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

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!