5/AE (given AD = 1.5 cm)
We get,
AE = 1.5√2 cm
11. Evaluate the following:
(a) sin 62°/cos 28°
(b) sec 34°/cosec 56°
(c) tan 12°/cot 78°
(d) sin 25° cos 43°/sin 47° cos 65°
(e) sec 32° cot 26°/tan 64° cosec 58°
(f) cos 34° cos 33°/ sin 57° sin 56°
Answer:
(a) sin 62°/cos 28°
This can be written as,
= sin (90° – 28°)/cos 28°
= cos 28°/cos 28°
We get,
= 1
(b) sec 34°/cosec 56°
This can be written as,
= sec (90° – 56°)/cosec 56°
= cosec 56°/cosec 56°
We get,
= 1
(c) tan 12°/cot 78°
This can be written as,
= tan (90° – 78°)/cot 78°
= cot 78°/cot 78°
We get,
= 1
(d) sin 25° cos 43°/sin 47° cos 65°
This can be written as,
= sin (90° – 65°) cos (90° – 47°)/sin 47° cos 65°
= cos 65° sin 47°/sin 47° cos 65°
We get,
= 1
(e) sec 32° cot 26°/tan 64° cosec 58°
This can be written as,
= sec (90° – 58°) cot (90° – 64°)/tan 64° cosec 58°
= cosec 58° tan 64° / tan 64° cosec 58°
We get,
= 1
(f) cos 34° cos 33°/sin 57° sin 56°
This can be written as,
= cos (90° – 56°) cos (90° – 57°)/sin 57° sin 56°
= sin 56° sin 57°/sin 57° sin 56°
We get,
= 1
12. Evaluate the following:
(a) sin 31° – cos 59°
(b) cot 27° – tan 63°
(c) cosec 54° – sec 36°
(d) sin 28° sec 62° + tan 49° tan 41°
(e) sec 16° tan 28° – cot 62° cosec 74°
(f) sin 22° cos 44° – sin 46° cos 68°
Answer
(a) sin 31° – cos 59°
= sin (90° – 59°) – cos 59°
= cos 59° – cos 59°
We get,
= 0
(b) cot 27° – tan 63°
= cot (90° – 63°) – tan 63°
= tan 63° – tan 63°
We get,
= 0
(c) cosec 54° – sec 36°
= cosec (90° – 36°) – sec 36°
= sec 36° – sec 36°
We get,
= 0
(d) sin 28° sec 62° + tan 49° tan 41°
= sin 28° sec (90° – 28°) + tan 49° tan (90° – 49°)
= sin 28° cosec 28° + tan 49° cot 49°
= sin 28° ×(1/sin 28°) + tan 49°×(1/tan 49°)
= 1 + 1
We get,
= 2
(e) sec 16° tan 28° – cot 62° cosec 74°
= sec (90° – 74°) tan (90° – 62°) – cot 62° cosec 74°
= cosec 74° cot 62° – cot 62° cosec 74°
We get,
= 0
(f) sin 22° cos 44° – sin 46° cos 68°
= sin (90° – 68°) cos (90° – 46°) – sin 46° cos 68°
= cos 68° sin 46° – sin 46° cos 68°
We get,
= 0
13. Evaluate the following:
(a) (sin 36°/cos 54°) + (sec 31°/cosec 59°)
(b) (tan 42°/cot 48°) – (cos 33°/sin 57°)
(c) (2 sin 28°/cos 62°) + (3 cot 49°/tan 41°)
(d) (5 sec 68°/cosec 22°) + (3 sin 52° sec 38°)/(cot 51° cot 39°)
Answer
(a) (sin 36°/cos 54°) + (sec 31°/cosec 59°)
This can be written as,
= {sin (90° – 54°)/cos 54°} + {sec (90° – 59°)/cosec 59°}
= (cos 54°/cos 54°) + (cosec 59°/cosec 59°)
= 1 + 1
We get,
= 2
(b) (tan 42°/cot 48°) – (cos 33°/sin 57°)
This can be written as,
= {tan (90° – 48°)/cot 48°} – {cos (90° – 57°)/sin 57°}
= (cot 48°/cot 48°) – (sin 57°/sin 57°)
= 1 – 1
We get,
= 0
(c) (2 sin 28°/cos 62°) + (3 cot 49°/tan 41°)
