Question

Solve triangle ABC which haveangleC=1250:431;a=4:2cmandc=8:2cm:Findb:

Answer 1

Answer on Question #51011 – Math – Trigonometry

Solve triangle ABC which has angle C=1250:431; a=4:2 cm and c=8:2 cm. Find b

Solution

Fig.1

The law of cosines (see Fig.1) tells us that

$$c ^ {2} = a ^ {2} + b ^ {2} - 2 a b \cos \angle C$$

which is equivalent to

$$b ^ {2} - 2 a b \cos \angle C + a ^ {2} - c ^ {2} = 0$$

After plugging all known values, obtain the following equation:

$$b ^ {2} - 2 \cdot \frac {4}{2} b \cos \frac {1 2 5 0}{4 3 1} + \left(\frac {4}{2}\right) ^ {2} - \left(\frac {8}{2}\right) ^ {2} = 0$$

that is,

$$b ^ {2} - 4 b \cos \frac {1 2 5 0}{4 3 1} - 1 2 = 0$$ $$\begin{array}{l} D = \left[ 4 \cos \left(\frac {1 2 5 0}{4 3 1}\right) \right] ^ {2} + 4 \cdot 1 2 \\ b _ {1, 2} = \frac {4 \cos \left(\frac {1 2 5 0}{4 3 1}\right) \pm \sqrt {1 6 \left[ \cos \left(\frac {1 2 5 0}{4 3 1}\right) \right] ^ {2} + 4 \cdot 1 2}}{2} \\ \end{array}$$

Because $b > 0$ as the length of triangle, then take

$$b = \frac {4 \cos \left(\frac {1250}{431}\right) + \sqrt {16 \left[ \cos \left(\frac {1250}{431}\right) \right] ^ {2} + 4 \cdot 12}}{2} = 2 \cos \left(\frac {1250}{431}\right) + 2 \sqrt {\left[ \cos \left(\frac {1250}{431}\right) \right] ^ {2} + 3} \approx 2.03 \text{cm}$$

Answer: $b = 2\cos \left(\frac{1250}{431}\right) + 2\sqrt{\left[\cos\left(\frac{1250}{431}\right)\right]^2 + 3} \approx 2.03\text{cm}$

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