Chapter 8 - Trigonometry
Practice Set 8.1
1. In the Fig.8.12, ∠R is the right angle of ∆PQR. Write the following ratios.
(i) sin P
Solution:
sin P = \( \large \frac{Opposite\, side\, of\, ∠P}{Hypotenuse} \)
∴ sin P = \( \large \frac{QR}{PQ} \)
(ii) cos Q
Solution:
cos Q = \( \large \frac{Adjacent\, side\, of\, ∠Q}{Hypotenuse} \)
∴ cos Q = \( \large \frac{QR}{PQ} \)
(iii) tan P
Solution:
tan P = \( \large \frac{Opposite\, side\, of\, ∠P}{Adjacent\, side\, of\, ∠P} \)
∴ tan P = \( \large \frac{QR}{PR} \)
(iv) tan Q
Solution:
tan Q = \( \large \frac{Opposite\, side\, of\, ∠Q}{Adjacent\, side\, of\, ∠Q} \)
∴ tan Q = \( \large \frac{PR}{QR} \)
2. In the right angled ∆XYZ, ∠XYZ = 90⁰ and a,b,c are the lengths of the sides as shown in the figure. Write the following ratios.
(i) sin X
Solution:
sin X = \( \large \frac{Opposite\, side\, of\, ∠X}{Hypotenuse} \)
∴ sin X = \( \large \frac{YZ}{XZ} \)
∴ sin X = \( \large \frac{a}{c} \)
(ii) tan Z
Solution:
tan Z = \( \large \frac{Opposite\, side\, of\, ∠Z}{Adjacent\, side\, of\, ∠Z} \)
∴ tan Z = \( \large \frac{XY}{YZ} \)
∴ tan Z = \( \large \frac{b}{a} \)
(iii) cos X
Solution:
cos X = \( \large \frac{Adjacent\, side\, of\, ∠X}{Hypotenuse} \)
∴ cos X = \( \large \frac{XY}{XZ} \)
∴ cos X = \( \large \frac{b}{c} \)
(iv) tan X
Solution:
tan X = \( \large \frac{Opposite\, side\, of\, ∠X}{Adjacent\, side\, of\, ∠X} \)
∴ tan X = \( \large \frac{YZ}{XY} \)
∴ tan X = \( \large \frac{a}{b} \)
3. In right angled ∆LMN, ∠LMN = 90⁰, ∠L = 50⁰ and ∠N = 40⁰, write the following ratios.
(i) sin 50⁰
Solution:
sin 50⁰ = \( \large \frac{Opposite\, side\, of\, 50⁰}{Hypotenuse} \)
∴ sin 50⁰ = \( \large \frac{MN}{LN} \)
(ii) cos 50⁰
Solution:
cos 50⁰ = \( \large \frac{Adjacent\, side\, of\, 50⁰}{Hypotenuse} \)
∴ cos 50⁰ = \( \large \frac{LM}{LN} \)
(iii) tan 40⁰
Solution:
tan 40⁰ = \( \large \frac{Opposite\, side\, of\, 40⁰}{Adjacent\, side\, of\, 40⁰} \)
∴ tan 40⁰ = \( \large \frac{LM}{MN} \)
(iv) cos 40⁰
Solution:
cos 40⁰ = \( \large \frac{Adjacent\, side\, of\, 40⁰}{Hypotenuse} \)
∴ cos 40⁰ = \( \large \frac{MN}{LN} \)
4. In the figure 8.15, ∠PQR = 90⁰, ∠PQS = 90⁰, ∠PRQ = α and ∠QPS = q. Write the following trigonometric ratios.
(i) sin α, cos α, tan α
Solution:
In ∆PQR,
sin α = \( \large \frac{Opposite\, side\, of\, α}{Hypotenuse} \)
∴ sin α = \( \large \frac{PQ}{PR} \)
cos α = \( \large \frac{Adjacent\, side\, of\, α}{Hypotenuse} \)
∴ cos α = \( \large \frac{RQ}{PR} \)
tan α = \( \large \frac{Opposite\, side\, of\, α}{Adjacent\, side\, of\, α} \)
∴ tan α = \( \large \frac{PQ}{RQ} \)
(ii) sin θ, cos θ, tan θ
Solution:
In ∆PQS,
sin θ = \( \large \frac{Opposite\, side\, of\, θ}{Hypotenuse} \)
∴ sin θ = \( \large \frac{QS}{PS} \)
cos θ = \( \large \frac{Adjacent\, side\, of\, θ}{Hypotenuse} \)
∴ cos θ = \( \large \frac{PQ}{PS} \)
tan θ = \( \large \frac{Opposite\, side\, of\, θ}{Adjacent\, side\, of\, θ} \)
∴ tan θ = \( \large \frac{QS}{PQ} \)
1. In the following table, a ratio is given in each column. Find the remaining two ratios in the column and complete the table.
