#### (a)From the figure (i0) given below, calculate all the six t-ratios for both acute………

#### (b)From the figure (ii) given below, find the values of x and y in terms of t-ratios

**Answer:**

**(a)** From right angled triangle ABC,

AC^{2} = AB^{2} + BC^{2}

⇒ AB^{2} = AC^{2} – BC^{2}

⇒ AB^{2} = (3)^{2} – (2)^{2}

⇒ AB^{2} = 9 – 4

⇒ AB^{2} = 5

⇒ AB = √5

**(i)** sin A = perpendicular/ hypotenuse

= BC/AC

= 2/3

**(ii)** cos A = base/ hypotenuse

= AB/AC

= √5/3

**(iii)** tan A = perpendicular/ base

= BC/AB

= 2/ √5

**(iv)** cot A = base/perpendicular

= AB/ BC

= √5/2

**(v)** sec A = hypotenuse/ base

= AC/AB

= 3/ √5

**(vi)** cosec A = hypotenuse/perpendicular

= AC/BC

= 3/2

**(b)** From right angled triangle ABC,

∠BAC = θ

cotθ = base/ perpendicular

= AB/ BC

= x/ 10

x = 10 cotθ

Also,

cosecθ = hypotenuse/ perpendicular

= AC/ BC

= y/ 10

y = 10 cosec θ

Hence, x = 10 cot θ and y = 10 cosec θ.

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