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Two points, A and B, are on opposite sides of a building. A surveyor chooses a third point, C, 60 yd from B and 105 yd from A, with angle ACB measuring 69.3 . How far apart are A and B (to the nearest yard)?

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Riia

In the given question it is given that , AC = 105 yd, BC = 60 yd and angle ACB= 69.3 degree.

We have to find the distance of AB, and for that we use law of cosine, which is

[tex] (AB)^2 = (AC)^2 +(BC)^2-2(AC)(BC) cos \theta [/tex]

Substituting the values of AC, BC and theta, we will get

[tex] (AB)^2 = 105^2 +60^2 -2(105)(60) cos 69.3 [/tex]

[tex] (AB)^2 = 10171.217 [/tex]

[tex] AB= 101 yard [/tex]

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