Question

Solve the triangle.

A = 51°, b = 11, c = 7

Answer 1

Answer on Question #53921 – Math – Geometry

Question:

Solve the triangle.

$$A = 51{}^{\circ}, b = 11, c = 7$$

Answer:

Law of cosines:

$$a^2 = b^2 + c^2 - 2cb \cdot \cos A$$

where $A$ denotes the angle contained between sides of lengths $a$ and $b$ and opposite the side of length $c$.

From cosine law we get:

$$a^2 = 11^2 + 7^2 - 2 \cdot 11 \cdot 7 \cdot \cos 51{}^{\circ} \approx 73$$

then $a = 8.55$

Law of sines:

$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$

where $a$, $b$, and $c$ are the lengths of the sides of a triangle, and $A$, $B$, and $C$ are the opposite angles.

From sine law we get:

$$\frac{8.55}{\sin 51} = \frac{11}{\sin B} = \frac{7}{\sin C}$$

then

$$B = 89.5{}^{\circ} \text{ and } C = 39.5{}^{\circ}$$

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