(a) Given:
cos θ = \( \large \frac{35}{37} \)
To find:
sin θ and tan θ
Solution:
In ∆ABC,
∠C = θ
We know that,
cos θ = \( \large \frac{Adjacent\, side\, of\, θ}{Hypotenuse} \)
\( \large \frac{35}{37} \) = \( \large \frac{BC}{AC} \) …[Given & from figure]
∴ \( \large \frac{BC}{AC} \) = \( \large \frac{35}{37} \)
Let the common multiple be k
∴ BC = 35k and AC = 37k
∵ ∆ABC is a right angled triangle
∴ AC² = AB² + BC² …[By applying Pythagoras Theorem]
∴ (37k)² = AB² + (35k)²
∴ 1369k² = AB² + 1225k²
∴ AB² = 1369k² – 1225k²
∴ AB² = 144k²
∴ \( \sqrt { { AC }^{ 2 } }\) = \( \sqrt { 144{ k }^{ 2 } }\) …[By Taking square root of both sides]
∴ AB = 12k
Now,
sin θ = \( \large \frac{Opposite\, side\, of\, θ}{Hypotenuse} \)
∴ sin θ = \( \large \frac{AB}{AC} \)
∴ sin θ = \( \large \frac{12k}{37k} \)
∴ sin θ = \( \large \frac{12}{37} \)
And,
tan θ = \( \large \frac{Opposite\, side\, of\, θ}{Adjacent\, side\, of\, θ} \)
∴ tan θ = \( \large \frac{AB}{BC} \)
∴ tan θ = \( \large \frac{12k}{35k} \)
∴ tan θ = \( \large \frac{12}{35} \)
Ans: sin θ = \( \large \frac{12}{37} \) and tan θ = \( \large \frac{12}{35} \)
(b) Given:
sin θ = \( \large \frac{11}{61} \)
To find:
cos θ and tan θ
Solution:
In ∆ABC,
∠C = θ
We know that,
sin θ = \( \large \frac{Opposite\, side\, of\, θ}{Hypotenuse} \)
\( \large \frac{11}{61} \) = \( \large \frac{AB}{AC} \) …[Given & from figure]
∴ \( \large \frac{AB}{AC} \) = \( \large \frac{11}{61} \)
Let the common multiple be k
∴ AB = 11k and AC = 61k
∵ ∆ABC is a right angled triangle
∴ AC² = AB² + BC² …[By applying Pythagoras Theorem]
∴ (61k)² = (11k)² + BC²
∴ 3721k² = 121k² + BC²
∴ BC² = 3721k² – 121k²
∴ BC² = 3600k²
∴ \( \sqrt { { BC }^{ 2 } }\) = \( \sqrt {3600{ k }^{ 2 } }\) …[By Taking square root of both sides]
∴ BC = 60k
Now,
cos θ = \( \large \frac{Adjacent\, side\, of\, θ}{Hypotenuse} \)
∴ cos θ = \( \large \frac{BC}{AC} \)
∴ cos θ = \( \large \frac{60k}{61k} \)
∴ cos θ = \( \large \frac{60}{61} \)
And,
tan θ = \( \large \frac{Opposite\, side\, of\, θ}{Adjacent\, side\, of\, θ} \)
∴ tan θ = \( \large \frac{AB}{BC} \)
∴ tan θ = \( \large \frac{11k}{60k} \)
∴ tan θ = \( \large \frac{11}{60} \)
Ans: cos θ = \( \large \frac{60}{61} \) and tan θ = \( \large \frac{11}{60} \)
(c) Given:
tan θ = 1
To find:
sin θ and cos θ
Solution:
In ∆ABC,
∠C = θ
We know that,
tan θ = \( \large \frac{Opposite\, side\, of\, θ}{Adjacent\, side\, of\, θ} \)
1 = \( \large \frac{AB}{BC} \) …[Given & from figure]
∴ \( \large \frac{AB}{BC} \) = 1
Let the common multiple be k
∴ AB = BC = k
∵ ∆ABC is a right angled triangle
∴ AC² = AB² + BC² …[By applying Pythagoras Theorem]
∴ AC² = k² + k²
∴ AC² = 2k²
∴ \( \sqrt { { AC }^{ 2 } }\) = \( \sqrt { 2{ k }^{ 2 } }\) …[By Taking square root of both sides]
∴ AC = \( \sqrt { 2 }\) k
Now,
sin θ = \( \large \frac{Opposite\, side\, of\, θ}{Hypotenuse} \)
∴ sin θ = \( \large \frac{AB}{AC} \)
∴ sin θ = \( \large \frac{k}{ \sqrt { 2 } k} \)
∴ sin θ = \( \large \frac{1}{ \sqrt { 2 }} \)
And,
cos θ = \( \large \frac{Adjacent\, side\, of\, θ}{Hypotenuse} \)
∴ cos θ = \( \large \frac{BC}{AC} \)
∴ cos θ = \( \large \frac{k}{ \sqrt { 2 } k} \)
∴ cos θ = \( \large \frac{1}{ \sqrt { 2 }} \)
Ans: sin θ = \( \large \frac{1}{ \sqrt { 2 }} \) and cos θ = \( \large \frac{1}{ \sqrt { 2 }} \)
