Question

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

Answer:

$Q=41\°$

Step-by-step explanation:

we know that

Applying the law of cosine

$c^{2} =a^{2}+b^{2}-2abcos (C)$

Solve for cos(C)

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

In this problem we have

$a=6\ units,b=5\ units, c=4\ units, C=Q$

substitute the values

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

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

$Q=cos^{-1}(\frac{45}{60}) =41\°$

Answer 2

Using SSS the given angles are 83, 56, and 41
41.41 is angle Q
41.41 so it's B