Question

Solve the triangle.

B = 36°, a = 38, c = 18

Help me please

Answer 1

Answer on question 42445 – Math - Geometry

Solve the triangle. $B = 36{}^{\circ}$, $a = 38$, $c = 18$.

Solution:

1) Use the Law of Cosines to find the side $b$:

$$\cos(36{}^{\circ}) = \frac{1 + \sqrt{5}}{4} \approx 0.809017$$ $$b^{2} = a^{2} + c^{2} - 2ac \cos(B) = 38^{2} + 18^{2} - 2 \cdot 38 \cdot 18 \cdot \cos(36{}^{\circ}) \approx 1444 + 324 - 1368 \cdot 0.809017 \approx 661.264744$$

$b \approx \sqrt{661.264744} \approx 25.715$ (rounded to 3 decimal places)

2) Use the Law of Cosines to find angle $A$:

$$\cos(A) = \frac{b^{2} + c^{2} - a^{2}}{2bc} \approx \frac{661.264744 + 324 - 1444}{2 \cdot 25.715 \cdot 18} \approx \frac{-458.735256}{925.74} \approx -0.4955$$

$A \approx \arccos(-0.4955) \approx 119.7{}^{\circ}$ (rounded to 1 decimal place)

3) Use the Law of Sine to find angle $C$:

$$\frac{25.715}{\sin(36{}^{\circ})} = \frac{18}{\sin(C)} \Rightarrow \sin(C) \approx \frac{18 \sin(36{}^{\circ})}{25.715} \approx 0.4114$$

$C \approx \arcsin(0.4114) \approx 24.3{}^{\circ}$

Answer: $b = 25.715$, $A = 119.7{}^{\circ}$, $C = 24.3{}^{\circ}$.

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