(d) Given:
sin θ = \( \large \frac{1}{2} \)
To find:
cos θ and tan θ
Solution:
In ∆ABC,
∠C = θ
We know that,
sin θ = \( \large \frac{Opposite\, side\, of\, θ}{Hypotenuse} \)
\( \large \frac{1}{2} \) = \( \large \frac{AB}{AC} \) …[Given & from figure]
∴ \( \large \frac{AB}{AC} \) = \( \large \frac{1}{2} \)
Let the common multiple be k
∴ AB = k and AC = 2k
∵ ∆ABC is a right angled triangle
∴ AC² = AB² + BC² …[By applying Pythagoras Theorem]
∴ (2k)² = (k)² + BC²
∴ 4k² = k² + BC²
∴ BC² = 4k² – k²
∴ BC² = 3k²
∴ \( \sqrt { { BC }^{ 2 } }\) = \( \sqrt { 3{ k }^{ 2 } }\) …[By Taking square root of both sides]
∴ BC = \( \sqrt { 3 }\) k
Now,
cos θ = \( \large \frac{Adjacent\, side\, of\, θ}{Hypotenuse} \)
∴ cos θ = \( \large \frac{BC}{AC} \)
∴ cos θ = \( \large \frac{ \sqrt { 3 } k}{2k} \)
∴ cos θ = \( \large \frac{ \sqrt { 3 }}{2} \)
And,
tan θ = \( \large \frac{Opposite\, side\, of\, θ}{Adjacent\, side\, of\, θ} \)
∴ tan θ = \( \large \frac{AB}{BC} \)
∴ tan θ = \( \large \frac{k}{ \sqrt { 3 } k} \)
∴ tan θ = \( \large \frac{1}{ \sqrt { 3 }} \)
Ans: cos θ = \( \large \frac{ \sqrt { 3 }}{2} \) and tan θ = \( \large \frac{1}{ \sqrt { 3 }} \)
(e) Given:
cos θ = \( \large \frac{1}{\sqrt { 3 }} \)
To find:
sin θ and tan θ
Solution:
In ∆ABC,
∠C = θ
We know that,
cos θ = \( \large \frac{Adjacent\, side\, of\, θ}{Hypotenuse} \)
\( \large \frac{1}{\sqrt { 3 }} \) = \( \large \frac{BC}{AC} \) …[Given & from figure]
∴ \( \large \frac{BC}{AC} \) = \( \large \frac{1}{\sqrt { 3 }} \)
Let the common multiple be k
∴ BC = k and AC = \( \sqrt { 3}\)k
∵ ∆ABC is a right angled triangle
∴ AC² = AB² + BC² …[By applying Pythagoras Theorem]
∴ (\( \sqrt { 3 }\)k)² = AB² + (k)²
∴ 3k² = AB² + k²
∴ AB² = 3k² – k²
∴ AB² = 2k²
∴ \( \sqrt { { AB }^{ 2 } }\) = \( \sqrt { 2{ k }^{ 2 } }\) …[By Taking square root of both sides]
∴ AB = \( \sqrt { 2 }k\)
Now,
sin θ = \( \large \frac{Opposite\, side\, of\, θ}{Hypotenuse} \)
∴ sin θ = \( \large \frac{AB}{AC} \)
∴ sin θ = \( \large \frac{ \sqrt { 2 }k}{ \sqrt { 3 }k} \)
∴ sin θ = \( \large \frac{ \sqrt { 2 }}{ \sqrt { 3 }} \)
And,
tan θ = \( \large \frac{Opposite\, side\, of\, θ}{Adjacent\, side\, of\, θ} \)
∴ tan θ = \( \large \frac{AB}{BC} \)
∴ tan θ = \( \large \frac{ \sqrt { 2 }k}{k} \)
∴ tan θ = \( \sqrt { 2 } \)
Ans: sin θ = \( \large \frac{ \sqrt { 2 }}{ \sqrt { 3 }} \) and tan θ = \( \sqrt { 2 } \)
(f) Given:
tan θ = \( \large \frac{21}{20} \)
To find:
sin θ and cos θ
Solution:
In ∆ABC,
∠C = θ
We know that,
tan θ = \( \large \frac{Opposite\, side\, of\, θ}{Adjacent\, side\, of\, θ} \)
\( \large \frac{21}{20} \) = \( \large \frac{AB}{BC} \) …[Given & from figure]
∴ \( \large \frac{AB}{BC} \) = \( \large \frac{21}{20} \)
Let the common multiple be k
∴ AB = 21k and BC = 20k
∵ ∆ABC is a right angled triangle
∴ AC² = AB² + BC² …[By applying Pythagoras Theorem]
∴ AC² = (21k)² + (20k)²
∴ AC² = 441k² + 400k²
∴ AC² = 841k²
∴ \( \sqrt { { AC }^{ 2 } }\) = \( \sqrt { 841{ k }^{ 2 } }\) …[By Taking square root of both sides]
∴ AC = 29k
Now,
sin θ = \( \large \frac{Opposite\, side\, of\, θ}{Hypotenuse} \)
∴ sin θ = \( \large \frac{AB}{AC} \)
∴ sin θ = \( \large \frac{21k}{29k} \)
∴ sin θ = \( \large \frac{21}{29} \)
And,