This can be written as,
= {2 sin (90° – 62°)/cos 62°} + {3 cot (90° – 41°)/tan 41°}
= (2 cos 62°/cos 62°) + (3 tan 41°/ tan 41°)
= 2 + 3
We get,
= 5
(d) (5 sec 68°/cosec 22°) + (3 sin 52° sec 38°)/(cot 51° cot 39°)
This can be written as,
= {5 sec (90° – 22°)/cosec 22°} + {3 sin 52° sec (90° – 52°)/cot 51° cot (90° – 51°)}
= (5 cosec 22°/cosec 22°) + (3 sin 52° cosec 52°/cot 51° tan 51°)
= 5 + {3 sin 52° ×(1/sin 52°)/cot 51° ×(1/cot 51°)}
= 5 + 3/1
= 5 + 3
We get,
= 8
14. Express each of the following in terms of trigonometric ratios of angles between 0°and 45°
(a) sin 65° + cot 59°
(b) cos 72° – cos 88°
(c) cosec 64° + sec 70°
(d) tan 77° – cot 63° + sin 57°
(e) sin 53° + sec 66° – sin 50°
(f) cos 84° + cosec 69° – cot 68°
Answer
(a) sin 65° + cot 59°
This can be written as,
= sin (90° – 25°) + cot (90° – 31°)
We get,
= cos 25° + tan 31°
(b) cos 72° – cos 88°
This can be written as,
= cos (90° – 18°) – cos (90° – 2°)
We get,
= sin 18° – sin 2°
(c) cosec 64° + sec 70°
This can be written as,
= cosec (90° – 26°) + sec (90° – 20°)
We get,
= sec 26° + cosec 20°
(d) tan 77° – cot 63° + sin 57°
This can be written as,
= tan (90° – 13°) – cot (90° – 27°) + sin (90° – 33°)
We get,
= cot 13° – tan 27° + cos 33°
(e) sin 53° + sec 66° – sin 50°
This can be written as,
= sin (90° – 37°) + sec (90° – 24°) – sin (90° – 40°)
We get,
= cos 37° + cosec 24° – cos 40°
(f) cos 84° + cosec 69° – cot 68°
This can be written as,
= cos (90° – 6°) + cosec (90° – 21°) – cot (90° – 22°)
We get,
= sin 6° + sec 21° – tan 22°
15. Evaluate the following:
(a) sin 35° sin 45° sec 55° sec 45°
(b) cot 20° cot 40° cot 45° cot 50° cot 70°
(c) cos 39° cos 48° cos 60° cosec 42° cosec 51°
(d) sin (35° + θ) – cos (55° – θ) – tan (42° + θ) + cot (48° – θ)
(e) tan (78° + θ) + cosec (42° + θ) – cot (12° – θ) – sec (48° – θ)
(f) (3 sin 37°/cos 53°) – (5 cosec 39°/sec 51°) + {(4 tan 23° tan 37° tan 67° tan 53°)/(cos 17° cos 67° cosec 73° cosec 23°)}
(g) (sin 0° sin 35° sin 55° sin 75°)/(cos 22° cos 64° cos 68° cos 90°)
(h) {(2 sin 25° sin 35° sec 55° sec 65°)/(5 tan 29° tan 45° tan 61°)} + {(3 cos 20° cos 50° cot 70° cot 40°)/(5 tan 20° tan 50° sin 70° sin 40°)}
(i) {(3 sin2 40°)/(4 cos2 50°)} – {(cosec2 28°)/(4 sec2 62°)} + {(cos 10° cos 25° cos 45° cosec 80°)/(2 sin 15° sin 45° sin 65° sec 75°)}
(j) {(5 cot 5° cot 15° cot 25° cot 35° cot 45°) / (7 tan 45° tan 55° tan 65° tan 75° tan 85°)} + {(2 cosec 12° cosec 24° cos 78° cos 66°)/(7 sin 14° sin 23° sec 76° sec 67°)}
Answer
(a) sin 35° sin 45° sec 55° sec 45°
This can be written as,
= sin (90° – 55°)×(1/√2)×(1/cos 55°)×(√2)
= cos 55° ×(1/cos 55°)×(1/√2)×(√2)
We get,
= 1
