Question

2. Find all the t-ratios of (i) 150° (ii) 870°.​

Answer 1

Step-by-step explanation:

150° is in second quadrant

in 2nd quadrant, x is negative, y is positive and hypotenuse is always positive

since angle is measured w.r.t x axis, x works as base and y as perpendicular

now 150° = 180 ° - 30°

we will think of a right angled triangle with angle 30° in which base is negative, perpendicular is positive and hypotenuse is always positive

now

sin150° = p/h = +/+ = + sin30° = 1/2

cos150° = b/h = -/+ = -cos30° = -√3/2

tan150° = p/b = +/- = -tan30° = -1/√3

cot150° = reciprocal of tan = -√3

sec150° = reciprocal of cos = -2/√3

cosec150° = reciprocal of sin = 2

similarly

870° = 870 - 2*360

= 870 -720

= 150°

hence the ratios will be same as above as adding or subtracting multiples of 360° from any angle does not make any difference in ratios