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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, sin^{2}θ, cos^{2}θ and tan^{2}θ.

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 = \(\frac{\mathrm{AB}}{\mathrm{BC}}\) = \(\frac { 5 }{ 12 }\)

sin^{2}θ = \(\left(\frac{5}{13}\right)^2\) = \(\frac { 25 }{ 169 }\)

cos^{2}θ = \(\left(\frac{12}{13}\right)^2\) = \(\frac { 144 }{ 169 }\)

tan^{2}θ = \(\left(\frac{5}{12}\right)^2\) = \(\frac { 25 }{ 144 }\)

Question 3.

From the figure, find the value of tanθ, tan^{2}θ, tan^{3}θ, sin^{2}θ and cos^{3}θ.

Solution:

In the given figure, ∠ABC = θ and ∠C = 90°

AB = 5, BC = 3 and AC = 4

∴ tan θ = \(\frac{\mathrm{AC}}{\mathrm{BC}}\) = \(\frac { 4 }{ 3 }\)

tan^{2}θ = \(\left(\frac{4}{3}\right)^2\) = \(\frac { 16 }{ 9 }\)

tan^{3}θ = \(\left(\frac{4}{3}\right)^3\) = \(\frac { 64 }{ 27 }\)

∵ sin θ = \(\frac{\mathrm{AC}}{\mathrm{AB}}\) = \(\frac { 4 }{ 5 }\)

∴ sin^{3}θ = \(\left(\frac{4}{5}\right)^3\) = \(\frac { 64 }{ 125 }\)

and cos θ = \(\frac{\mathrm{BC}}{\mathrm{AB}}\) = \(\frac { 3 }{ 5 }\)

∴ cos^{3}θ = \(\left(\frac{3}{5}\right)^3\) = \(\frac{27}{125}\)

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 = \(\frac{\mathrm{OQ}}{\mathrm{OR}}\)