cos θ = \( \large \frac{Adjacent\, side\, of\, θ}{Hypotenuse} \)
∴ cos θ = \( \large \frac{BC}{AC} \)
∴ cos θ = \( \large \frac{20k}{29k} \)
∴ cos θ = \( \large \frac{20}{29} \)
Ans: sin θ = \( \large \frac{21}{29} \) and cos θ = \( \large \frac{20}{29} \)
(g) Given:
tan θ = \( \large \frac{8}{15} \)
To find:
sin θ and cos θ
Solution:
In ∆ABC,
∠C = θ
We know that,
tan θ = \( \large \frac{Opposite\, side\, of\, θ}{Adjacent\, side\, of\, θ} \)
\( \large \frac{8}{15} \) = \( \large \frac{AB}{BC} \) …[Given & from figure]
∴ \( \large \frac{AB}{BC} \) = \( \large \frac{8}{15} \)
Let the common multiple be k
∴ AB = 8k and BC = 15k
∵ ∆ABC is a right angled triangle
∴ AC² = AB² + BC² …[By applying Pythagoras Theorem]
∴ AC² = (8k)² + (15k)²
∴ AC² = 64k² + 225k²
∴ AC² = 289k²
∴ \( \sqrt { { AC }^{ 2 } }\) = \( \sqrt { 289{ k }^{ 2 } }\) …[By Taking square root of both sides]
∴ AC = 17k
Now,
sin θ = \( \large \frac{Opposite\, side\, of\, θ}{Hypotenuse} \)
∴ sin θ = \( \large \frac{AB}{AC} \)
∴ sin θ = \( \large \frac{8k}{17k} \)
∴ sin θ = \( \large \frac{8}{17} \)
And,
cos θ = \( \large \frac{Adjacent\, side\, of\, θ}{Hypotenuse} \)
∴ cos θ = \( \large \frac{BC}{AC} \)
∴ cos θ = \( \large \frac{15k}{17k} \)
∴ cos θ = \( \large \frac{15}{17} \)
Ans: sin θ = \( \large \frac{8}{17} \) and cos θ = \( \large \frac{15}{17} \)
(h) Given:
sin θ = \( \large \frac{3}{5} \)
To find:
cos θ and tan θ
Solution:
In ∆ABC,
∠C = θ
We know that,
sin θ = \( \large \frac{Opposite\, side\, of\, θ}{Hypotenuse} \)
\( \large \frac{3}{5} \) = \( \large \frac{AB}{AC} \) …[Given & from figure]
∴ \( \large \frac{AB}{AC} \) = \( \large \frac{3}{5} \)
Let the common multiple be k
∴ AB = 3k and AC = 5k
∵ ∆ABC is a right angled triangle
∴ AC² = AB² + BC² …[By applying Pythagoras Theorem]
∴ (5k)² = (3k)² + BC²
∴ 25k² = 9k² + BC²
∴ BC² = 25k² – 9k²
∴ BC² = 16k²
∴ \( \sqrt { { BC }^{ 2 } }\) = \( \sqrt { 16{ k }^{ 2 } }\) …[By Taking square root of both sides]
∴ BC = 4k
Now,
cos θ = \( \large \frac{Adjacent\, side\, of\, θ}{Hypotenuse} \)
∴ cos θ = \( \large \frac{BC}{AC} \)
∴ cos θ = \( \large \frac{4k}{5k} \)
∴ cos θ = \( \large \frac{4}{5} \)
And,
tan θ = \( \large \frac{Opposite\, side\, of\, θ}{Adjacent\, side\, of\, θ} \)
∴ tan θ = \( \large \frac{AB}{BC} \)
∴ tan θ = \( \large \frac{3k}{4k} \)
∴ tan θ = \( \large \frac{3}{4} \)
Ans: cos θ = \( \large \frac{4}{5} \) and tan θ = \( \large \frac{3}{4} \)
(i) Given:
tan θ = \( \large \frac{1}{2\sqrt {2}} \)
To find:
sin θ and cos θ
Solution:
In ∆ABC,
∠C = θ
We know that,
tan θ = \( \large \frac{Opposite\, side\, of\, θ}{Adjacent\, side\, of\, θ} \)
\( \large \frac{1}{2\sqrt {2}} \) = \( \large \frac{AB}{BC} \) …[Given & from figure]
∴ \( \large \frac{AB}{BC} \) = \( \large \frac{1}{2\sqrt {2}} \)
Let the common multiple be k
∴ AB = k and BC = \(2\sqrt {2}\)k
∵ ∆ABC is a right angled triangle
∴ AC² = AB² + BC² …[By applying Pythagoras Theorem]
∴ AC² = (k)² + ( \(2\sqrt {2}\)k )²
∴ AC² = k² + 8k²
∴ AC² = 9k²
∴ \( \sqrt { { AC }^{ 2 } }\) = \( \sqrt { 9{ k }^{ 2 } }\) …[By Taking square root of both sides]
∴ AC = 3k
Now,
sin θ = \( \large \frac{Opposite\, side\, of\, θ}{Hypotenuse} \)
∴ sin θ = \( \large \frac{AB}{AC} \)
∴ sin θ = \( \large \frac{k}{3k} \)
