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Jobisdone [24]
3 years ago
14

Find b, given that a = 20, angle A = 30°, and angle B = 45° in triangle ABC

Mathematics
2 answers:
Kamila [148]3 years ago
6 0
<span>The side b is opposite to the angle B, applying the law of the sines, we have:

</span>\frac{a}{sinA} = \frac{b}{sinB}
\frac{20}{sin30^0} = \frac{b}{sin45^0}
\frac{20}{ \frac{1}{2} } = \frac{b}{ \frac{ \sqrt{2} }{2} }
20* \frac{ \sqrt{2} }{2}  = b* \frac{1}{2}
\frac{20 \sqrt{2} }{2} = \frac{b}{2}
2*b =2*20 \sqrt{2}
2b = 40 \sqrt{2}
b =  \frac{40 \sqrt{2} }{2}
\boxed{b = 20 \sqrt{2} }
Inga [223]3 years ago
4 0

Answer:

b = 20√2

Step-by-step explanation:

Given: ΔABC

           a = 20 , m∠A =  30° and m∠B = 45°

To find: value of b.

We use Sine result, which state that

\frac{a}{sin\,A}=\frac{b}{sin\,B}

Substituting given values we, get

\frac{20}{sin\,30^{\circ}}=\frac{b}{sin\,45^{\circ}}

we know that sin\,30^{\circ}=\frac{1}{2}\:and\:sin\,45^{\circ}=\frac{1}{\sqrt{2}}, we get

\frac{20}{\frac{1}{2}}=\frac{b}{\frac{1}{\sqrt{2}}}

20{\times2}=b\times\sqrt{2}

b\times\sqrt{2}=40

b=\frac{40}{\sqrt{2}}

b=20\sqrt{2}

Therefore, b = 20√2

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Given: Two Isosceles triangle ABC and Δ PBC having same base BC. AD is the median of Δ ABC and PD is the median of Δ PBC.

To prove: Point A,D,P are collinear.

Proof:

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In Δ ABD and Δ ACD

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Similarly, Δ PBD ≅ Δ PCD [By SSS]

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But,  ∠1+∠2+ ∠ 3 + ∠4 =360° [At a point angle formed is 360°]

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But ,∠2 and ∠ 4 forms a linear pair i.e Point D is common point of intersection of median AD and PD of ΔABC and ΔPBC respectively.

So, point A, D, P lies on a line.

CASE 2.

When ΔABC and ΔPBC lie on same side of Base BC.

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∠PDB=∠PDC [CPCT]

Similarly, By proving ΔADB≅ΔADC we will get,  ∠ADB=∠ADC[CPCT]

As PD and AD are medians to same base BC of ΔPBC and ΔABC.

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