Question

In triangle ABC m(<A)=30. M(<C)=80. a+2b=15 cm,find the value of c and the length of the radius of the circle

Answer 1

Answer on Question #47781 – Math – Geometry

In triangle ABC $m(<A)=30$. $M(<C)=80$, $a+2b=15$ cm. Find the value of $c$ and the length of the radius of the circle

Solution:

Given:

$$\begin{array}{l} \angle A = 30{}^{\circ} \\ \angle C = 80{}^{\circ} \\ a + 2b = 15; \quad a = 15 - 2b \\ \end{array}$$

c, R–?

We can find the angle $\angle B$ of the triangle:

$$\begin{array}{l} \angle A + \angle B + \angle C = 180{}^{\circ} \\ \angle B = 180{}^{\circ} - \angle A - \angle C = 180{}^{\circ} - 80{}^{\circ} - 30{}^{\circ} = 70{}^{\circ} \\ \end{array}$$

Law of sines for the triangle ABC:

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

Plug(1) into (2):

$$\begin{array}{l} \frac{15 - 2b}{\sin A} = \frac{b}{\sin B} \\ (15 - 2b)\sin B = b\sin A \\ 15 \cdot \sin B = b(\sin A + 2\sin B) \\ b = \frac{15 \cdot \sin B}{\sin A + 2 \sin B} = \frac{15 \, \text{cm} \cdot \sin 70{}^{\circ}}{\sin 30{}^{\circ} + 2 \sin 70{}^{\circ}} = 5.9 \, \text{cm} \\ a = 15 \, \text{cm} - 2 \cdot 5.9 \, \text{cm} = 3.2 \, \text{cm} \\ \end{array}$$

Law of sines for the triangle ABC:

$$c = \frac{b \sin C}{\sin B} = \frac{\frac{b}{\sin B} = \frac{c}{\sin C}}{\frac{5.9 \, \text{cm} \cdot \sin 80{}^{\circ}}{\sin 70{}^{\circ}}} = 6.2 \, \text{cm}$$

Length of the radius of the circle Inscribed within a triangle ABC.

$$R = \frac{\text{Triangle area}}{k} = \frac{\sqrt{k(k - a)(k - b)(k - c)}}{k},$$ $$\text{where } k = \frac{1}{2}(a + b + c) = \frac{1}{2}(3.2 \, \text{cm} + 5.9 \, \text{cm} + 6.2 \, \text{cm}) = 7.65 \, \text{cm}$$ $$R = \frac{\sqrt{7.65 \, \text{cm}(7.65 \, \text{cm} - 3.2 \, \text{cm})(7.65 \, \text{cm} - 5.9 \, \text{cm})(7.65 \, \text{cm} - 6.2 \, \text{cm})}}{7.65 \, \text{cm}} = 1.2 \, \text{cm}$$

Answer: $c = 6.2 \, \text{cm}; R = 1.2 \, \text{cm}$

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