Trigonometric Ratios

 Trigonometric Ratios

As we have seen that a right-angled triangle consists of 3 sides, the perpendicular, the base and the hypotenuse. As we talk about ratios, 6 ratios can be obtained from these 3 sides which are as follows:

(i).  p/h            (ii).  b/h           (iii).  p/b          (iv).  b/p          (v). h/b            (vi).  h/p

These six ratios are called trigonometric ratios and they are given certain names.  

Now, let’s introduce the names for these ratios.

S. No.	Ratio	Nomenclature	Abbreviation
1.	p/h	sine	sin
2.	b/h	cosine	cos
3.	p/b	tangent	tan
4.	b/p	cotangent	cot
5.	h/b	secant	sec
6.	h/p	cosecant	Cosec/csc

  1.  In ∆ ABC given alongside find all the trigonometric rations for the reference angle θ.

     Solution,

     Here In, ∆ ABC, <B = 90˚

     And < ACB = θ is the reference angle.

     So, AB = perpendicular, BC = base and AC = hypotenuse.

sinθ = p/h       =          AB/AC

cosθ =  b/h      =          BC/AC

tanθ = p/b        =          AB/BC

Secθ = h/b       =          AC/BC

cosecθ = h/p    =          AC/AB

cotθ =  b/p      =         BC/AB

2.  From the adjoining figure, prove tanθ = 12/5

    

Solution,

In right-angled ∆ ABD,

(AD)² = (BD)² + (AB)²

Or,       (15)² = (9)² + (AB)²

Or,       (AB)² = 225 – 81

Or,       (AB)² = 144

Or,       (AB)² = (12)²

            AB = 12

Again, In right-angled ∆ ABC,

(AC)² = (AB)² + (BC)²

Or,       (13)² = (12)² + (BC)²

Or,       169 = 144 + (BC)²

Or,       (BC)² = 169 – 144

Or,       (BC)² = 25

Or,       (BC)² = (5)²

            BC = 5

Now,   tanθ = AB/BC

            tanθ = 12/5 proved.