∴ sin θ = \( \large \frac{1}{3} \)
And,
cos θ = \( \large \frac{Adjacent\, side\, of\, θ}{Hypotenuse} \)
∴ cos θ = \( \large \frac{BC}{AC} \)
∴ cos θ = \( \large \frac{2\sqrt {2}k}{3k} \)
∴ cos θ = \( \large \frac{2\sqrt {2}}{3} \)
Ans: sin θ = \( \large \frac{1}{3} \) and cos θ = \( \large \frac{2\sqrt {2}}{3} \)
2. Find the values of –
(i) 5 sin 30⁰ + 3 tan 45⁰
Solution:
We know that,
sin 30⁰ = \( \large \frac{1}{2} \)
tan 45⁰ = 1
∴ 5 sin 30⁰ + 3 tan 45⁰ = 5 ( \large \( \frac{1}{2} \) ) + 3(1)
∴ 5 sin 30⁰ + 3 tan 45⁰ = \( \large \frac{5}{2} \) + 3
∴ 5 sin 30⁰ + 3 tan 45⁰ = \( \large \frac{5}{2} \) + \( \large \frac{3\, ×\, 2}{2} \) …[Equalising the denominators]
∴ 5 sin 30⁰ + 3 tan 45⁰ = \( \large \frac{5\, +\, 6}{2} \)
∴ 5 sin 30⁰ + 3 tan 45⁰ = \( \large \frac{11}{2} \)
Ans: 5 sin 30⁰ + 3 tan 45⁰ = \( \large \frac{11}{2} \)
(ii) \( \frac{4}{5}\) tan² 60⁰ + 3 sin² 60⁰
Solution:
We know that,
tan 60⁰ = \( \large \sqrt {3} \)
sin 60⁰ = \( \large \frac{\sqrt {3}}{2}\)
∴ \( \large \frac{4}{5}\) tan² 60⁰ + 3 sin² 60⁰ = \( \large \frac{4}{5} (\sqrt {3})^2 \) + 3 \( (\large \frac{\sqrt {3}}{2}) ^2\)
∴ \( \large \frac{4}{5}\) tan² 60⁰ + 3 sin² 60⁰ = \( \large \frac{4}{5}\) × 3 + 3 × \( \large \frac{3}{4} \)
∴ \( \large \frac{4}{5}\) tan² 60⁰ + 3 sin² 60⁰ = \( \large \frac{12}{5}\) + \( \large \frac{9}{4} \)
∴ \( \large \frac{4}{5}\) tan² 60⁰ + 3 sin² 60⁰ = \( \large \frac{12}{5}\) × \( \large \frac{4}{4} \) + \( \large \frac{9}{4} \) × \( \large \frac{5}{5} \) …[Equalising the denominators]
∴ \( \large \frac{4}{5}\) tan² 60⁰ + 3 sin² 60⁰ = \( \large \frac{48}{20}\) + \( \large \frac{45}{20} \)
∴ \( \large \frac{4}{5}\) tan² 60⁰ + 3 sin² 60⁰ = \( \large \frac{48\, +\, 45}{20}\)
∴ \( \large \frac{4}{5}\) tan² 60⁰ + 3 sin² 60⁰ = \( \large \frac{93}{20}\)
Ans: \( \large \frac{4}{5}\) tan² 60⁰ + 3 sin² 60⁰ = \( \large \frac{93}{20}\)
(iii) 2 sin 30⁰ + cos 0⁰ + 3 sin 90⁰
Solution:
We know that,
sin 30⁰ = \( \large \frac{1}{2}\)
cos 0⁰ = 1
sin 90⁰ = 1
∴ 2 sin 30⁰ + cos 0⁰ + 3 sin 90⁰ = 2 (\( \large \frac{1}{2}\)) + 1 + 3(1)
∴ 2 sin 30⁰ + cos 0⁰ + 3 sin 90⁰ = \( \large \frac{2}{2}\) + 1 + 3
∴ 2 sin 30⁰ + cos 0⁰ + 3 sin 90⁰ = 1 + 1 + 3
∴ 2 sin 30⁰ + cos 0⁰ + 3 sin 90⁰ = 5
Ans: 2 sin 30⁰ + cos 0⁰ + 3 sin 90⁰ = 5
(iv) \( \large \frac{tan\, 60⁰}{sin\, 60⁰\, +\, cos\, 60⁰}\)
Solution:
We know that,
tan 60⁰ = \(\sqrt {3}\)
sin 60⁰ = \( \large \frac{\sqrt {3}}{2} \)
cos 60⁰ = \( \large \frac{1}{2} \)
∴ \( \large \frac{tan\, 60⁰}{sin\, 60⁰\, +\, cos\, 60⁰} \) = \( \large \frac{\sqrt {3}}{\large \frac{\sqrt {3}}{2}\, +\, \large \frac{1}{2}} \)
∴ \( \large \frac{tan\, 60⁰}{sin\, 60⁰\, +\, cos\, 60⁰} \) = \( \large \frac{\sqrt {3}}{\large \frac{\sqrt {3}\, +\, 1}{2}} \)
∴ \( \large \frac{tan\, 60⁰}{sin\, 60⁰\, +\, cos\, 60⁰} \) = \( \large \frac{\sqrt {3}\, ×\, 2}{\sqrt {3}\, +\, 1} \)
∴ \( \large \frac{tan\, 60⁰}{sin\, 60⁰\, +\, cos\, 60⁰} \) = \( \large \frac{2 \sqrt {3}}{\sqrt {3}\, +\, 1} \)
Ans: \( \large \frac{tan\, 60⁰}{sin\, 60⁰\, +\, cos\, 60⁰} \) = \( \large \frac{2 \sqrt {3}}{\sqrt {3}\, +\, 1} \)
(v) cos² 45⁰ + sin² 30⁰
Solution:
We know that,
cos 45⁰ = \( \large \frac{1}{\sqrt {2}} \)
sin 30⁰ = \( \large \frac{1}{2} \)
∴ cos² 45⁰ + sin² 30⁰ = \( (\large \frac{1}{\sqrt {2}}) ^2\) + \( (\large \frac{1}{2})^2 \)
