Consider the quadrilateral shown below that has vertices A, B, C, and D.

Suppose we know the following measurements:
¯¯¯¯¯¯¯¯AB=5.8 cm
¯¯¯¯¯¯¯¯BC=5.4 cm
¯¯¯¯¯¯¯¯¯CD=3.3 cm
¯¯¯¯¯¯¯¯¯AD=6.2 cm
m∠A=27∘
What is the degree measure of ∠C?
m∠C=  °

Consider the quadrilateral shown below that has vertices A B C and DSuppose we know the following measurementsAB58 cmBC54 cmCD33 cmAD62 cmmA27What is the degree class=

Respuesta :

Answer:

C = 25.94°

Step-by-step explanation:

By applying cosine rule in ΔABD,

BD² = AB² + AD² - 2(AB)(BD)cosA

BD² = (5.8)² + (6.2)² - 2(5.8)(6.2)(cos27°)

BD² = 33.64 + 38.44 - 64.08

BD² = 8

BD = √8

By applying cosine rule in ΔBCD,

BD² = BC² + CD²- 2(BC)(CD)cosC

(√8)² = (5.4)²+ (3.3)² - 2(5.4)(3.3)cosC

8 = 29.16 + 10.89 - 35.64cosC

35.64cosC = 32.05

cosC = 0.89927

C = [tex]\text{cos}^{-1}(0.89927)[/tex]

C = 25.94°

By using cosine formula of triangles we got that measure of ∠C is 25.94 °

What is a quadrilateral ?

Quadrilateral is a closed shape with 4 sides and 4 corners.

Here given that

AB=5.8 cm

BC=5.4 cm

CD=3.3 cm

AD=6.2 cm

and [tex]\angle A=27[/tex] °

in [tex]\triangle \mathrm{ABD}[/tex]

Applying cosine formula

[tex]$\begin{aligned}&\mathrm{BD}^{2}=\mathrm{AB}^{2}+A D^{2}-2(\mathrm{AB})(\mathrm{BD}) \cos A \\\\&\mathrm{BD}^{2}=(5.8)^{2}+(6.2)^{2}-2(5.8)(6.2)\left(\cos 27^{\circ}\right) \\\\&\mathrm{BD}^{2}=33.64+38.44-64.08\\ \\&\mathrm{BD}^{2}=8 \\\\&\mathrm{BD}=\sqrt{8}\end{aligned}$[/tex]

Now in [tex]\triangle B C D[/tex]

Applying cosine formula

[tex]$\begin{aligned}&\mathrm{BD}^{2}=\mathrm{BC}^{2}+\mathrm{CD}^{2}-2(\mathrm{BC})(\mathrm{CD}) \cos \mathrm{C} \\\\&(\sqrt{8})^{2}=(5.4)^{2}+(3.3)^{2}-2(5.4)(3.3) \cos \mathrm{C} \\\\&8=29.16+10.89-35.64 \cos \mathrm{C}\\ \\&35.64 \cos \mathrm{C}=32.05\\ \\&\cos \mathrm{C}=0.89927 \\\\&\mathrm{C}=\cos ^{-1}(0.89927) \\&\end{aligned}$[/tex]

C=25.94 °

By using cosine formula of triangles we got that measure of ∠C is 25.94 °

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