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4 years ago

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BC is tangent to circle A at B and to circle D at C. AB=9, BC=26, and DC=8. Find AD to the nearest tenth.

1 answer:

Ksju [112]4 years ago

3 0

There are two circles with center A and D. The tangent line touches both point B and C. The given measurements are enough to solve for the missing value and the solution is shown below:
AB=9
BC=26
DC=8
Solve for the measurement of AC which the hypotenuse of legs AB and BC by Pythagorean theorem:
c²=a²+b²
c²=9²+26²
c=AC=27.51
Solve for angle of A
sin A=26/27.51
A=70.93°
Finally, we solve for the length of AD using SOH
sin 70.93°=AD/27.51
AD=26
The answer is 26.

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Answer:

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Step-by-step explanation:

Given triangle ABC with right angle at C.

And AB = AD + 6 .

Now, consider the triangle ABC.

⇒ cos(∠BAC) = $\frac{AC}{AB}$ (cosФ = adj/hyp)

cos(20) = $\frac{AC}{AB}$ .

0.9397 = $\frac{AD+CD}{AD + 6}$

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⇒ 0.9397 AD + 5.6382 = AD + CD

⇒ CD = 0.0603 AD + 5.6382. →→→→→ (1)

⇒ sin(∠BAC) = $\frac{BC}{AB}$ (sinФ = opp/hyp)

sin(20) = $\frac{BC}{AB}$ .

⇒ BC = AB sin(20) . →→→→→(2)

Now, consider the triangle BCD,

sin(∠BDC) = $\frac{BC}{CD}$

⇒ sin(80) = $\frac{BC}{CD}$

CD = $\frac{BC}{sin(80)}$

From (2), CD = $\frac{AB sin(20)}{sin(80)}$ .

⇒ CD = AB (0.3473)

⇒ CD = (AD + 6) (0.3473)

⇒ CD = 0.3473 AD + 2.0838 →→→→→→(3)

Now, (1) →→ CD = 0.0603 AD + 5.6382

(3) →→ CD = 0.3473 AD + 2.0838

⇒ 0.0603 AD + 5.6382 = 0.3473 AD + 2.0838

0.287 AD = 3.5544.

⇒ AD = 12.3847

⇒ From (1), CD = 0.0603(12.3847) + 5.6382

⇒ CD = 6.385 units

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Step-by-step explanation:

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