contestada

GEOMETRY NEED HELP ASAP WILL ASSIGN BRAINLIEST!!!

11. In triangle DEF, mD = 43 degrees, mE = 62 degrees, and EF = 22 in. What is DE to the nearest tenth of an inch?

Respuesta :

smmwaite
Answer
31.12 in

Explanation
To get the length required, lets first find angle mF,
mF = 180 – (43 + 62) =75o
Now we can use the sin rule to find length DE.
a/sin⁡A =b/sin⁡B Where a and b are the length opposite to angle A and B respectively.
DE/sin75=22/sin⁡43
DE=(22 sin⁡75)/sin⁡43
DE=31.12 in  

Answer:

31.2 inches.

Step-by-step explanation:

Given,

In triangle DEF,

m∠D = 43°,

m∠E = 62°

EF = 22 in

Since, the sum of all interior angle of a triangle is 180°,

⇒ m∠D + m∠E + m∠F = 180°

⇒ 43° + 62° + m∠F = 180°

⇒ 105° + m∠F = 180° ⇒ m∠F = 180° - 105° = 75°,

Now, by the law of sine,

[tex]\frac{sin D}{EF}=\frac{sin F}{DE}[/tex]

[tex]DE\times sin D = EF\times sin F[/tex]  ( Cross multiplication ),

[tex]DE=\frac{EF sin F}{sin D}[/tex]

By substituting values,

[tex]DE=\frac{22\times sin 75^{\circ}}{sin 43^{\circ}}=\frac{21.2503681784}{0.68199836006}=31.1589725473\approx 31.2[/tex]

Thus, the measure of side DE is 31.2 inches.

Otras preguntas