Question

Determine whether each triangle should be solved by beginning with the Law of Sines or the Law of Cosines. Then solve each triangle. Round measures of sides and angles to the nearest tenth after calculating. a = 8, b = 7, c = 4 Question 3 options: Law of Cosines; A ≈ 89°, B ≈ 61°, C ≈ 30° Law of Sines; A ≈ 89°, B ≈ 61°, C ≈ 30° Law of Sines; A ≈ 30°, B ≈ 61°, C ≈ 89° Law of Cosines; A ≈ 61°, B ≈ 89°, C ≈ 30°

Answer 1

Answer:

Law of Cosines; A ≈ 61°, B ≈ 89°, C ≈ 30°

Step-by-step explanation:

In this problem the given values are the length sides of the triangle, therefore, the triangle should be solved by beginning with the Law of Cosines

step 1

Applying the law of cosines find the value of angle C

we know that

$c^{2}=a^{2}+b^{2}-2(a)(b)cos(C)$

we have

$a = 8, b = 7, c = 4$

substitute the values and solve for cos(C)

$4^{2}=8^{2}+7^{2}-2(8)(7)cos(C)$

$16=64+49-112cos(C)$

$16=113-112cos(C)$

$112cos(C)=113-16$

$cos(C)=97/112$

$C=arccos(97/112)=30\°$

step 2

Applying the law of cosines find the value of angle B

we know that

$b^{2}=a^{2}+c^{2}-2(a)(c)cos(B)$

we have

$a = 8, b = 7, c = 4$

substitute the values and solve for cos(B)

$7^{2}=8^{2}+4^{2}-2(8)(4)cos(B)$

$49=64+16-64cos(B)$

$49=80-64cos(B)$

$64cos(B)=80-49$

$cos(B)=31/64$

$B=arccos(31/64)=61\°$

step 3

Find the measure of angle A

we know that

The sum of the interior angles of a triangle must be equal to 180 degrees

so

$A+B+C=180\°$

we have

$C=30\°$

$B=61\°$

substitute and solve for A

$A+61\°+30\°=180\°$

$A+91\°=180\°$

$A=180\°-91\°=89\°$