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Ah pwease help! It’s due tonight!!
Geometry!!

Ah pwease help! It’s due tonight!! Geometry!!-example-1
Ah pwease help! It’s due tonight!! Geometry!!-example-1
Ah pwease help! It’s due tonight!! Geometry!!-example-2
Ah pwease help! It’s due tonight!! Geometry!!-example-3

1 Answer

2 votes

Answer 1:

FE is 53.55 cm long

Explanation:

In this figure DF is hypotenuse , FE is perpendicular and DE is base

So we will use TanQ = perpendicular/Base

where Q is theta which is 57 here,

tanQ = FE/ DE,

Here DE is 35

tan(57) = FE/35

1.53 =FE/35

FE=1.53 X35

FE = 53.55

FE is 53.55 cm long.

Answer 2:

DF = 63.9

Explanation:

By using pythagoras theorem

( H)^2 = (P)^2 + (B)^2

(DF) ^2 = (FE)^2 + (DE)^2

(DF)^2 = (53.55)^2 + (35)^2

(DF) ^2 =2867.6 + 1225

(DF)^2 = 4092.6

DF = 63.9

Answer 3:

RP = 53.3mm

Explanation:

Given in the question as following,

Q (Q) = 53 QR =41mm

QR is base,

RP is perpendicular,

PQ is hypotenuse

TanQ = perpendicuar/base

Tan(53) =RP/QR

1.3 =RP/41

RP = 53.3mm

Answer 4:

(PQ) =75.16mm

Explanation:

(PQ)^2 = (PR)^2+(QR)^2

(PQ)^2 = (53.3)^2+(53)^2

(PQ)^2 =2840.8 +2809

(PQ)^2 = 5649.8

(PQ) =75.16mm

Answer 5:

P(Q) =37

Explanation:

Given values in the figure that,

Q (Q) = 53

Equation of that make a combination of 180 degrees.

R(Q)+P(Q) +Q(Q) =180

90 +P(Q)+53 =180

P(Q) = 180 -143

P(Q) =37

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