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In ΔABC,

∠BAC = 50° AB = 10cm BC = 9cm
Given that ∠BCA = x°
find the two possible values, to one decimal place, of x

1 Answer

1 vote

Answer:

86.3

Explanation:

To find the two possible values of x, we can use the law of sines, which states that:

asinA​=bsinB​=csinC​

where A, B, and C are the angles of the triangle and a, b, and c are the opposite sides.

In this case, we are given A = 50°, a = 10 cm, and b = 9 cm. We can use the law of sines to find sin B:

10sin50°​=9sinB​

sinB=109sin50°​≈0.6904

Now, we can use the inverse sine function to find B:

B=sin−1(0.6904)≈43.7°

However, this is not the only possible value for B. Since the sine function is positive in both quadrant I and quadrant II, there is another angle with the same sine value but in quadrant II. This angle is the supplement of B:

B′=180°−B≈180°−43.7°=136.3°

This means that there are two possible triangles that satisfy the given information: one with B = 43.7° and one with B = 136.3°.

To find x, we need to find C for each triangle using the fact that the sum of angles in a triangle is 180°:

C=180°−A−B≈180°−50°−43.7°=86.3°

C′=180°−A−B′≈180°−50°−136.3°=−6.3°

However, C’ is not a valid angle for a triangle because it is negative. Therefore, we can ignore this possibility and conclude that there is only one possible value for x:

x=C≈86.3°

User PiotrDomo
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