Triangle E F G is shown. Angle G F E is a right angle. The length of G F is 3 and the length of E F is 4.6. Which equation could be used to find m∠E in △EFG? m∠E = cos–1(StartFraction 3 Over 4.6 EndFraction) m∠E = cos–1(StartFraction 4.6 Over 3 EndFraction) m∠E = tan–1(StartFraction 3 Over 4.6 EndFraction) m∠E = tan–1(StartFraction 4.6 Over 3 EndFraction)

m∠E = tan–1(StartFraction 3 Over 4.6 EndFraction)

Step-by-step explanation:

Given the following

m<GFE = 90 degrees

GF = 3

EF = 4.6

The side opposite to m<E is GF

Adjacent side is EF

Using the SOH CAH TOA identity

tan theta = opp/adj

tan m<E =GF/EF

Substitute the given values

tan m<E  = 3/4.6

m<E = tan^-1 (3/4.6)

Hence the required result is m∠E = tan–1(StartFraction 3 Over 4.6 EndFraction)

The answer is c

Step-by-step explanation:

d. bc = 12 and ef = 16

step-by-step explanation:

we are given that δabc : δdef such that .

also, it is given that

so, from the options, we need that .

a.

b.

c.  

d.

as, we can see that only option d gives the same ratio as i.e. .

hence, the possible lengths are bc = 12 and ef = 16.

i = 2 / - 5

step-by-step explanation:

evaluation of ( 2 - 4 ) ( p + q ) when p = 2 and q = 5 i, by putting values of p and q,