Find all the missing elements.
Round to the nearest tenth.

a = 4
b = 5
c = 7


A=? °

B= ? °

C= ? °

Answer:

[tex]4^2=5^2+7^2-2(5)(7) cos A\\\\[/tex]

[tex]A=34.0^o[/tex]

-----------------

[tex]\frac{4}{Sin34^o} =\frac{5}{SinB} \\\\[/tex]

[tex]B=44.4^o[/tex]

--------------

[tex]7^2=4^2+5^2-2(4)(5)CosC[/tex]

[tex]C= 101.5^o[/tex]

[tex]---------[/tex]

hope it helps...

have a great day!!

All the missing elements of the triangle by the Cosine rule are A = 40°, B = 45°, and C = 102°.

Given that,

Three sides on a triangle are,

a = 4

b = 5

c = 7

Used the formula of the Cosine rule which states that,

a² = b² + c² - 2bc cos A

b² = a² + c² - 2ac cos B

c² = a² + b² - 2ab cos C

Where, a, b, and c are three sides and A, B, and C are angles of a triangle.

Now, apply the Cosine rule to find the all angles of a given triangle,

a² = b² + c² - 2bc cos A

Substitute a = 4, b = 5 and c = 7,

4² = 5² + 7² - 2 × 5 × 7 cos A

16 = 25 + 49 - 70 cos A

70 cos A = 74 - 16

70 cos A = 58

Divide both sides by 58,

cos A = 58/70

cos A = 0.83

A = cos ⁻¹ (0.83)

A = 33.9°

Round to the whole number,

A = 40°

b² = a² + c² - 2ac cos B

Substitute a = 4, b = 5 and c = 7,

5² = 4² + 7² - 2 × 4 × 7 cos B

25 = 16 + 49 - 56 cos B

25 = 65 - 56 cos B

56 cos B = 65 - 25

56 cos B = 40

cos B = 40/56

cos B = 0.71

B = cos ⁻¹ (0.71)

B = 44.7°

Round to the whole number,

B = 45°

c² = a² + b² - 2ab cos C

Substitute a = 4, b = 5 and c = 7,

7² = 4² + 5² - 2 × 4 × 5 cos C

49 = 16 + 25 - 40 cos C

49 = 41 - 40 cos C

40 cos C = 41 - 49

40 cos C = - 8

cos C = - 8/40

cos C = - 1/5

cos C = - 0.2

C = cos ⁻¹ (0.2)

C = 101.5°

Round to the whole number,

C = 102°

To learn more about the triangle visit;

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