Q:

# Find the solutions for a triangle with a =11.4, b =13.7, and c =12.2.

Accepted Solution

A:
Answer:$$\large \boxed{\angle A = 51.83 ^{\circ}; \, \angle B = 70.88 ^{\circ}; \, \angle C = 57.29 ^{\circ}}$$Step-by-step explanation:You use the Law of Cosines when you know all three sides and want to find the angles of a triangle. For example, if you want to find ∠A, you use the formula $$\cos A = \dfrac{b^{2} + c^{2} - a^{2}}{2bc}$$1. ∠ A $$\begin{array}{rcl}\cos A &=& \dfrac{b^{2} + c^{2} - a^{2}}{2bc}\\\\& = & \dfrac{13.7^{2} + 12.2^{2} - 11.4^{2}}{2\times 13.7 \times 12.2}\\\\& = & \dfrac{187.69 + 148.84 - 129.96}{334.28}\\\\&=& \dfrac{206.57}{334.28}\\\\& = & 0.6180\\A& = & \arccos 0.6180\\& = & \mathbf{51.83 ^{\circ}}\\\end{array}$$ 2. ∠B $$\begin{array}{rcl}\cos B &=& \dfrac{a^{2} + c^{2} - b^{2}}{2ac}\\\\& = & \dfrac{11.4^{2} + 12.2^{2} - 13.7^{2}}{2\times 11.4 \times 12.2}\\\\& = & \dfrac{129.96 + 148.84 - 187.69}{278.16}\\\\&=& \dfrac{91.11}{278.16}\\\\& = & 0.3275\\B& = & \arccos 0.3275\\& = & \mathbf{70.88 ^{\circ}}\\\end{array}$$ 3. ∠C $$\begin{array}{rcl}\cos C &=& \dfrac{a^{2} + b^{2} - c^{2}}{2bc}\\\\& = & \dfrac{11.4^{2} + 13.7^{2} - 12.2^{2}}{2\times 11.4 \times 13.7}\\\\& = & \dfrac{129.96 + 187.69 - 148.84}{312.36}\\\\&=& \dfrac{168.81}{312.36}\\\\& = & 0.5404\\C& = & \arccos 0.5404\\& = & \mathbf{57.29 ^{\circ}}\\\end{array}\\\text{The three angles are \large \boxed{\mathbf{\angle A = 51.83 ^{\circ}; \, \angle B = 70.88 ^{\circ}; \, \angle C = 57.29 ^{\circ}}}}$$