If H is the orthocentre of - ABC- then AH is equal toA- c AB- b AC- a cot BD- a cot A

The correct option is D a cot A From Δ AHB, AH/sin(90^0 A)=c/sin(A+B) ⇒AH=c cos A/sin(180^0 C)=c cos A/sin C=a/sin Acos A [since a/sinA=c/sinC] =a cot A ...

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Standard XI

Mathematics

Sin(A+B)Sin(A-B)

If H is the o...

Question

If H is the orthocentre of $Δ$ ABC, then AH is equal to

c cot A

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b cot A

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a cot B

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a cot A

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Solution

The correct option is D
a cot A

$From Δ A H B, \frac{A H}{s i n (90^{0} - A)} = \frac{c}{s i n (A + B)} \Rightarrow A H = \frac{c c o s A}{s i n (180^{0} - C)} = \frac{c c o s A}{s i n C} = \frac{a}{s i n A} c o s A [s i n c e \frac{a}{s i n A} = \frac{c}{s i n C}] = a c o t A$

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If H is the orthocentre of Δ ABC, then AH is equal to

The correct option is D a cot A From ΔAHB,AHsin(900−A)=csin(A+B)⇒AH=c cosAsin(1800−C)=c cosAsinC=asin Acos A[since asinA=csinC]=acotA

If H is the orthocentre of $Δ$ ABC, then AH is equal to

The correct option is D
a cot A

$From Δ A H B, \frac{A H}{s i n (90^{0} - A)} = \frac{c}{s i n (A + B)} \Rightarrow A H = \frac{c c o s A}{s i n (180^{0} - C)} = \frac{c c o s A}{s i n C} = \frac{a}{s i n A} c o s A [s i n c e \frac{a}{s i n A} = \frac{c}{s i n C}] = a c o t A$