(b) cot 20° cot 40° cot 45° cot 50° cot 70°
This can be written as,
= cot (90° – 70°)×cot (90° – 50°)× 1×cot 50° cot 70°
= tan 70° × tan 50° × cot 50° × cot 70°
= tan 70° × cot 70° × tan 50° × cot 50°
= tan 70° × (1/tan 70°) × tan 50° × (1/tan 50°)
We get,
= 1
(c) cos 39° cos 48° cos 60° cosec 42° cosec 51°
This can be written as,
= cos (90° – 51°) ×cos (90° – 42°) ×(1/2) ×(1/sin 42°)× (1/sin 51°)
= sin 51° ×sin 42° × (1/2)×(1/sin 42°)×(1/sin 51°)
We get,
= 1/2
(d) sin (35° + θ) – cos (55° – θ) – tan (42° + θ) + cot (48° – θ)
This can be written as,
= sin {90° – (55° – θ)} – cos (55° – θ) – tan {90° – (48° – θ)} + cot (48° – θ)
= cos (55° – θ) – cos (55° – θ) – cot (48° – θ) + cot (48° – θ)
We get,
= 0
(e) tan (78° + θ) + cosec (42° + θ) – cot (12° – θ) – sec (48° – θ)
This can be written as,
= tan {90° – (12° – θ)} + cosec {90° – (48° – θ)} – cot (12° – θ) – sec (48° – θ)
= cot (12° – θ) + sec (48° – θ) – cot (12° – θ) – sec (48° – θ)
We get,
= 0
(f) (3 sin 37°/cos 53°) – (5 cosec 39°/sec 51°) + {(4 tan 23° tan 37° tan 67° tan 53°)/(cos 17° cos 67° cosec 73° cosec 23°)}
This can be written as,
= {3 sin (90° – 53°)/cos 53°} – {5 cosec (90° – 51°)/sec 51°} + [{4 tan (90° – 67°) tan (90° – 53°) x (1/cot 67°) x (1/cot 53°)}]/{cos (90° – 73°) cos (90° – 23°) ×(1/sin 73°) x (1/sin 23°)}
= (3 cos 53°/cos 53°) – (5 sec 51°/sec 51°) + [4 cot 67° cot 53°×(1/cot 67°) x (1/cot 53°]/{sin 73° sin 23° ×(1/sin 73°)×(1/sin 23°)}
On calculating further, we get,
= 3 – 5 + 4
= 2
(g) (sin 0° sin 35° sin 55° sin 75°)/(cos 22° cos 64° cos 68° cos 90°)
= 0 ×sin 35° sin 55° sin 75°)/(cos 22° cos 64° cos 68° ×0)
(∵ sin 0° = 0 and cos 90° = 0)
We get,
= 0
(h) {(2 sin 25° sin 35° sec 55° sec 65°)/(5 tan 29° tan 45° tan 61°)} + {(3 cos 20° cos 50° cot 70° cot 40°)/(5 tan 20° tan 50° sin 70° sin 40°)}
This can be written as,
= {2 sin (90° – 65°) sin (90° – 55°) sec 55° sec 65°}/{5 tan (90° – 61°)×1× tan 61°} + {3 cos (90° – 70°) cos (90° – 40°) cot (90° – 20°) cot (90° – 50°)}/(5 tan 20° tan 50° sin 70° sin 40°)
= {2 cos 65° cos 55°× (1/cos 55°)× (1/cos 65°)}/{5 cot 61° ×1× (1/cot 61°)} + {3 sin 70° sin 40° tan 20° tan 50°}/(5 tan 20° tan 50° sin 70° sin 40°)
On calculating further, we get,
= (2/5) + (3/5)
= (2 + 3)/5
= 5/5
= 1
(i) {(3 sin2 40°)/(4 cos2 50°)} – {(cosec2 28°)/(4 sec2 62°)} + {(cos 10° cos 25° cos 45° cosec 80°)/(2 sin 15° sin 45° sin 65° sec 75°)}
= {3 sin2 (90° – 50°)/4 cos2 50°} – {cosec2 (90° – 62°)/4 sec2 62°} + {cos (90° – 80°) cos 25° ×(1/ √2) ×(1/sin 80°}/{2 sin (90° – 75°) ×(1/√2) ×sin (90°– 25°)× (1/cos 75°)
= (3 cos2 50°/4 cos2 50°) – (sec2 62°/4 sec2 62°) + (sin 80°×cos 25°×(1/sin 80°)/{2 cos 75° x cos 25° x (1/cos 75°)}
