Question

What is the measurement of the largest angle in ΔABC such that a = 15 cm, b = 41 cm, and c = 29 cm? 135.0° 106.2° 50.0° 30.0°

Answer 1

From the given question

There are given that the sides of the triangle:

$a=15\operatorname{cm},\text{ b=41 cm, c=29 cm}$

Now,

Find the first angle by using the cosine rule

So,

For angle A,

$a^2=b^2+c^2-2bc\cos A$

Then,

$\begin{gathered} a^2=b^2+c^2-2bc\cos A \\ 15^2=41^2+29^2-2\times41\times29\times\cos A \\ 225^{}=1681+841-2378\times\cos A \\ 225=2522-2378\cos A \\ -2297=-2378\cos A \\ \cos A=\frac{2297}{2378} \\ \cos A=0.96 \end{gathered}$

Then,

$\begin{gathered} A=\cos ^{-1}(0.96) \\ A=16.26 \end{gathered}$

Now,

for the second angle B,

Use sine law:

So,

$\begin{gathered} \frac{\sin A}{a}=\frac{\sin B}{b} \\ \frac{\sin16.26}{15}=\frac{\sin B}{41} \\ \sin B\times15=\sin 16.26\times41 \\ \sin B=\frac{\sin16.26\times41}{15} \\ \sin B=\frac{11.5}{15} \\ \sin B=0.76 \end{gathered}$

Then,

$\begin{gathered} \sin B=0.76 \\ B=135.0 \end{gathered}$

Now,

For the third angle:

$\begin{gathered} \angle A+\angle B+\angle C=180^{\circ} \\ 16.26+135+\angle C=180^{\circ} \\ \angle C=28.74 \end{gathered}$

Hence, the correct option is A.