OP Malhotra Class 9 Maths Solutions Chapter 19 Trigonometrical Ratios Ex 19(A)

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S Chand Class 9 ICSE Maths Solutions Chapter 19 Trigonometrical Ratios Ex 19(A)

Question 1.
Complete the following:

Solution:
We know that tan θ = \(\frac{\text { Perpendicular }}{\text { Base }}\)
cos θ = \(\frac{\text { Base }}{\text { Hypotenuse }}\),
sin θ = \(\frac{\text { Perpendicular }}{\text { Hypotenuse }}\)
cot θ = \(\frac{1}{\tan \theta}\) = \(\frac{\text { Base }}{\text { Perpendicular }}\)
sec θ = \(\frac{1}{\cos \theta}\) = \(\frac{\text { Hypotenuse }}{\text { Base }}\)
and cosec θ = \(\frac{1}{\sin \theta}\) = \(\frac{\text { Hypotenuse }}{\text { Perpendicular }}\)
Now, we shall complete the given

Question 2.
From the figure, find the value of sin θ, cos θ, tan θ, sin²θ, cos²θ and tan²θ.

Solution:
In the given figure,
AB = 5, BC = 12 and hyp. AC = 13 (units)
and ∠ACB = θ
∴ sin θ = \(\frac{\mathrm{AB}}{\mathrm{AC}}\) = \(\frac { 5 }{ 13 }\)
cos θ = \(\frac{\mathrm{BC}}{\mathrm{AC}}\) = \(\frac { 12 }{ 13 }\)
tan θ = \(\frac{\mathrm{AB}}{\mathrm{BC}}\) = \(\frac { 5 }{ 12 }\)
sin²θ = \(\left(\frac{5}{13}\right)^2\) = \(\frac { 25 }{ 169 }\)
cos²θ = \(\left(\frac{12}{13}\right)^2\) = \(\frac { 144 }{ 169 }\)
tan²θ = \(\left(\frac{5}{12}\right)^2\) = \(\frac { 25 }{ 144 }\)

Question 3.
From the figure, find the value of tanθ, tan²θ, tan³θ, sin²θ and cos³θ.

Solution:
In the given figure, ∠ABC = θ and ∠C = 90°
AB = 5, BC = 3 and AC = 4
∴ tan θ = \(\frac{\mathrm{AC}}{\mathrm{BC}}\) = \(\frac { 4 }{ 3 }\)
tan²θ = \(\left(\frac{4}{3}\right)^2\) = \(\frac { 16 }{ 9 }\)
tan³θ = \(\left(\frac{4}{3}\right)^3\) = \(\frac { 64 }{ 27 }\)
∵ sin θ = \(\frac{\mathrm{AC}}{\mathrm{AB}}\) = \(\frac { 4 }{ 5 }\)
∴ sin³θ = \(\left(\frac{4}{5}\right)^3\) = \(\frac { 64 }{ 125 }\)
and cos θ = \(\frac{\mathrm{BC}}{\mathrm{AB}}\) = \(\frac { 3 }{ 5 }\)
∴ cos³θ = \(\left(\frac{3}{5}\right)^3\) = \(\frac{27}{125}\)

Question 4.
In the figure, ∠OMP, ∠ORQ, ∠OQM are right angles. Write the values of the following t-ratios:
(a) sin RQM
(b) sin QMP
(c) sin OQR
(d) cos QMP
(e) tan RQM
(f) cot MOP
(g) sec ROQ

Solution:
In the figure, ∠OMP, ∠ORQ and ∠OQM are right angles. Therefore,
(a) sin RQM = \(\frac{\mathrm{RM}}{\mathrm{QM}}\)
(b) sin QMP = \(\frac{\mathrm{PQ}}{\mathrm{PM}}\)
(c) sin OQR = \(\frac{\mathrm{OR}}{\mathrm{OQ}}\)
(d) cos QMP = \(\frac{\mathrm{QM}}{\mathrm{PM}}\)
(e) tan RQM = \(\frac{\mathrm{RM}}{\mathrm{QR}}\)
(f) cot MOP = \(\frac{\mathrm{OM}}{\mathrm{MP}}\)
(g) sec ROQ = \(\frac{\mathrm{OQ}}{\mathrm{OR}}\)