In ∆abc, if sin a = 4/5 and tan a = 4/3 , then what is cos a? a. 3/5 b. 4/5 c. 3/4 d. 5/3

A = 3/5

Step-by-step explanation:

At the outset, the question doesn't give us a figure to refer to nor does it tell us if this is a right angled triangle. However we observe that both sin A and tan A have the numbers 3, 4 and 5. We recognize this to be consistent with  3-4-5 standard right angled triangle.

Hence we can guess that it is probably a right angled triangle, but we should do the following to confirm:

Knowning that for a right angled triangle,

sin A = opposite / hypotenuse = 4/5

tan A = opposite / adjacent = 4/3

From this we can surmise that

Opposite = 4

hypotenuse = 5

adjacent = 3

Assemble the triangle to see if this works (see attached). We can futher verify that 3-4-5 works using the Pythagorean theorem.

Now that we have determined that the triangle is a 3-4-5 right angled triangle,

cos A = adjacent / hypotenuse = 3/5

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?