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the question is in the picture below, if you could show all your work so i can see how to do it, that would be appreciated.

the question is in the picture below, if you could show all your work so i can see-example-1

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m∠R = 27.03°

Solution:

Given In ΔQRP, p = 28 km, q = 17 km, r = 15 km

To find the measure of angle R:

Law of cosine formula for ΔQRP:


r^(2)=p^(2)+q^(2)-2 pq \cos R

Substitute the given values in the above formula.


(15)^(2)=(28)^(2)+(17)^(2)-2 (28)(17) \cos R


225=784+289-952 \cos R


225=1073-952 \cos R

Switch the given equation.


1073-952 \cos R=225

Subtract 1073 from both side of the equation.


-952 \cos R=-848

Divide by –952 on both sides.


$ \cos R=(106)/(119)


$ R=\cos^(-1) \left((106)/(119)\right)


R=27.03^\circ

Hence m∠R = 27.03°.

User Nexneo
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