In a triangle ABC- if sin A sin B - ab - c 2- then the triangle isA- isoscelesB- right angledC- obtuse angledD- equilateral

The correct option is B right angled Given, sin A sin B=ab/c^2 ⇒ c^2=ab/sin A sin B= (a/sin A ) (b/sin b ) ⇒ c^2= (c/sin C )^2 ∵ (a/sin A=b/sin B=c/sin ...

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Byju's Answer

Standard XI

Mathematics

L'Hospital Rule to Remove Indeterminate Form

In a triangle...

Question

In a triangle ABC, if sin $A s i n B = \frac{a b}{c^{2}},$ then the triangle is

equilateral

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isosceles

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right angled

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obtuse angled

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Solution

The correct option is C
right angled

$G i v e n, s i n A s i n B = \frac{a b}{c^{2}} \Rightarrow c^{2} = \frac{a b}{s i n A s i n B} = (\frac{a}{s i n A}) (\frac{b}{s i n b}) \Rightarrow c^{2} = {(\frac{c}{s i n C})}^{2} ∵ (\frac{a}{s i n A} = \frac{b}{s i n B} = \frac{c}{s i n C} =) \Rightarrow s i n^{2} C = 1 \Rightarrow C = 90^{0} H e n c e, Δ A B C is a right angled triangle.$

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In a triangle ABC, if sin A sin B=abc2, then the triangle is

The correct option is C right angled Given, sin A sin B=abc2⇒c2=absin A sin B=(asin A)(bsin b)⇒c2=(csin C)2 ∵(asin A=bsin B=csin C=)⇒sin2C=1⇒C=900Hence,ΔABCis a right angled triangle.

In a triangle ABC, if sin $A s i n B = \frac{a b}{c^{2}},$ then the triangle is