sin B = 12/13 ,
then find
cotB.
Answers
Answer:
Step-by-step explanation:
Given \: sinB = \frac{12}{13}GivensinB=
13
12
\implies \frac{1}{cosec B } = \frac{12}{13}⟹
cosecB
1
=
13
12
\implies Cosec B = \frac{13}{12} \: --(1)⟹CosecB=
12
13
−−(1)
\implies cot B = \sqrt{ cosec^{2} B - 1}⟹cotB=
cosec
2
B−1
= \sqrt{ ( \frac{13}{12})^{2} - 1}=
(
12
13
)
2
−1
= \sqrt{ ( \frac{169}{144}) - 1}=
(
144
169
)−1
= \sqrt{\frac{169-144}{144}}=
144
169−144
=\sqrt{\frac{25}{144}}=
144
25
= \sqrt{(\frac{5}{12})^{2}}=
(
12
5
)
2
\green {=\frac{5}{12}}=
12
5
Therefore.,
\red { Value \: of \: Cot B }\green {=\frac{5}{12}}ValueofCotB=
12
5
•••♪
Answer:
5/12
Step-by-step explanation:
sin B=12/13
means perpendicular/hypotenuse=12/13
by phthagoras theorem
therefore base=√(hypotenuse²-perpendicular²)
base=√(13²-12²)=√25=5
cot B=base/perpendicular=5/12