Question

Suppose that triangle ABC has m< C = 90o, AC = 7 and AB = 9. Find cot(A) and csc(B).

Answer 1

you need the other side, which you find by pythagoras
\[7^2+(BC)^2=9^2\]
\[BC=\sqrt{9^2-7^2}=\sqrt{81-49}=\sqrt{32}=4\sqrt{2}\]

Answer 2

SOH CAH TOA

Answer 3

\[\cot(A) = \frac{7}{4\sqrt{2}}\] because it is adjacent over opposite

Answer 4

find cb my pythagoras theorem
u get cb=root of 32=4 root 2
now cot A=7/4 root 2
cosec b=9/7

Answer 5

and \[csc(B)= \frac{9}{7}\] because it is hypotenuse over opposite

Answer 6

what neel said

Answer 7

i cant get u wat ur askin ????

Answer 8

Sat are you sre for cot (A)?

Answer 9

isn't it (7/8)(sqrt2)

Answer 10

no my friend.
root 32=root of 16*2
which equals 4root 2

Answer 11

r u geting it ?????

Answer 12

yup. thanks!

Answer 13

most welcum

Answer 14

just wasnt one of my answer options

Answer 15

ok....
dont go on options be syre of ur method......

Answer 16

sorry friend
m not able to see th attachment

Answer 17

ok: here the options

Answer 18

a) cot =(7/8) sqrt 2, csc 9/7)

Answer 19

b) cot = 9/7, csc = (4/7) sqrt2

Answer 20

c) cot = (9/8) sqrt2, csc = 7/9

Answer 21

d) cot = sqrt130, csc = (1/8)sqrt2

Answer 22

e) cot + 4sqrt2, csc = 9sqrt130

Answer 23

option A

Answer 24

a

Answer 25

we wrote
\[\cot(A)=\frac{7}{4\sqrt{2}}\]

Answer 26

oh, i see the diff

Answer 27

if you multiply top and bottom by
\[\sqrt{2}\] you get
\[\frac{7\sqrt{2}}{8}\]

Answer 28

i had A - so yay! thanks

Answer 29

yeah just did that

Answer 30

some people don't like radicals in the denominator, so you get rid of them