Calculate all the six t-ratios for both acute………

(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,

AC² = AB² + BC²

⇒ AB² = AC² – BC²

⇒ AB² = (3)² – (2)²

⇒ AB² = 9 – 4

⇒ AB² = 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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