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प्रश्न
In triangle ABC, AB = AC = 15 cm and BC = 18 cm. Find:
- cos B
- sin C
- tan2 B - sec2 B + 2
उत्तर
In the isosceles ΔABC, the perpendicular drawn from angle A to the side BC divides the side BC into two equal parts BD = DC = 9 cm
Since ∠ADB = 90°
⇒ AB2 = AD2 + BD2 ...(AB is hypotenuse in ΔABD)
⇒ AD2 = 152 – 92
⇒ AD2 = 144
⇒ AD = `sqrt144`
⇒ AD = 12
(i) cos B = `"base"/"hypotenue" = "BD"/"AB" = (9)/(15) = (3)/(5)`
(ii) sin C = `"perpendicular"/"hypotenuse" = "AD"/"AB" = (12)/(15) = (4)/(5)`
(iii) tan B = `"perpendicular"/"base" = "AD"/"BD" = (12)/(9) = (4)/(3)`
sec B = `"hypotenuse"/"base" = "AB"/"BD" = (15)/(9) = (5)/(3)`
Therefore,
tan2 B – sec2 B + 2
= `(4/3)^2 – (5/3)^2+2`
= `16/9 - 25/9 + 2`
= `16/9 - 25/9 + (2 xx 9)/(1 xx 9)`
= `(16 – 25 + 18)/(9)`
= `9/9`
= 1
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