How do you solve this? My algebra 2 teacher taught it to us in class but it still confuses me! Thank you!
1 answer:
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For non-right triangles you must use the "Law of Cosines" and then, the "Law of Sines" to solve this<span>.
a= </span> 8.25m<span>
b=</span> 10.4m<span>
c= </span>3.16m
∠<span>A= UNKNOWN
</span>∠<span>B= UNKNOWN
</span>∠<span>C=UNKNOWN
Law of Cosines:
c</span>²= a²+b²-2abCos(C)
(3.16)²= (8.25)²+(10.4)²- 2(8.25)(10.4)(cos(C))
9.9856 = 68.0625 + (108.16) - (171.6)(cos(C)
9.9856 = 176.2225- 171.6 cos C
-166.2369= - (171.6(cosC))
cosC= 0.968746503
<span>Take the inverse cosine of that to get the measure of angle C
</span>∠C= 15.95813246°
<span>
Now Use law of sines to find </span>∠B:
(take the inverse sine to get the measure of ∠B)
∠B= 60.8040992°
<span>
Answer:
The angle measures approximately 60.80</span>°.<span>
</span>
a= </span> 8.25m<span>
b=</span> 10.4m<span>
c= </span>3.16m
∠<span>A= UNKNOWN
</span>∠<span>B= UNKNOWN
</span>∠<span>C=UNKNOWN
Law of Cosines:
c</span>²= a²+b²-2abCos(C)
(3.16)²= (8.25)²+(10.4)²- 2(8.25)(10.4)(cos(C))
9.9856 = 68.0625 + (108.16) - (171.6)(cos(C)
9.9856 = 176.2225- 171.6 cos C
-166.2369= - (171.6(cosC))
cosC= 0.968746503
<span>Take the inverse cosine of that to get the measure of angle C
</span>∠C= 15.95813246°
<span>
Now Use law of sines to find </span>∠B:
(take the inverse sine to get the measure of ∠B)
∠B= 60.8040992°
<span>
Answer:
The angle measures approximately 60.80</span>°.<span>
</span>