CBSE > Class 09 > Mathematics | CBSE Board | Homework Help

In the given triangleABC,
Angle B = 90°
Hence Angle A + Angle C = 180°-90°.....[Sum of all angles of the triangle are 180°.]

Therefore Angle B is the greater angle than Angle A and Angle C. [Angle A>Angle C ]

We know that.. Side opposite to greater angle is always longer.
Therefore.. Side opposite to angle B is AC
[AC > BC and AC > AB]

That is AC is the longest side..
[AC is the hypotenuse]

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CBSE > Class 09 > Mathematics

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CBSE > Class 09 > Mathematics

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CBSE > Class 09 > Mathematics

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CBSE > Class 09 > Mathematics

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CBSE > Class 09 > Mathematics

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CBSE > Class 09 > Mathematics

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CBSE > Class 09 > Mathematics

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CBSE > Class 09 > Mathematics

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CBSE > Class 09 > Mathematics

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CBSE > Class 09 > Mathematics

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CBSE > Class 09 > Mathematics

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CBSE > Class 09 > Mathematics

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Posted by Arshdeep Virk 5 years, 10 months ago

CBSE > Class 09 > Mathematics

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If y3 +ay2 by +6 is divisible by y_2 and leaves the remainder 3 when divided by y_3 find the value of a and b

Posted by Shagandeep Singh 5 years, 10 months ago

CBSE > Class 09 > Mathematics

1 answers

Sia ? 5 years, 10 months ago

Let $p\left( y \right){\text{ }} = {\text{ }}{y^3} + {\text{ }}a{y^2} + {\text{ }}by{\text{ }} + {\text{ }}6$
p(y) is divisible by y – 2
Then P (2) = 0
${2^3} + a \times {2^2} + b \times 2 + 6 = 0$
$8 + 4a + 2b + 6 = 0$
${\text{4a + 2b = }} - {\text{14}}$
${\text{2a + b = }} - {\text{7 (i)}}$
If p (y) is divided by y-3 remainder is 3
$\therefore {\text{ p (3) = 3}}$
${{\text{3}}^3} + a \times {3^2} + b \times 3 + 6 = 3$
9a+3b=-30
3a+b=-10 ---(ii)
Subtracting (i) from (ii)
-a = 3 and a = -3
Put a = -3 in eq (i)
$2 \times - 3{\text{ }} + {\text{ }}b{\text{ }} = {\text{ }} - 7$
-6+b=-7
b=-7+6
b=-1