On further calculation, we get,
= (3/4) – (1/4) + (1/2)
= (1/2) + (1/2)
= 1
(j) {(5 cot 5° cot 15° cot 25° cot 35° cot 45°)/(7 tan 45° tan 55° tan 65° tan 75° tan 85°)} + {(2 cosec 12° cosec 24° cos 78° cos 66°)/(7 sin 14° sin 23° sec 76° sec 67°)}
= {5 cot (90° – 85°) cot (90° – 75°) cot (90° – 65°) cot (90° – 55°) ×1}/(7×1× tan 55° tan 65° tan 75° tan 85°) + {2 cosec (90° – 78°) cosec (90° – 66°) cos 78° cos 66°}/{7 sin (90° – 76°) sin (90° – 67°) sec 76° sec 67°}
= (5 tan 85° tan 75° tan 65° tan 55°)/(7× tan 55° tan 65° tan 75° tan 85°) + {(2 sec 78° sec 66° ×(1/sec 78°)×(1/sec 66°)}/{7 cos 76° cos 67°×(1/cos 76°) ×(1/cos 67°)}
On further calculation, we get,
= (5/7) + (2/7)
= (5 + 2)/7
= 7/7
= 1
16. (a) Solve for ‘θ’ :
sin θ/3 = 1
(b) Solve for ‘θ’:
cot2 (θ – 5)° = 3
(c) Solve for ‘θ‘ ;
sec (θ/2 + 10°) = 2/√3
Answer
(a)
17. Find the value of x in the following:
(i) 2 sin 3x = √3
(ii) 2 sin x/2 = 1
(iii) √3 sin x = cos x
(iv) tan x = sin 45° cos 45° + sin 30°
(v) √3 tan 2x = cos 60° + sin 45° cos 45°
(vi) cos 2x = cos 60° cos 30° + sin 60° sin 30°
Answer
(i)
18. If sin θ = cos θ and 0° < θ < 90°, find the value of ‘θ’.
Answer
19. If tan θ = cot θ and 0° ≤ θ ≤ 90°, find the value of ‘θ’.
Answer
20. If √2 = 1.414 and √3 = 1.732, find the value of the following correct to two decimal places.
Answer
(b) If θ = 30°, verify that: sin 2θ = 2 tan θ/(1 + tan2 θ)
(c) cos 2θ = (1 – tan2 θ)/(1 + tan2 θ) = cos4 θ – sin4 θ = 2 cos2 θ - 1
= 1 – 2 sin2 θ
If A = 30°, verify that:
(d) If θ = 30°, verify that :
sin 3θ = 4 sin θ. sin (60° + θ)
(e) If θ = 30°, verify that:
1 – sin 2θ = (sin θ – cos θ)2
Answer
(a)
22. Evaluate the following:
(a) (sin 3θ – 2 sin 4θ)/(cos 3 θ – 2 cos 4 θ), when 2 θ = 30°
(b) (1 – cos θ)(1 + cos θ)/(1 – sin θ)(1 + sin θ), if θ = 30°
Answer
(a)
(b)
23. If θ = 15° , find the value of: 3/2 cos 3θ – sin 6θ + 3 sin (5θ + 15°)- 2 tan2 3θ
Answer
24. If A = B = 60° , verify that:
(i) cos (A – B) = cos A cos B + sin A sin B
(ii) sin (A – B) = sin A cos B - cos A sin B
(iii) tan (A – B) = (tan A – tan B)/(1 + tan A tan B)
Answer
(i)
(ii)
sin (A + B) = sin A cos B + cos A sin B
(b) If A = 30° and B = 60°, verify that :
cos (A + B) = cos A cos B – sin A sin B
(c) If A = 30° B = 60°, verify that :
sin (A + B)/(cos A. cos B) = tan A + tan B
(d) If A = 30° and B = 60°, verify that:
sin (A + B)/(sin A. sin B) = cot B – cot A
Answer
(a)
26. (a) If A = B = 45°, verify that
Sin (A – B) = sin A. cos B = cos A. sin B
(b) If A = B = 45°, verify that
Cos (A – B) = cos A. cos B + sin A. sin B
Answer
(a)
(b)
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 value of sin 15° and cos 15°.
