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In △ABC, AC=10 in., BC=2.5 in., and m∠C=63°.

What is the area of △ABC?

Enter your answer as a decimal in the box. Round only your final answer to the nearest hundredth.

User Mhsekhavat
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2 Answers

3 votes
Answer:

Area = 78.98 square inches

Explanation:

We can use the formula for the area of a triangle, which is:

Area = (1/2) * base * height

where the base and height are both unknown. To find them, we can use the law of sines, which states that for any triangle ABC:

a/sin(A) = b/sin(B) = c/sin(C)

where a, b, and c are the side lengths and A, B, and C are the angles opposite those sides.

In this case, we know that AC = 10, BC = 2.5, and C = 63°. Let's call AB = c, and let h be the height from C to AB. Then we have:

10/sin(A) = c/sin(B)
2.5/sin(A) = c/sin(180° - B - A) = c/sin(C)

Using the first two equations, we can solve for c and sin(A) in terms of sin(B):

c = (10sin(B))/sin(A)
2.5 = (csin(A))/sin(B)
= (10*sin(A))/sin(C)

Substituting c from the first equation into the second equation, we get:

2.5 = (10*sin(A)^2)/(sin(B)*sin(C))

Solving for sin(A), we get:

sin(A) = sqrt((2.5*sin(B)sin(C))/10)
= sqrt(0.157sin(B)*sin(C))

Now we can use the formula for the area of a triangle:

Area = (1/2) * AB * h

where h is the height from C to AB. We can find h using sin(A):

h = ACsin(A) = 10sin(A)

Substituting the expression for sin(A), we get:

h = 10sqrt(0.157sin(B)*sin(C))

Finally, we can find the area:

Area = (1/2) * AB * h = (1/2) * c * h = (1/2) * (10sin(B))/sin(A) * 10sqrt(0.157*sin(B)*sin(C))

Area = 78.98 square inches

Therefore, the area of △ABC is approximately 78.98 square inches, rounded to the nearest hundredth.
User Amos Joshua
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8.4k points
4 votes

Answer:

Explanation:

In △ABC, AC=10 in., BC=2.5 in., and m∠C=63°. What is the area of △ABC? Enter your-example-1
User Mozzbozz
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8.5k points