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Consider a triangle ABC like the one below. Suppose that A = 31°, C = 52°, and b = 29. (The figure is not drawn to scale.) Solve the triangle.Round your answers to the nearest tenth.If there is more than one solution, use the button labeled "or".

Consider A Triangle ABC Like The One Below Suppose That A 31 C 52 And B 29 The Figure Is Not Drawn To Scale Solve The TriangleRound Your Answers To The Nearest class=

Sagot :

a = 15, b = 29 and c = 23

A= 31º B= 97º C= 52º

1) To solve a triangle is to find out its angles and measures. Since we have angle A= 31º, C =52º, and b= 29 we can solve this by using the Sum of Interior angles and the Law of sines:

[tex]\frac{a}{\sin(A)}=\frac{b}{\sin(B)}=\frac{c}{\sin(C)}[/tex]

2) The sum of the interior angles of a Triangle is 180º So, we can find angle B this way

∠A +∠B+∠C= 180º

31º +∠B +52º = 180º

∠B + 83º = 180º Subtract 83 from both sides

∠B =97º

[tex]\begin{gathered} \frac{29}{\sin(97)}=\frac{c}{\sin(52)} \\ \\ c=\frac{29\cdot\sin(52)}{\sin(97)}\Rightarrow c\approx23 \end{gathered}[/tex]

2.2) Now we can find the leg "a "

[tex]\begin{gathered} \frac{a}{\sin (31)}=\frac{29}{\sin (97)} \\ a\cdot\sin (97)\text{ = 29 }\cdot\sin (31) \\ a=\frac{29\cdot\sin (31)}{\sin (97)}\Rightarrow a\approx15 \end{gathered}[/tex]

3) Hence, the answer is

a = 15, b = 29 and c = 23

A= 31º B= 97º C= 52º

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