Given 15 cot A =8 find sin a and sec a
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Answered by
24
cot A =8/15 (cot A= b/p)
so, base is 8 and prependicular is 15
then , hypotenous is 17
so sinA is 15/17 (sinA =p/h)
and sec is 17/8 (secA=h/b)
so, base is 8 and prependicular is 15
then , hypotenous is 17
so sinA is 15/17 (sinA =p/h)
and sec is 17/8 (secA=h/b)
namo17:
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Answered by
10
cot A = 8/15
and we know the relation between cot and sin
( 1+cot²a = (1/sin²a))
so 1+(64/225) = (1/sin²a)
sina= 15/17
similarly (1+ tan²a = (1/cos²a))
i.e 1+(1/cot²a) = sec²a
1+(225/64)= sec²a
sec a = 17/8
answered.
and we know the relation between cot and sin
( 1+cot²a = (1/sin²a))
so 1+(64/225) = (1/sin²a)
sina= 15/17
similarly (1+ tan²a = (1/cos²a))
i.e 1+(1/cot²a) = sec²a
1+(225/64)= sec²a
sec a = 17/8
answered.
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