Answer
28. (a) If θ < 90°, find the value of:
sin2 θ + cos2 θ
(b) If θ < 90°, find the value of :
tan2 θ – 1/cos2 θ
(c) If √3 sec 2θ = 2 and θ < 90°, find the value of cos2 (30°+ θ) + sin2 (45° - θ)
Answer
(a)
(b)
29. In the given figure, PQ = 6 cm, RQ = x cm and RP = 10 cm, find
(a) cos θ
(b) sin2 θ – cos2 θ
(c) Use tan θ to find the value of RQ
(a) cos 2θ
(b) sin 2θ
Answer
(a) cos 2θ = cos 2 × 30° = cos 60° = 1/2
(b) sin 3θ = sin 3 × 30° = sin 90° = 1
31. If sin (A + B) = 1 and cos (A – B) = 1, find A and B.
Answer
32. If tan (A – B) = 1/√3 and tan (A + B) = √3, find A and B.
Answer
33. If sin (A – B) = 1/2 and cos (A + B) = 1/2, find A and B.
Answer
34. In △ABC right angled at B, ∠A = ∠C. Find the value of :
(i) sin A cos C + cos A sin C
(ii) sin A sin B + cos A cos B
Answer
(ii)
35. If tan A = 1/2, tan B = 1/3 and tan (A + B) = (tan A + tan B)/(1 – tan A tan B), find A + B.
Answer
Exercise 27.2
1. In a right triangle ABC, right angled at C, if ∠B = 60° and AB = 15 units, find the remaining angles and sides.
Answer
2. If △ABC is a right triangle such that ∠C = 90°, ∠A = 45° and BC = 7 units, find ∠B, AB and AC.
Answer
3. In a rectangle ABCD, AB = 20 cm, ∠BAC = 60°, calculate side BC and diagonals AC and BD.
Answer
4. Find:
(a) BC
(b) AD
(c) AC
(a)
5. (a) Find the value ‘x’, if:
(a)
6. (a) Find the value of ‘y’ if √3 = 1.723.
Given your answer correct to 2 decimal places.
Given your answer correct to 2 decimal places.
(a)
7. In the given figure, if tan θ = 5/13, tan α = 3/5 and RS = 12 m, find the value of ‘h’.
8. (a) Find x and y, in each of the following figure:
(a)
(b)
9. If tan x° = 5/12. tan y° = 3/4 and AB = 48 m; find the length CD.
10. In a right-angled triangle ABC; ∠B = 90°. Find the magnitude of angle A, if :
(a) AB is √3 times of BC.
(b) BC is √3 times of BC.
Answer
Consider the following figure,
11. A ladder is placed against a vertical tower. If the ladder makes an angle of 30° with the ground and reaches upto a height of 18 m of the tower, find the length of the ladder.
Answer
12. The perimeter of a rhombus is 100 cm and obtuse angle of it is 120°. Find the lengths of its diagonals.
Answer
Consider the following figure,
13. In the given figure; ∠B = 90°, ∠ADB = 45° and AB = 24 m. Find the length of CD.
14. In the given figure, a rocket is fired vertically upwards from its launching pad P. It first rises 20 km vertically upward and then 20 km at 60° to the vertical. PQ represents the first stage of the journey and QR the second. S is a point vertically below R on the horizontal level as P, find:
(a) the height of the rocket when it is at point R.
(b) the horizontal distance of point S from P.
