Question

Law of cosines: a2 = b2 + c2 – 2bccos(A) 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°

Answer 1

a² = b² + c² – 2bccos(A)
The smallest angle is across the smallest side, so
4²=5²+6²-2*5*6*cos(Q)
16=25+36-60*cos(Q)
16-36-25=-60*cos(Q)
45=60*cos(Q)
cos(Q)=45/60
Q=cos⁻¹(45/60) ≈41⁰

Answer 2

Answer:The smallest angle of the triangle will be nearest to 41°.

Step-by-step explanation:

since, the sides of the triangle are 4,5 and 6.

when we talk about the smallest angle then it must be the opposite angle of side 4.

Thus according to the law of cosines,

`4^2=5^2+6^2-2*5*6* cos(A)`(where A(let) be the smallest angle of the triangle.)

⇒16=25+36-60cos(A)

⇒16-61=-60cos(A)

⇒-45=-60cos(A)

⇒
`cos(A)=3/4\implies A = cos^(-1)(3/4)` ⇒A=41.4096221093 degree
`\approx` 41°