Find the measure of Q, the smallest angle in a triangle whose sides have lengths 4, 5, and 6. Round the measure to the nearest whole degree.


34°

41°

51°

56°

Find the measure of Q the smallest angle in a triangle whose sides have lengths 4 5 and 6 Round the measure to the nearest whole degree 34 41 51 56 class=

Respuesta :

Answer:

[tex]Q=41\°[/tex]

Step-by-step explanation:

we know that

Applying the law of cosine

[tex]c^{2} =a^{2}+b^{2}-2abcos (C)[/tex]

Solve for cos(C)

[tex]cos (C)=[a^{2}+b^{2}-c^{2}]/[2ab][/tex]

In this problem we have

[tex]a=6\ units,b=5\ units, c=4\ units, C=Q[/tex]

substitute the values

[tex]cos (Q)=[6^{2}+5^{2}-4^{2}]/[2*6*5][/tex]

[tex]cos (Q)=[45]/[60][/tex]

[tex]Q=cos^{-1}(\frac{45}{60}) =41\°[/tex]

The measure of the ∠Q is 41°.

What is the Law of Cosine?

The law of cosine helps us to know the third side of a triangle when two sides of the triangle are already known the angle opposite to the third side is given. It is given by the formula,

[tex]c =\sqrt{a^2 + b^2 -2ab\cdot Cos\theta}[/tex] c =\sqrt{a^2 + b^2 -2ab\cdot Cos\theta}

where

c is the third side of the triangle

a and b are the other two sides of the triangle,

and θ is the angle opposite to the third side, therefore, opposite to side c.

Given to us

  • PR = 4
  • RQ = 6
  • PQ = 5

Law of Cosine

According to the law of cosine,

[tex]c =\sqrt{a^2 + b^2 -2ab\cdot Cos\theta}[/tex]

Substituting the values,

[tex]PR =\sqrt{PQ^2 + RQ^2 -2(PQ)(RQ)\cdot Cos(\angle Q)}[/tex]

[tex]4 =\sqrt{5^2 + 6^2 -2\times 5 \times 6\cdot Cos(\angle Q)}\\\\4^2 =5^2 + 6^2 -2\times 5 \times 6\cdot Cos(\angle Q)\\\\16 = 25 + 36 - 60 Cos(\angle Q)\\\\16 -25 -36= -60 Cos(\angle Q)\\\\-45 = -60 Cos(\angle Q)\\\\\dfrac{45}{60} = Cos(\angle Q)\\\\(\angle Q) = Cos^{-1} (\dfrac{45}{60} )\\\\(\angle Q) = 41.4^o \approx 41^o[/tex]

Hence, the measure of the ∠Q is 41°.

Learn more about the Law of Cosine:

https://brainly.com/question/17289163