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प्रश्न
From the following figure, find the values of
(i) sin B
(ii) tan C
(iii) sec2 B - tan2B
(iv) sin2C + cos2C
उत्तर
Given angle ADB = 90° and ADC = 90°
⇒ AB2 = AD2 + BD2 ...( AB is hypotenuse in ΔABD)
⇒ 132 = AD2 + 52
∴ AD2 = 169 – 25 = 144 and AD = 12
⇒ AC2 = AD2 + DC2 ...( AC is hypotenuse in ΔADC)
⇒ AC2 = 122 + 162
∴ AC2 = 144 + 256 = 400 and AC = 20
(i) sin B = `"perpendicular"/"hypotenuse" = "AD"/"AB" = (12)/(13)`
(ii) tan C = `"perpendicular"/"base" = "AD"/"DC" = (12)/(16) = (3)/(4) `
(iii) sec B = `"hypotenuse"/"base" = "AB"/"BD" = (13)/(5)`
tan B = `"perpendicular"/"base" = "AD"/"BD" = (12)/(5) `
sec2 B – tan2 B = `(13/5)^2 – (12/5)^2`
= `(169 – 144)/( 25)`
= `(25)/(25)`
= 1
(iv) sin C = `"perpendicular"/"hypotenuse" = "AD"/"AC" = (12)/(20) = (3)/(5)`
cos C = `"base"/"hypotenuse" = "DC"/"AC" = (16)/(20) = (4)/(5)`
sin2 C + cos2 C = `(3/5)^2 + (4/5)^2`
= `(9 + 16 )/(25)`
= `(25)/(25)`
= 1
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