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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 = $$\frac{\mathrm{OQ}}{\mathrm{OR}}$$