∴ cos² 45⁰ + sin² 30⁰ = \( \large \frac{1}{2} \) + \(\large \frac{1}{4} \)
∴ cos² 45⁰ + sin² 30⁰ = \( \large \frac{1}{2} × (\large \frac{2}{2}) \) + \( \large \frac{1}{4} \) …[Equalising the denominators]
∴ cos² 45⁰ + sin² 30⁰ = \( \large \frac{2}{4} + \large \frac{1}{4} \)
∴ cos² 45⁰ + sin² 30⁰ = \( \large \frac{2\, +\, 1}{4} \)
∴ cos² 45⁰ + sin² 30⁰ = \( \large \frac{3}{4} \)
Ans: cos² 45⁰ + sin² 30⁰ = \( \large \frac{3}{4} \)
(vi) cos 60⁰ × cos 30⁰ + sin 60⁰ × sin 30⁰
Solution:
We know that,
cos 60⁰ = \( \large \frac{\sqrt {3}}{2} \)
cos 30⁰ = \( \large \frac{1}{2} \)
sin 60⁰ = \( \large \frac{\sqrt {3}}{2} \)
sin 30⁰ = \( \large \frac{1}{2} \)
∴ cos 60⁰ × cos 30⁰ + sin 60⁰ × sin 30⁰ = \( \large \frac{\sqrt {3}}{2} \) + \( \large \frac{1}{2} \) + \( \large \frac{\sqrt {3}}{2} \) + \( \large \frac{1}{2} \)
∴ cos 60⁰ × cos 30⁰ + sin 60⁰ × sin 30⁰ = \( \large \frac{\sqrt {3}}{2} \) × \( \large \frac{1}{2} \) + \( \large \frac{\sqrt {3}}{2} \) × \( \large \frac{1}{2} \)
∴ cos 60⁰ × cos 30⁰ + sin 60⁰ × sin 30⁰ = \( \large \frac{\sqrt {3}}{4} \) + \( \large \frac{\sqrt {3}}{4} \)
∴ cos 60⁰ × cos 30⁰ + sin 60⁰ × sin 30⁰ = \( \large \frac{\sqrt {3} + \sqrt {3}}{4} \)
∴ cos 60⁰ × cos 30⁰ + sin 60⁰ × sin 30⁰ = \( \large \frac{2\sqrt {3}}{4} \)
∴ cos 60⁰ × cos 30⁰ + sin 60⁰ × sin 30⁰ = \( \large \frac{\sqrt {3}}{2} \)
Ans: cos 60⁰ × cos 30⁰ + sin 60⁰ × sin 30⁰ = \( \large \frac{\sqrt {3}}{2} \)
3. If sin θ = \( \large \frac{4}{5}\) then find cos θ
Given:
sin θ = \( \large \frac{4}{5} \)
To find:
cos θ
Solution:
In ∆ABC,
∠C = θ
We know that,
sin θ = \( \large \frac{Opposite\, side\, of\, θ}{Hypotenuse} \)
\( \large \frac{4}{5} \) = \( \large \frac{AB}{AC} \) …[Given & from figure]
∴ \( \large \frac{AB}{AC} \) = \( \large \frac{4}{5} \)
Let the common multiple be k
∴ AB = 4k and AC = 5k
∵ ∆ABC is a right angled triangle
∴ AC² = AB² + BC² …[By applying Pythagoras Theorem]
∴ (5k)² = (4k)² + BC²
∴ 25k² = 16k² + BC²
∴ BC² = 25k² – 16k²
∴ BC² = 9k²
∴ \( \sqrt { { BC }^{ 2 } }\) = \( \sqrt { 9{ k }^{ 2 } }\) …[By Taking square root of both sides]
∴ BC = 3k
Now,
cos θ = \( \large \frac{Adjacent\, side\, of\, θ}{Hypotenuse} \)
∴ cos θ = \( \large \frac{BC}{AC} \)
∴ cos θ = \( \large \frac{3k}{2k} \)
∴ cos θ = \( \large \frac{3}{2} \)
Ans: cos θ = \( \large \frac{3}{2} \)
4. If cos θ = \( \large \frac{15}{17} \), then find sin θ
Given:
cos θ = \( \large \frac{15}{17} \)
To find:
sin θ
Solution:
In ∆ABC,
∠C = θ
We know that,
cos θ = \( \large \frac{Adjacent\, side\, of\, θ}{Hypotenuse} \)
\( \large \frac{15}{17} \) = \( \large \frac{BC}{AC} \) …[Given & from figure]
∴ \( \large \frac{BC}{AC} \) = \( \large \frac{15}{17} \)
Let the common multiple be k
∴ BC = 15k and AC = 17k
∵ ∆ABC is a right angled triangle
∴ AC² = AB² + BC² …[By applying Pythagoras Theorem]
∴ (17k)² = AB² + (15k)²
∴ 289k² = AB² + 225k²
∴ AB² = 289k² – 225k²
∴ AB² = 64k²
∴ \( \sqrt { { AC }^{ 2 } }\) = \( \sqrt { 64{ k }^{ 2 } }\) …[By Taking square root of both sides]
∴ AB = 8k
Now,
sin θ = \( \large \frac{Opposite\, side\, of\, θ}{Hypotenuse} \)
∴ sin θ = \( \large \frac{AB}{AC} \)
∴ sin θ = \( \large \frac{8k}{17k} \)
∴ sin θ = \( \large \frac{8}{17} \)
Ans: sin θ = \( \large \frac{8}{17} \)
