With these questions, i don't understand them!

• arc PS = 40°

• arc UV = 24°

Step-by-step explanation:

There are relationships between the arcs intercepted by secant lines and the angle the secant lines make with each other. In these problems, you are expected to make use of these relationships, along with others you have learned about triangles.

The relationships are basically these:

• when the secants intersect inside the circle, the angle between them is half the sum of the intercepted arcs.

• when the secants intersect outside the circle, the angle between them is half the difference of the intercepted arcs.

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First problem:

The angle between the secants QS and PR is shown to be 50°. Intercepted arc QR is shown to be 60°. You are asked to find the other intercepted arc, PS. Based on the above, we know ...

∠POS = (1/2)(arc PS + arc QR)

50° = (1/2)(arc PS + 60°)

Multiplying by 2, we get ...

100° = arc PS + 60°

Subtracting 60°, gives ...

40° = arc PS

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Second problem:

We need to name a couple of points so we can describe more clearly what is going on. Call the point on arc PS where line QT intersects it point R. Call the point where line QT crosses line SU point X. (Point X is the vertex of the 99° angle.)

The relations described above tell us ...

angle W = (1/2)(arc PS - arc UV)

In this equation, we only know the value of arc PS = arc PR + arc RS = 20° + 94° = 114°.

But, we know two of the angles in triangle QWX. They are angle Q = 36° and angle X = 99°. Then angle W must be ...

angle W = 180° -36° -99° = 45°

Now, we can finish the above equation involving arc UV:

45° = (1/2)(arc PS - arc UV) . . . . . put the values we know in the secant relation

90° = 114° -arc UV

arc UV = 114° -90° = 24° . . . . . . . .solve for arc UV

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation: