Notes: `A = 120^@, a = b = 25` Use the law of sines to solve (if possible) the triangle. If two solutions exist, find both. Round your answer to...

Saturday, 29 October 2016

$\displaystyle{A}={{120}}^{\circ},{a}={b}={25}$ Use the law of sines to solve (if possible) the triangle. If two solutions exist, find both. Round your answer to...

The given in the triangle are A=120^o, a=25 and b=25.

To solve using Sine Law, let's use the formula:

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

Plug-in the values of the sides a and b. Also, plug-in the value of angle A.

$\displaystyle\frac{{\sin{{\left({{120}}^{{o}}\right)}}}}{{25}}=\frac{{{\sin{{B}}}}}{{25}}$

Then, simplify the equation. To simplify, cancel the denominators.

$\displaystyle\frac{{\sin{{\left({{120}}^{{o}}\right)}}}}{{25}}\cdot{25}=\frac{{{\sin{{B}}}}}{{25}}\cdot{25}$

$\displaystyle{\sin{{\left({{120}}^{{o}}\right)}}}={\sin{{B}}}$

$\displaystyle{{120}}^{{o}}={B}$

Now two angles of the triangle are known, which are $\displaystyle{A}={{120}}^{{o}}$ and $\displaystyle{B}={{120}}^{{o}}$ .

Take note that the sum of three angles of the triangle...

The given in the triangle are A=120^o, a=25 and b=25.

To solve using Sine Law, let's use the formula:

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

Plug-in the values of the sides a and b. Also, plug-in the value of angle A.

$\displaystyle\frac{{\sin{{\left({{120}}^{{o}}\right)}}}}{{25}}=\frac{{{\sin{{B}}}}}{{25}}$

Then, simplify the equation. To simplify, cancel the denominators.

$\displaystyle\frac{{\sin{{\left({{120}}^{{o}}\right)}}}}{{25}}\cdot{25}=\frac{{{\sin{{B}}}}}{{25}}\cdot{25}$

$\displaystyle{\sin{{\left({{120}}^{{o}}\right)}}}={\sin{{B}}}$

$\displaystyle{{120}}^{{o}}={B}$

Now two angles of the triangle are known, which are $\displaystyle{A}={{120}}^{{o}}$ and $\displaystyle{B}={{120}}^{{o}}$ .

Take note that the sum of three angles of the triangle is $\displaystyle{{180}}^{{o}}$ .

$\displaystyle{A}+{B}+{C}={{180}}^{{o}}$

However, the sum of these two angles A and B is:

$\displaystyle{A}+{B}={{120}}^{{o}}+{{120}}^{{o}}={{240}}^{{o}}$

which is larger than $\displaystyle{{180}}^{{o}}$ .

Therefore, it is not possible to form a triangle with the given measures $\displaystyle{A}={{120}}^{{o}}$ , $\displaystyle{a}={25}$ and $\displaystyle{b}={25}$ .

Notes

Saturday, 29 October 2016

$\displaystyle{A}={{120}}^{\circ},{a}={b}={25}$ Use the law of sines to solve (if possible) the triangle. If two solutions exist, find both. Round your answer to...

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Is there any personification in "The Tell-Tale Heart"?

Saturday, 29 October 2016

Use the law of sines to solve (if possible) the triangle. If two solutions exist, find both. Round your answer to...

No comments:

Post a Comment

Is there any personification in &quot;The Tell-Tale Heart&quot;?

$\displaystyle{A}={{120}}^{\circ},{a}={b}={25}$ Use the law of sines to solve (if possible) the triangle. If two solutions exist, find both. Round your answer to...

Is there any personification in "The Tell-Tale Heart"?