Questions: Solve this triangle A = (Round to one decimal place as needed.) Triangle ABC B = 5 C = 7 A = 10

Solution Steps

We are given a triangle with sides a = 7, b = 5, and c = 10. We are asked to solve the triangle, which means finding all missing angles.

The Law of Cosines states that $a^2 = b^2 + c^2 - 2bc \cos(A)$. Plugging in the given values, we get:

$7^2 = 5^2 + 10^2 - 2(5)(10) \cos(A)$ $49 = 25 + 100 - 100 \cos(A)$ $49 = 125 - 100 \cos(A)$ $100 \cos(A) = 125 - 49$ $100 \cos(A) = 76$ $\cos(A) = \frac{76}{100} = 0.76$ $A = \arccos(0.76) \approx 40.54^\circ$

$b^2 = a^2 + c^2 - 2ac \cos(B)$ $5^2 = 7^2 + 10^2 - 2(7)(10) \cos(B)$ $25 = 49 + 100 - 140 \cos(B)$ $25 = 149 - 140 \cos(B)$ $140 \cos(B) = 149 - 25$ $140 \cos(B) = 124$ $\cos(B) = \frac{124}{140} \approx 0.8857$ $B = \arccos(0.8857) \approx 27.66^\circ$

Final Answer: