Questions: （2）△ABC で，C=30°, c=5 のとき，この三角形の外接円の半径を求めなさい。 （3）次の値を求めなさい。 （1）右図の △ABC で，b=2, c=3, A=60° のとき，辺 BC の長さは a=√ （2）右図の △ABC で，a=√13, b=3, c=4 のとき，LA の大きさは

Solution Steps

We are given $b=2$, $c=3$, and $A=60^{\circ}$. We can use the cosine rule to find the length of side $a$: $a^2 = b^2 + c^2 - 2bc\cos{A}$ $a^2 = 2^2 + 3^2 - 2(2)(3)\cos{60^{\circ}}$ $a^2 = 4 + 9 - 12(1/2)$ $a^2 = 13 - 6$ $a^2 = 7$ $a = \sqrt{7}$

We are given $a=\sqrt{13}$, $b=3$, and $c=4$. We can use the cosine rule to find the cosine of angle A: $a^2 = b^2 + c^2 - 2bc\cos{A}$ $13 = 3^2 + 4^2 - 2(3)(4)\cos{A}$ $13 = 9 + 16 - 24\cos{A}$ $13 = 25 - 24\cos{A}$ $24\cos{A} = 12$ $\cos{A} = \frac{12}{24} = \frac{1}{2}$ Since $\cos{A} = \frac{1}{2}$, $A = 60^{\circ}$.

We are given $C=30^{\circ}$ and $c=5$. We can use the sine rule to find the radius (R) of the circumcircle: $\frac{c}{\sin{C}} = 2R$ $\frac{5}{\sin{30^{\circ}}} = 2R$ $\frac{5}{1/2} = 2R$ $10 = 2R$ $R = 5$

Final Answer