Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1

Students can download Maths Chapter 6 Trigonometry Ex 6.1 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 6 Trigonometry Ex 6.1

Question 1.
From the given figure, find all the trigonometric ratios of angle B.
Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1 1
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1 2

Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1

Question 2.
From the given figure, find the values of
Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1 3
(i) sin B
(ii) sec B
(iii) cot B
(v) tan C
(vi) cosec C
Solution:
In the right ΔABD,
AD² = AB² – BD²
= 13² – 5²
= 169 – 25
= 144
AD = \(\sqrt{144}\)
= 12
In the right ΔADC,
AC² = AD² + DC²
= 12² + 16²
= 144 + 256
= 400
AC = \(\sqrt{400}\)
= 20
Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1 4

Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1

Question 3.
If 2 cos θ = √3, then find all the trigonometric ratios of angle θ.
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1 5
2 cos θ = √3 ⇒ cos θ = \(\frac{√3}{2}\)
AB² = AC² – BC²
= 2² – (√3)² ⇒ = 4 – 3 = 1
AB = √1 = 1
Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1 6

Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1

Question 4.
If cos A =\(\frac{3}{5}\), then find the value of \(\frac{sin A-cos A}{2 tan A}\)
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1 7
cos A = \(\frac{3}{5}\)
ΔABC
BC² = AC² – AB²
= 5² – 3²
= 25 – 9
= 16
BC = \(\sqrt{16}\) = 4
sin A = \(\frac{4}{5}\); tan A = \(\frac{4}{3}\)
Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1 8
∴ The value of \(\frac{sin A-cos A}{2 tan A}\) = \(\frac{3}{40}\)

Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1

Question 5.
If cos A = \(\frac{2x}{1+x_{2}}\) then find the values of sin A and tan A in terms of x.
Solution:
cos A = \(\frac{2x}{1+x_{2}}\)
In the triangle ABC
Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1 9
BC² = AC² – AB²
= (1 + x²)² – (2x)²
= 1 + x4 + 2x² – 4x²
= x4 – 2x² + 1
= (x² – 1)² (or) (1 – x²)² (using (a – b)²)
BC = \(\sqrt{(x^{2}-1)^{2}}\) (or) \(\sqrt{(1-x^{2})^{2}}\)
BC = x² – 1
The value of sin A = \(\frac{BC}{AC}\) = \(\frac{x²-1}{x²+1}\)
tan A = \(\frac{BC}{AB}\) = \(\frac{x²-1}{2x}\)
and
BC = 1 – x²
The value of sin A = \(\frac{1-x²}{1+x²}\)
tan A = \(\frac{1-x²}{2x}\)

Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1

Question 6.
If sin θ = \(\frac{a}{\sqrt{a²+b²}}\) then show that b sin θ = a cos θ.
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1 10
sin θ = \(\frac{a}{\sqrt{a²+b²}}\)
In the triangle ΔABC
BC² = AC² – AB²
Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1 11
L. H. S = R. H. S

Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1

Question 7.
If 3 cot A = 2, then find the value of \(\frac{4sin A-3cos A}{2 sin A+3 Cos A}\)
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1 12
3 cot A = 2 ⇒ cot A = \(\frac{2}{3}\)
AC² = AB² + BC²
= 3² + 2²
= 9 + 4
AC = \(\sqrt{13}\)
cos A = \(\frac{AB}{AC}\) = \(\frac{3}{\sqrt{13}}\)
sin A = \(\frac{BC}{AC}\) = \(\frac{2}{\sqrt{13}}\)
The value of \(\frac{4sin A-3cos A}{2 sin A+3 Cos A}\)
Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1 13
The value is (\(\frac{-1}{13}\))

Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1

Question 8.
If cos θ : sin θ = 1 : 2, then find the value of \(\frac{8cos θ-2cos θ}{4 cos θ+2 sin θ}\)
Solution:
cos θ : sin θ = 1 : 2
Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1 14
Aliter:
cos θ : sin θ = 1 : 2
2 cos θ = sin θ ⇒ 2 = \(\frac{sin θ}{cos θ}\) ⇒ 2 = tan θ
Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1 15
∴ The value is \(\frac{1}{2}\)

Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1

Question 9.
From the given figure, prove that θ + ∅ = 90°. Also prove that there are two other right angled triangles. Find sin α, cos β and tan ∅.
Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1 16
Solution:
In the ΔABC,
AB = 9 + 16 = 25
AC = 15; BC = 20
AB² = 25²
= 625 ……. (1)
AC² + BC² = 15² + 20²
= 225 + 400
= 625 …….. (2)
From (1) and (2) we get
AB² = AC² + BC²
ABC is a right angle triangle at C (Pythagoras theorem)
∴ ∠C = 90°
θ + ∅ = 90°
Also ADC is a right angle triangle ∠ADC = 90° (Given)
BDC is also a right angle triangle ∠BDC = 90° (since ADB is a straight line sum of the two angle is 180°)
From the given diagram
Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1 17

Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1

Question 10.
A boy standing at a point O finds his kite flying at a point P with distance OP = 25 m. It is at a height of 5 m from the ground. When the thread is extended by 10 m from P, it reaches a point Q. What will be the height QN of the kite from the ground? (use trigonometric ratios).
Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1 18
Solution:
Let the angle O be “θ”
In ΔONQ
Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1 19
In ΔOMP
sin θ = \(\frac{PM}{OP}\) ⇒ sin θ = \(\frac{5}{25}\)
sin θ = \(\frac{1}{5}\) ……… (2)
From (1) and (2) we get
\(\frac{h}{35}\) = \(\frac{1}{5}\)
5 h = 35 ⇒ h = \(\frac{35}{5}\) = 7
The height of the kite from the ground is 7m.

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