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प्रश्न
Use the given figure to find :
(i) sin xo
(ii) cos yo
(iii) 3 tan xo - 2 sin yo + 4 cos yo.
उत्तर
Consider the given figure : 
Since the triangle is a right-angled triangle, so using Pythagorean Theorem
AD2 = 82 + 62
AD2 = 64 + 36 = 100
AD = 10
Also
BC2 = AC2 – AB2
BC2 = 172 – 82 = 225
BC = 15
(i) sin x° = `"perpendicular"/"hypotenuse" = (8)/(17)`
(ii) cos y° = `"base"/"hypotenuse" = (6)/(10) =(3)/(5)`
(iii) sin y° = `"perpendicular"/"hypotenuse" = "AB"/"AD" = (8)/(10) = (4)/(5)`
cos y° = `"base"/"hypotenuse" = (6)/(10) = (3)/(5)`
tan x° = `"perpendicular"/"base" = "AB"/"BC" = (8)/(15)`
Therefore
3tan x° – 2sin y° + 4 cos y°
= `3 (8/15) – 2 (4/5) + 4 (3/5)`
= `(8)/(5) – (8)/(5) + (12)/(5)`
= `2(2)/(5)`
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