1. In the Fig.8.12, \( \angle \text R \) is the right angle of \( \triangle \)PQR.
    Write the following ratios:
    1. sin P
    2. cos Q
    3. tan P
    4. tan Q
Solution:
Fig. 8.12 : Textbook Page 104

(i) sin P = \(\displaystyle \frac {\text{Opposite Side of }\angle \text P}{\text {Hypotenuse}}\)

∴ sin P = \(\displaystyle \frac {\text{QR}}{\text {PQ}}\) ... (i)


(ii) cos Q = \(\displaystyle \frac {\text{Adjacent Side of }\angle \text Q}{\text {Hypotenuse}}\)

∴ cos Q = \(\displaystyle \frac {\text{QR}}{\text {PQ}}\) ... (ii)


(iii) tan P = \(\displaystyle \frac {\text{Opposite Side of }\angle \text P}{\text {Adjacent Side of }\angle \text P}\)

∴ tan P = \(\displaystyle \frac {\text{QR}}{\text {PR}}\) ... (iii)


(iv) tan Q = \(\displaystyle \frac {\text{Opposite Side of }\angle \text Q}{\text {Adjacent Side of }\angle \text Q}\)

∴ tan Q = \(\displaystyle \frac {\text{PR}}{\text {QR}}\) ... (iv)

  1. In the right angled \( \triangle \)XYZ, \( \angle \text {XYZ} \) = 90° and a, b, c are the lengths of the sides as shown in the figure.
    Write the following ratios:
    1. sin X
    2. tan Z
    3. cos X
    4. tan Y
Solution:
Fig. 8.13 : Textbook Page 104

(i) sin X = \(\displaystyle \frac {\text{Opposite Side of }\angle \text X}{\text {Hypotenuse}}\)

∴ sin X = \(\displaystyle \frac {\text{YZ}}{\text {XZ}}\)

∴ sin X = \(\displaystyle \frac {a}{c}\) ... (i)


(ii) tan Z = \(\displaystyle \frac {\text{Opposite of }\angle \text Z}{\text {Adjacent Side of }\angle \text Z}\)

∴ tan Z = \(\displaystyle \frac {\text{XY}}{\text {YZ}}\)

∴ tan Z = \(\displaystyle \frac {b}{a}\) ... (ii)


(iii) cos X = \(\displaystyle \frac {\text{Adjacent Side of }\angle \text X}{\text {Hypotenuse}}\)

∴ cos X = \(\displaystyle \frac {\text{XY}}{\text {XZ}}\)

∴ cos X = \(\displaystyle \frac {b}{c}\) ... (iii)


(iv) tan X = \(\displaystyle \frac {\text{Opposite Side of }\angle \text X}{\text {Adjacent Side of }\angle \text X}\)

∴ tan X = \(\displaystyle \frac {\text{YZ}}{\text {XY}}\)

∴ tan X = \(\displaystyle \frac {a}{b}\) ... (iv)

  1. In the right angled \( \triangle \)LMN, \( \angle \text {LMN} \) = 90°, \( \angle \text L \) = 50° and \( \angle \text N \) = 40°.
    Write the following ratios:
    1. sin 50°
    2. cos 50°
    3. tan 40°
    4. cos 40°
Solution:
Fig. 8.14 : Textbook Page 104

(i) sin 50° = \(\displaystyle \frac {\text{Opposite Side of }\angle \text 50°}{\text {Hypotenuse}}\)

∴ sin 50° = \(\displaystyle \frac {\text{MN}}{\text {LN}}\) ... (i)


(ii) cos 50° = \(\displaystyle \frac {\text{Adjacent Side of }\angle \text 50°}{\text {Hypotenuse}}\)

∴ cos 50° = \(\displaystyle \frac {\text{LM}}{\text {LN}}\) ... (ii)


(iii) tan 40° = \(\displaystyle \frac {\text{Opposite Side of }\angle \text 40°}{\text {Adjacent Side of }\angle \text 40°}\)

∴ tan 40° = \(\displaystyle \frac {\text{LM}}{\text {MN}}\) ... (iii)


(iv) cos 40° = \(\displaystyle \frac {\text{Adjacent Side of }\angle \text 40°}{\text {Hypotenuse}}\)

∴ cos 40° = \(\displaystyle \frac {\text{MN}}{\text {LN}}\) ... (iv)

  1. In Fig. 8.15, \( \angle \text {PQR} \) = 90°, \( \angle \text {PQS} \) = 90°, \( \angle \text {PRQ} \) = α and \( \angle \text {QPS} \) = θ.
    Write the following ratios:
    1. sin α, cos α, tan α
    2. sin θ, cos θ, tan θ
Solution:
Fig. 8.15 : Textbook Page 104

(i) In right \( \triangle \text {PQR} \),

 sin α = \(\displaystyle \frac {\text{Opposite Side of }\angle \alpha}{\text {Hypotenuse}}\)

∴ sin α = \(\displaystyle \frac {\text{PQ}}{\text {PR}}\) ... (I)


 cos α = \(\displaystyle \frac {\text{Adjacent Side of }\angle \alpha}{\text {Hypotenuse}}\)

∴ cos α = \(\displaystyle \frac {\text{QR}}{\text {PR}}\) ... (II)


 tan α = \(\displaystyle \frac {\text{Opposite Side of }\angle \alpha}{\text{Adjacent Side of }\angle \alpha}\)

∴ tan α = \(\displaystyle \frac {\text{PQ}}{\text {QR}}\) ... (III)



(ii) Also, in right \( \triangle \text {PQS} \),

 sin θ = \(\displaystyle \frac {\text{Opposite Side of }\angle \theta}{\text {Hypotenuse}}\)

∴ sin θ = \(\displaystyle \frac {\text{QS}}{\text {PS}}\) ... (IV)


 cos θ = \(\displaystyle \frac {\text{Adjacent Side of }\angle \theta}{\text {Hypotenuse}}\)

∴ cos θ = \(\displaystyle \frac {\text{PQ}}{\text {PS}}\) ... (V)


 tan θ = \(\displaystyle \frac {\text{Opposite Side of }\angle \theta}{\text{Adjacent Side of }\angle \theta}\)

∴ tan θ = \(\displaystyle \frac {\text{QS}}{\text {PQ}}\) ... (VI)

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