In this figure, mBDA=_ and mBCA=_.

Mathematics

2 answers:

Serhud [2]4 years ago

5 0

∠BDA is inscribed angle
inscribed angle = 1/2 × central angle
the central angle is small ∠BOA, use ∠BOA that face the same direction as ∠BDA

Find the central angle ∠BOA
central angle = 360° - large ∠BOA
central angle = 360° - 250°
central angle = 110°

Find ∠BDA
inscribed angle = 1/2 × central angle
∠BDA = 1/2 × small ∠BOA
∠BDA = 1/2 × 110°
∠BDA = 55°

To find ∠BCA, use tetragon BOAC to solve the problem
The sum of interior angles in tetragon is 360°. Because CA and CB is tangent of the circle, the angles which is formed by the tangent is 90°. So ∠CAO and ∠CBO are both equal to 90°
∠BCA + ∠CAO + ∠CBO + small ∠BOA = 360°
∠BCA + 90° + 90° + 110° = 360°
∠BCA + 290° = 360°
∠BCA = 360° - 290°
∠BCA = 70°

THE ANSWER
m∠BDA = 55°
m∠BCA = 70°

san4es73 [151]4 years ago

4 0

MBDA= ADB and mBCA= ACB

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The equation of a circle is (x + 6)^2 + (y - 4)^2 = 16. The point (-6, 8) is on the circle.

pantera1 [17]

Answer:

y = 8 is the equation of tangent.

Step-by-step explanation:

The equation of the tangent to the circle at (-6,8) is of the form:

y = mx + c

where m is the slope of the tangent and c is the y-intercept.

The point (-6,8) lies on the circle and the tangent line as well.

Hence (-6,8) satisfies the line equation:

8 = m(-6) + c ⇒ c-6m = 8 -------------1

We know that slope of two perpendicular lines are related as:

$m_{1}\times m_{2}=-1$

At any point on the circle, the normal line at a point is always perpendicular to the tangent line at that point.

Hence :

$m_{normal} \times m_{tangent}=-1$

We can find the slope of the normal at point (-6,8) as it passes through the centre of the circle (-6,4) by using the two-points formula for slope.

$m=\frac{y_2-y_1}{x_2-x_1}$