Use the law of cosine to determine lengths and angles Geometry

According to the law of cosine, for a triangle ABC with AB=x and AC=y we have:

BC = \sqrt{x^2+y^2-2xy\cos\left(\widehat{A}\right)}

Here, we have:

• AB=2
• AC=3
• \angle A=60°

Therefore:

c = \sqrt{2^2 + 3^2 -\left(2\times 2\times 3 \times\cos\left(60^\circ\right)\right)}

c=\sqrt{4+9 -\left( 12 \times{\dfrac{1}{2}}\right)}

According to the law of cosine, for a triangle ABC with AB=x, AC=y and BC=z, we have:

z^2 = x^2+y^2-2xy\cos\left(\widehat{A}\right)

Here, we have:

• AB=\sqrt8
• AC=\sqrt{5}
• BC=3

Therefore:

3^2 = \left({\sqrt 5}\right)^2 + \left({\sqrt 8}\right)^2 -\left(2\times {\sqrt 5}\times {\sqrt 8} \times\cos\left(\theta\right)\right)

9 = 5 + 8 -2 {\sqrt 40}\cos\left(\theta\right)

-4 = -2 {\sqrt 40}\cos\left(\theta\right)

\cos\left(\theta\right) =\dfrac{2}{\sqrt{40}} = \dfrac{2}{\sqrt{4}\times\sqrt{10}} =\dfrac{1}{\sqrt{10}}= \dfrac{\sqrt{10}}{10}

According to the law of cosine, for a triangle ABC with AB=x, AC=y and BC=z, we have:

z^2 = x^2+y^2-2xy\cos\left(\widehat{A}\right)

Here, we have:

• AB=\sqrt3
• AC=4
• BC=\sqrt7

Therefore:

\left(\sqrt7\right)^2 = \left({\sqrt 3}\right)^2 + \left({4}\right)^2 -\left(2\times {\sqrt 3}\times {4} \times\cos\left(\widehat{A}\right)\right)

7 = 3 + 16 -\left(8\times {\sqrt 3}\times\cos\left(\widehat{A}\right)\right)

-12= -\left(8\times {\sqrt 3}\times\cos\left(\widehat{A}\right)\right)

\cos\left(\widehat{A}\right)= \dfrac{12}{8\times {\sqrt 3}}= \dfrac{\sqrt 3}{2}

\widehat{A} = 30^\circ

According to the law of cosine, for a triangle ABC with AB=x, BC=y and AC=z, we have:

z^2 = x^2+y^2-2xy\cos\left(\widehat{B}\right)

Here, we have:

• AC=\sqrt{2}
• BC=2

Therefore:

AB= \sqrt{\left({\sqrt 2}\right)^2 + \left(2\right)^2 -\left(2\times {\sqrt 2}\times {2} \times\cos\left(\widehat{B}\right)\right)}

= \sqrt{2 + 4 -\left(2\times {\sqrt 2}\times {2} \times\dfrac{\sqrt{2}}{2}\right)}=\sqrt{2}

Hence:

AB=\sqrt{2}

According to the law of cosine, for a triangle ABC with AB=x, AC=y and BC=z, we have:

z^2 = x^2+y^2-2xy\cos\left(\widehat{A}\right)

Here, we have:

• AB=3
• AC=4

Therefore:

3^2 = \left({3}\right)^2 + \left({4}\right)^2 -\left(2\times {3}\times {4} \times\cos\left(60\right)\right)

BC =\sqrt{9 + 16 -\left(2\times3\times 4\times \dfrac{1}{2}\right)} = \sqrt{13}

According to the law of cosine, for a triangle ABC with AB=x, AC=y and BC=z, we have:

z^2 = x^2+y^2-2xy\cos\left(\widehat{A}\right)

Here, we have:

• AB=\sqrt2
• AC=\sqrt{2}
• BC=2

Therefore:

2^2 = \left({\sqrt 2}\right)^2 + \left({\sqrt 2}\right)^2 -\left(2\times {\sqrt 2}\times {\sqrt 2} \times\cos\left(\widehat{A}\right)\right)

4 = 2 + 2 -4\cos\left(\widehat{A}\right)

4\cos\left(\widehat{A}\right)=0

\widehat{A}=90^\circ

According to the law of cosine, for a triangle ABC with AB=x, AC=y and BC=z, we have:

z^2 = x^2+y^2-2xy\cos\left(\widehat{A}\right)

Here, we have:

• BC=\sqrt{21}
• AC=4

Therefore:

\left({\sqrt 21}\right)^2 = 4^2 + x^2 -\left(2\times 4 \times x\times\cos\left(60^\circ\right)\right)

21= \left(16 + x^2 -\left(8\times x\times \dfrac{1}{2}\right)\right)

x^2 -4 x-5=0

\left(x+1\right)\left(x-5\right)=0

• x=-1
• x=5

Of course x=-1 is not acceptable. Hence x=5.