Answer
Exercise 27.3
1. (a) Evaluate the following:
sin 62°/cos 28°
(b) Evaluate the following:
sec 34°/cosec 56°
(c) Evaluate the following:
tan 12°/cot 78°
(d) Evaluate the following:
sin 25° cos 43°
sin 47° cos 65°
(e) Evaluate the following:
sec 32° cot 26°/(tan 64° cosec58°)
(f) Evaluate the following:
(cos 34° cos 33°)/(sin 57° sin 56°)
Answer
(a)
(b)
(c)
(d)
sin 31° - cos 59°
(b) Evaluate the following:
cot 27° - tan 63°
(c) Evaluate the following:
cosec 54° - sec 36°
(d) Evaluate the following:
sin 28° sec 62° + tan 49° tan 41°
(e) Evaluate the following:
sec 16° tan 28° - cot 62° cosec 74°
(f) Evaluate the following:
sin 22° cos 44° - sin 46° cos 68°
Answer
(a)
sin 36°/cos 54° + sec 31°/cosec 59°
(b) Evaluate the following:
tan 42°/cot 48° - cos 33°/sin 57°
(c) Evaluate the following:
2 sin 28°/cos 62° + 3 cot 49°/tan 41°
(d) Evaluate the following:
5 sec 68°/cosec 22° + (3 sin 52° sec 38°)/(cot 51° cot 39°)
Answer
(a)
(b)
sin 65° + cot 59°
(b) Express each of the following in terms of trigonometric ratios of angles between 0° and 45°:
cos 72° - cos 88°
(c) Express each of the following in terms of trigonometric ratios of angles between 0° and 45°:
cosec 64° + sec 70°
(d) Express each of the following in terms of trigonometric ratios of angles between 0° and 45°:
tan 77° - cot 63° + sin 57°
(e) Express each of the following in terms of trigonometric ratios of angles between 0° and 45°:
sin 53° + sec 66° - sin 50°
(f) Express each of the following in terms of trigonometric ratios of angles between 0° and 45°:
cos 84° + cosec 69° - cot 68°
Answer
(a)
(b)
5. (a) Evaluate the following:
sin 35° sin 45° sec 55° sec 45°
(b) Evaluate the following:
cot 20° cot 40° cot 45° cot 50° cot 70°
(c) Evaluate the following:
cos 39° cos 48°cos 60° cosec 42° cosec 51°
(d) Evaluate the following:
sin (35° + θ) – cos (55° - θ) – tan (42° + θ) + cot (48° - θ)
(e) Evaluate the following:
tan (78° + θ) + cosec (42° + θ) – cot (12° - θ) – sec (48° - θ)
(f) Evaluate the following:
(3 sin 37°/cos 53°) – (5 cosec 39°/sec 51°) + 4 tan 23° tan 37° tan 67° tan 53°/cos 17° cos 67° cosec 73° cosec 23°)
(g) Evaluate the following:
(sin 0° sin 35° sin 55° sin 75°)/(cos 22° cos 64° cos 68° cos 90°)
(h) Evaluate the following:
(2 sin 25° sin 35° sec 55° sec 65°)/(5 tan 29° tan 45° tan 61°) + (3 cos 20° cos 50° cot 70° cot 40°)/(5 tan 20° tan 50° sin 70° sin 40°)
(i) Evaluate the following:
(3 sin2 40° /4 cos2 50°) – (cosec2 28°/ 4 sec2 62°) + (cos 10° cos 25° cos 45° cosec 80°)/(2 sin 15° sin 25° sin 45° sin 65° sec 75°)
(j) Evaluate the following:
(5 cot 5° cot 15° cot 25° cot 35° cot 45°)/(7 tan 45° tan 55° tan 65° tan 75° tan 85°) + (2 cosec 12° cosec 24° cos 78° cos 66°)/(7 sin 14° sin 23° sec 76° sec 67°)
Answer:
(a)
(b)
6. If cos 3θ = sin (θ - 34°), find the value of ° if 3θ is an acute angle.
Answer
7. If 4θ = cot (θ + 20°), find the value of ° if 4° is an acute angle.
Answer
8. If sec 2θ = cosec 3θ, find the value of θ if it is known that both 2θ and 3θ are acute angles.
Answer
9. If sin (θ - 15°) = cos (θ - 25°), find the value of (θ - 15°) and (θ - 25°) are acute angles.
Answer
10. If A, B and C are interior angles of △ABC, Prove that sin (A + B)/2 = cos C/2
Answer
11. If P, Q and R are the interior angles of △PQR, prove that cot (Q + R)/2 = tan P/2
Answer
12. If cos θ = sin 60° and θ is an acute angle find the value of 1 – 2 sin2 θ
Answer
13. If sec θ = cosec 30° and θ is an acute angle, find the value of 4 sin2θ – 2 cos2θ.
Answer
14. Prove the following :
(a) tan θ tan (90° – θ) = cot θ cot (90° – θ)
(b) Prove the following :
sin 58° sec 32° + cos 58° cosec 32° = 2
(c) Prove the following :
tan (90° - θ) cot θ/(cosec2 θ) = cos2 θ
(d) Prove the following :
sin2 30° + cos2 30° = 1/2 sec 60°
Answer
(a)
15. If A + B = 90°, prove that
(tan A tan B + tan A cot B)/(sin A sec B) – sin2 B/cos2 A = tan2 A
Answer
