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Kalahira
11.9 Actually providing the diagram would be useful. Also, distinguishing between upper and lower case would also be useful. But given the convention that angles and the opposite side of the triangle are both given the same letter, but different case, I will assume that angle Q is opposite side q and that you have a situation of SAS. So using the law of cosines we have: c^2 = a^2 + b^2 - 2ab cos C Substitute the known values: c^2 = a^2 + b^2 - 2ab cos C c^2 = 20^2 + 30^2 - 2*20*30*cos 15 c^2 = 400 + 900 - 1200*0.965925826 c^2 = 400 + 900 - 1159.110992 c^2 = 140.8890085 c = 11.86966758 Rounded to the nearest tenth, gives 11.9
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Answer:

[tex]q=11.9[/tex]

Step-by-step explanation:

It is given that the value of r is [tex]20[/tex], s is [tex]30[/tex] and the angle q is [tex]15^{\circ}[/tex].

In order to find the value of the length of q, we use the Law of cosines, thus

[tex]q^2=r^2+s^2-2rscosq^{\circ}[/tex]

Substituting the given values, we get

[tex]q^2=(20)^+(30)^2-2(20)(30)cos15^{\circ}[/tex]

[tex]q^2=400+900-1200(0.965)[/tex]

[tex]q^2=1300-1158.6[/tex]

[tex]q^2=141.4[/tex]

[tex]q=11.9[/tex]

which is the required vale of the length of q.

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