Problem Set 8
1. Choose the correct alternative answer for following multiple choice questions.
(i) Which of the following statements is true ?
(A) sin θ = cos (90 – θ)
(B) cos θ = tan (90 – θ)
(C) sin θ = tan (90 – θ)
(D) tan θ = tan (90 – θ)
OPTION (A) : sin θ = cos (90 – θ)
(ii) Which of the following is the value of sin 90° ?
(A) \( \large \frac{\sqrt{3}}{2} \)
(B) 0
(C) \( \large \frac{1}{2} \)
(D) 1
OPTION (D) : 1
(iii) 2 tan 45° + cos 45° – sin 45° = ?
(A) 0
(B) 1
(C) 2
(D) 3
Solution:
We know that,
tan 45° = 1
cos 45⁰ = \( \large \frac{1}{2} \)
sin 45° = \( \large \frac{1}{2} \)
∴ 2 tan 45° + cos 45° – sin 45° = 2(1) + \( \large \frac{1}{2} – \( \large \frac{1}{2} \)
∴ 2 tan 45° + cos 45° – sin 45° = 2 + \( \large \frac{1\, -\, 1}{2} \)
∴ 2 tan 45° + cos 45° – sin 45° = 2 + 0
∴ 2 tan 45° + cos 45° – sin 45° = 2
Option (C) : 2
(iv) \( \large \frac{cos\, 28⁰}{sin\, 62⁰} \) = ?
(A) 2
(B) -1
(C) 0
(D) 1
Solution:
We know that,
sin θ = cos (90 – θ)
∴ sin 62⁰ = cos (90 – 62)
∴ sin 62⁰ = cos 28⁰ …(i)
Now,
\( \large \frac{cos\, 28⁰}{sin\, 62⁰} \) = \( \large \frac{cos\, 28⁰}{cos\, 28⁰} \) …[From (i)]
∴ \( \large \frac{cos\, 28⁰}{sin\, 62⁰} \) = 1
OPTION (D) : 1
2. In right angled △TSU, TS = 5, ∠S = 90°, SU = 12 then find sin T, cos T, tan T. Similarly find sin U, cos U, tan U.
Given:
△TSU is a right angled triangle
TS = 5
∠S = 90°
SU = 12
To find:
sin T, cos T, tan T, sin U, cos U and tan U.
Solution:
△TSU is a right angled triangle …[Given]
∴ (TU)² = (TS)² + (SU)² …[By applying Pythagoras Theorem]
∴ (TU)² = (5)² + (12)²
∴ (TU)² = 25 + 144
∴ (TU)² = 169
∴ \( \sqrt { { TU }^{ 2 } }\) = \( \sqrt {169}\) …[By Taking square root of both sides]
∴ TU = 13
We know that,
(i) sin θ = \( \large \frac{Opposite\, side\, of\, θ}{Hypotenuse} \)
(ii) cos θ = \( \large \frac{Adjacent\, side\, of\, θ}{Hypotenuse} \)
(iii) tan θ = \( \large \frac{Opposite\, side\, of\, θ}{Adjacent\, side\, of\, θ} \)
∴ sin T = \( \large \frac{SU}{TU} \)
∴ sin T = \( \large \frac{12}{13} \)
∴ sin U = \( \large \frac{TS}{TU} \)
∴ sin U = \( \large \frac{5}{13} \)
∴ cos T = \( \large \frac{TS}{TU} \)
∴ cos T = \( \large \frac{5}{13} \)
∴ cos U = \( \large \frac{SU}{TU} \)
∴ cos U = \( \large \frac{12}{13} \)
∴ tan T = \( \large \frac{SU}{TS} \)
∴ tan T = \( \large \frac{12}{5} \)
∴ tan U = \( \large \frac{TS}{SU} \)
∴ tan U = \( \large \frac{5}{12} \)
Ans:
sin T = \( \large \frac{12}{13} \)
sin U = \( \large \frac{5}{13} \)
cos T = \( \large \frac{5}{13} \)
cos U = \( \large \frac{12}{13} \)
tan T = \( \large \frac{12}{5} \)
tan U = \( \large \frac{5}{12} \)
3. In right angled △YXZ, ∠X = 90°, XZ = 8 cm, YZ = 17 cm, find sin Y, cos Y, tan Y, sin Z, cos Z, tan Z.
Given:
△YXZ is a right angled triangle
∠X = 90
XZ = 8 cm
YZ = 17 cm
To find:
sin Y, cos Y, tan Y, sin Z, cos Z and tan Z.
