Respuesta :

Step 1:

Calculate the measure of angle ∠ABC

[tex]\angle DBC+\angle ABC=180(\text{ sum of angles on a straight line)}[/tex][tex]\angle ABC=65^0[/tex][tex]\begin{gathered} \angle DBC+\angle ABC=180 \\ \angle DBC+65^0=180^0 \\ \angle DBC=180^0-65^0 \\ \angle DBC=115^0 \end{gathered}[/tex]

From the triangle in the question,

[tex]a=10\operatorname{km},c=15\operatorname{km},B=115^0[/tex]

Step 2:

Calculate the value of AB using the cosine rule below

[tex]b^2=a^2+c^2-2\times a\times c\times\cos B[/tex]

By substituting the values, we will have

[tex]\begin{gathered} b^2=a^2+c^2-2\times a\times c\times\cos B \\ b^2=10^2+15^2-2\times10\times15\times\cos 115^0 \\ b^2=100+225-300\times(-0.4226) \\ b^2=325+126.78 \\ b^2=451.78 \\ \text{Square root both sides} \\ \sqrt[]{b^2}=\sqrt[]{451.78} \\ b=21.26\operatorname{km} \end{gathered}[/tex]

Hence,

The distance of point A to point C is = 21.26km

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