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alina1380 [7]
3 years ago
14

An angle bisector of a triangle divides the opposite side of the triangle into segments 5 cm and 3 cm long. A second side of the

triangle is 7.6 cm long. Find the longest and shortest possible lengths of the third side of the triangle. Round answers to the nearest tenth of a centimeter.
a. 12.7 cm, 4.6 cm
b. 15 cm, 4.6 cm
c. 38 cm, 2 cm
d. 38 cm, 12.7 cm

Mathematics
1 answer:
attashe74 [19]3 years ago
6 0
There is a little-known theorem to solve this problem.

The theorem says that
In a triangle, the angle bisector cuts the opposite side into two segments in the ratio of the respective sides lengths.

See the attached triangles for cases 1 and 2.  Let x be the length of the third side.

Case 1:
Segment 5cm is adjacent to the 7.6cm side, then
x/7.6=3/5  => x=7.6*3/5=4.56 cm

Case 2:
Segment 3cm is adjacent to the 7.6 cm side, then
x/7.6=5/3 => x=7.6*5/3=12.67 cm

The theorem can be proved by considering the sine rule on the adjacent triangles ADC and BDC with the common side CD and equal angles ACD and DCB.

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Option A

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<em><u>Solution:</u></em>

Given that we have to rewrite with only sin x and cos x

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We know that,

\cos (a+b)=\cos a \cos b-\sin a \sin b

Therefore,

\cos (x+2 x)=\cos x \cos 2 x-\sin x \sin 2 x  ---- eqn 1

We know that,

\sin 2 x=2 \sin x \cos x

\cos 2 x=\cos ^{2} x-\sin ^{2} x

Substituting these values in eqn 1

\cos (x+2 x)=\cos x\left(\cos ^{2} x-\sin ^{2} x\right)-\sin x(2 \sin x \cos x)  -------- eqn 2

We know that,

\cos ^{2} x-\sin ^{2} x=1-2 \sin ^{2} x

Applying this in above eqn 2, we get

\cos (x+2 x)=\cos x\left(1-2 \sin ^{2} x\right)-\sin x(2 \sin x \cos x)

\begin{aligned}&\cos (x+2 x)=\cos x-2 \sin ^{2} x \cos x-2 \sin ^{2} x \cos x\\\\&\cos (x+2 x)=\cos x-4 \sin ^{2} x \cos x\end{aligned}

\cos (x+2 x)=\cos x-4 \cos x \sin ^{2} x

Therefore,

\cos 3 x=\cos x-4 \cos x \sin ^{2} x

Option A is correct

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3 years ago
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5 0
3 years ago
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Effectus [21]
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2 years ago
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lyudmila [28]

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4 0
3 years ago
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Montano1993 [528]

Answer:

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Step-by-step explanation:

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