Solution:
△YXZ is a right angled triangle …[Given]
∴ (YZ)² = (XZ)² + (XY)² …[By applying Pythagoras Theorem]
∴ (17)² = (8)² + (XY)²
∴ 289 = 64 + (XY)²
∴ (XY)² = 289 – 64
∴ (XY)² = 225
∴ \( \sqrt { { XY }^{ 2 } }\) = \( \sqrt {225}\) …[By Taking square root of both sides]
∴ XY = 15 cm
We know that,
(i) sin θ = \( \large \frac{Opposite\, side\, of\, θ}{Hypotenuse} \)
(ii) cos θ = \( \large \frac{Adjacent\, side\, of\, θ}{Hypotenuse} \)
(iii) tan θ = \( \large \frac{Opposite\, side\, of\, θ}{Adjacent\, side\, of\, θ} \)
∴ sin Y = \( \large \frac{XZ}{YZ} \)
∴ sin Y = \( \large \frac{8}{17} \)
∴ sin Z = \( \large \frac{XY}{YZ} \)
∴ sin Z = \( \large \frac{15}{17} \)
∴ cos Y = \( \large \frac{XY}{YZ} \)
∴ cos Y = \(\large \frac{15}{17} \)
∴ cos Z = \( \large \frac{XZ}{YZ} \)
∴ cos Z = \( \large \frac{8}{17} \)
∴ tan Y = \( \large \frac{XZ}{XY} \)
∴ tan Y = \( \large \frac{8}{15} \)
∴ tan Z = \( \large \frac{XY}{XZ} \)
∴ tan Z = \( \large \frac{15}{8} \)
Ans:
sin Y = \( \large \frac{8}{17} \)
sin Z = \( \large \frac{15}{17} \)
cos Y = \( \large \frac{15}{17} \)
cos Z = \( \large \frac{8}{17} \)
tan Y = \( \large \frac{8}{15} \)
tan Z = \( \large \frac{15}{8} \)
4. In right angled △LMN, if ∠N = θ, ∠M = 90°, cos θ = \( \large \frac{24}{25} \), find sin θ and tan θ. Similarly, find (sin² θ) and (cos² θ).
Given:
△LMN is a right angled triangle
∠N = θ
∠M = 90°
cos θ = \( \large \frac{24}{25} \)
To find:
sin θ, tan θ, sin² θ and cos² θ
Solution:
We know that,
cos θ = \( \large \frac{Adjacent\, side\, of\, θ}{Hypotenuse} \)
\( \large \frac{24}{25} \) = \( \large \frac{MN}{LN} \) …[Given & from figure]
∴ \( \large \frac{MN}{LN} \) = \( \large \frac{24}{25} \)
Let the common multiple be k
∴ MN = 24k and LN = 25k
∵ ∆LMN is a right angled triangle
∴ LN² = LM² + MN² …[By applying Pythagoras Theorem]
∴ (25k)² = LM² + (24k)²
∴ 625k² = LM² + 576k²
∴ LM² = 625k² – 576k²
∴ LM² = 49k²
∴ \( \sqrt { { LM }^{ 2 } }\) = \( \sqrt {49}\) …[By Taking square root of both sides]
∴ LM = 7
Now,
sin θ = \( \large \frac{Opposite\, side\, of\, θ}{Hypotenuse} \)
∴ sin θ = \( \large \frac{LM}{LN} \)
∴ sin θ = \( \large \frac{7}{25} \)
And,
tan θ = \( \large \frac{Opposite\, side\, of\, θ}{Adjacent\, side\, of\, θ} \)
∴ tan θ = \( \large \frac{LM}{MN} \)
∴ tan θ = \( \large \frac{7}{24} \)
sin² θ = (sin θ)²
∴ sin² θ = (\( \large \frac{7}{25} \))²
∴ sin² θ = \( \large \frac{49}{625} \)
cos² θ = (cos θ)²
But, cos θ = \( \large \frac{24}{25} \) …[Given]
∴ cos² θ = (\( \large \frac{24}{25} \))²
∴ cos² θ = \( \large \frac{576}{625} \)
Ans: sin θ = \( \large \frac{7}{25} \), tan θ = \( \large \frac{7}{24} \), sin² θ = \( \large \frac{49}{625} \) and cos² θ = \( \large \frac{576}{625} \)
5. Fill in the blanks.
(i) sin 20° = cos __°
Solution:
We know that,
sin θ = cos (90 – θ)
∴ sin 20⁰ = cos (90 – 20)⁰
∴ sin 20⁰ = cos 70⁰
Ans: sin 20⁰ = cos 70⁰
(ii) tan 30° × tan __° = 1
Solution:
We know that,
tan θ × tan (90 – θ) = 1
∴ tan 30° × tan (90 – 30)° = 1
∴ tan 30⁰ × tan 60⁰ = 1
Ans: tan 30⁰ × tan 60⁰ = 1
(iii) cos 40° = sin __°
Solution:
We know that,
cos θ = sin (90 – θ)
∴ cos 40⁰ = sin (90 – 40)⁰
∴ cos 40⁰ = sin 50⁰
Ans: cos 40⁰ = sin 50⁰