Question

Suppose a triangle has vertices AA, BB and CC and has the following measurements:¯¯¯¯¯¯¯¯AB=23.5 cm ¯¯¯¯¯¯¯¯AC=87.2 cm ¯¯¯¯¯¯¯¯BC=64.8 cmStart by drawing a diagram of this triangle and labeling the known values.What is the degree measure of ∠C?m∠C=  °   What is the degree measure of ∠A?m∠A=  °   What is the degree measure of ∠B?m∠B=  °

Answer 1

First let's draw this triangle:

First let's compare the square of the bigger side with the sum of squares of the other two sides, this way we can find out if the triangle is acute, right or obtuse:

`\begin{gathered} 87.2^2=23.5^2+64.8^2 \\ 7603.84=522.25+4199.04 \\ 7603.84=4721.29 \end{gathered}`

Since the square of the bigger side is greater than the sum of squares, the triangle is obtuse.

In order to calculate the first angle, let's use the law of cosines, using angle A:

`\begin{gathered} 64.8^2=23.5^2+87.2^2-2\cdot23.5\cdot87.2\cdot\cos (A) \\ 4199.04=522.25+7603.84-4098.4\cos (A) \\ \cos (A)=(3927.05)/(4098.4)=0.958_{} \\ A=16.63\degree \end{gathered}`

Now we can use the law of sines for angle B:

`\begin{gathered} (64.8)/(\sin(16.63\degree))=(87.2)/(\sin (B)) \\ \sin (B)=(0.286\cdot87.2)/(64.8)=0.385 \\ B=157.36\degree \end{gathered}`

Finally, for angle C:

`\begin{gathered} (64.8)/(\sin(16.63\degree))=(23.5)/(\sin (C)) \\ \sin (C)=(0.286\cdot23.5)/(64.8)=0.104 \\ C=5.95\degree \end{gathered}`