Question

URGENT!!
Find the measure of angle C, to the nearest tenth of a degree, on a triangle with side lengths of a=5, b=12, c=13

A. 78.0°
B. 87.3°
C. 90.0°
D. 115.2°

Answer 1

Answer:

Correct answer is option C. $90.0^\circ$

Step-by-step explanation:

We are given a $\triangle ABC$ and  

the side lengths as following:

$a=5,\\b=12, \\c=13$

We have to find the $\angle C$ i.e. the angle which is opposite to side c.

Formula for cosine rule:

$cos C = \dfrac{a^{2}+b^{2}-c^{2}}{2ab}$

Where  

a is the side opposite to $\angle A$

b is the side opposite to $\angle B$

c is the side opposite to $\angle C$

$\Rightarrow cos C = \dfrac{5^{2}+12^{2}-13^{2}}{2 \times 5 \times 12}\\\Rightarrow cos C = \dfrac{25+ 144 -169}{120} \\\Rightarrow cos C = \dfrac{169 -169}{120} \\\Rightarrow cos C = 0\\\Rightarrow C = 90^\circ$

Please refer to the attached image for labeling and better understanding of the question.

Hence, it is a right angled triangle with $\angle C = 90^\circ$.

Correct answer is option C. $90.0^\circ$