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If angle A is acute and cos A = 8/17 then cot A is :
- a)8/15
- b)17/8
- c)15/8
- d)17/15
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
If angle A is acute and cos A =8/17 then cot A is :a)8/15b)17/8c)15/8d...
Cos A=8/17=B/H
base=8x, hypotenuse=17x
By pythagoras theorem,
H2 =P2 + B2
289x2 = P2 + 64x2
Cot A=B/P=8x/15x=8/15
base=8x, hypotenuse=17x
By pythagoras theorem,
H2 =P2 + B2
289x2 = P2 + 64x2
Cot A=B/P=8x/15x=8/15
Most Upvoted Answer
If angle A is acute and cos A =8/17 then cot A is :a)8/15b)17/8c)15/8d...
Solution:
Given, angle A is acute and cos A = 8/17
We need to find the value of cot A.
We know that,
cot A = 1/tan A
And,
tan A = sin A/cos A
We have the value of cos A as 8/17.
Now, let's find the value of sin A.
We know that,
sin² A + cos² A = 1
On substituting cos A = 8/17, we get
sin² A + (8/17)² = 1
sin² A = 1 - (64/289)
sin² A = (289-64)/289
sin² A = 225/289
Taking the square root on both sides, we get
sin A = 15/17
Now, let's find the value of tan A.
tan A = sin A/cos A
tan A = (15/17)/(8/17)
tan A = 15/8
Therefore,
cot A = 1/tan A
cot A = 1/(15/8)
cot A = 8/15
Hence, the correct answer is option 'A'.
Given, angle A is acute and cos A = 8/17
We need to find the value of cot A.
We know that,
cot A = 1/tan A
And,
tan A = sin A/cos A
We have the value of cos A as 8/17.
Now, let's find the value of sin A.
We know that,
sin² A + cos² A = 1
On substituting cos A = 8/17, we get
sin² A + (8/17)² = 1
sin² A = 1 - (64/289)
sin² A = (289-64)/289
sin² A = 225/289
Taking the square root on both sides, we get
sin A = 15/17
Now, let's find the value of tan A.
tan A = sin A/cos A
tan A = (15/17)/(8/17)
tan A = 15/8
Therefore,
cot A = 1/tan A
cot A = 1/(15/8)
cot A = 8/15
Hence, the correct answer is option 'A'.
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Community Answer
If angle A is acute and cos A =8/17 then cot A is :a)8/15b)17/8c)15/8d...
given cos A=8/17 let us assume a triangle ABC and it is right amgled at B now we know cos of any angle is adjacent side by hypotenuse so here adjacent side is 8 units and hypotenuse will be 17 units AB=8units and AC=17 units from Pythagoras theorem we know that hypotenuse ^2=side^2+side^2 AC^2=AB^2+BC^2 17^2=8^2+BC^2 289=64+BC^2 BC^2=289-64 BC^2=225 BC=15 units now we know cot of any angle is adjacent side by opposite side so cot A=AB/BC Cot A=8/15
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