Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 month ago

# I need help with this math question? Relevance

Referring to the given conditions, the angle falls in the Second Quadrant, where tan is " Negative "

We have BC² = AC² - AB²

=>  BC² = 25² - 7² = 576

=> BC  =  24

Now  tan x  =  Perpendicular/Base  =  7/24

But since tan has to be negative,

tan x  =  - 7/24 ............... Answer • The angle x must fall in the 1 st quadrant, for
otherwise, cos(x)=-24/25. If x falls in the 4 th
quadrant, then sin(x)=-7/25. Now they are both +ve..

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• Given that sin x = 7/25 and cos x = 24/25,

find the value of (a) tan x (b) cosec x.

(a) tan x = 7/24

(b) cosec x = 25/7

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• (a) tan x=sin x/cos x=(7/25)(25/24)=7/24.

(b) csc x=1/sin x=1/(7/25)=25/7.

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• a)

tan(x) = sin(x)/cos(x)

tan(x) = (7/25)/(24/25)

= 7/25 * 25/24

b)  csc(x) = 1/sin(x)

= 1/7/25

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• just use the formulas for tangent and cosecant

tan(x) = sin(x)/cos(x) =  (7/25)/ (24/25) = 7/24

cosec(x) =  1/sin(x)  =  1/ (7/25) = 25/7

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• cosec = 1/sin

tan = sin/cos

Plug in the values....

• Remember that cosec, sin & cos etc. are nonsense. It must be csc x, sin x & cos x etc.

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