Answer:
Sure. Here are the steps on how to solve for the sides and angles of triangle ABC using the law of cosines:
Draw a sketch of triangle ABC.
Label the sides and angles of triangle ABC.
Label the known and unknown values.
Use the law of cosines to solve for the unknown side or angle.
In this case, we know the following:
Side b = 60
Side c = 30
Angle A = 70°
We need to solve for side a and angles B and C.
The law of cosines states that
a^2 = b^2 + c^2 - 2bc cos A
where a, b, and c are the sides of the triangle and A is the angle opposite side a.
Plugging in the known values, we get
a^2 = 60^2 + 30^2 - 2(60)(30) cos 70°
a^2 = 900 + 900 - 3600 cos 70°
a^2 = 1800 - 3600 cos 70°
a = sqrt(1800 - 3600 cos 70°)
a = 57
Therefore, side a is 57.
To solve for angle B, we can use the law of sines. The law of sines states that
\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}
Plugging in the known values, we get
\frac{57}{\sin 70°} = \frac{60}{\sin B}
\sin B = \frac{57 \sin 70°}{60}
\sin B = 0.774
B = \sin^{-1}(0.774)
B = 81°
Therefore, angle B is 81°.
To solve for angle C, we can use the fact that the sum of the angles of a triangle is 180°.
A + B + C = 180°
70° + 81° + C = 180°
C = 29°
Therefore, angle C is 29°.
Therefore, the sides and angles of triangle ABC are a = 57, b = 60, c = 30, A = 70°, B = 81°, and C = 29°.
Explanation: