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3 years ago

In △ABC , m∠A=53°,m∠B=17°, and a=27. Find the perimeter of the triangle.

1 answer:

6 0

Using the Law of Sines (sinA/a=sinB/b=sinC/c) and the fact that all triangles have a sum of 180° for their angles.

The third angle is C is 180-53-17=110°

27/sin53=b/sin17=c/sin110

b=27sin17/sin53, c=27sin110/sin53

And the perimeter is a+b+c so

p=27+27sin17/sin53+27sin110/sin53 units

p≈68.65 units (to nearest hundredth of a unit)

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Consider the property, opposite angle of a cyclic quadrilateral are supplementary

hence

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Recall that

$\begin{gathered} \angle NMP=\angle M=(8x-24)^0 \\ \text{and } \\ \angle NOP=\angle O=(4x)^0 \end{gathered}$

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$\begin{gathered} \angle M+\angle O=180^0(opposite\text{ angles of a cyclic quadrilateral)} \\ (8x-24)^0+4x^0=180^0 \end{gathered}$

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$\begin{gathered} 8x-24+4x=180 \\ 8x+4x-24=180 \\ 12x-24=180 \\ \text{Add 24 t o both sides } \\ 12x-24+24=180+24 \\ 12x=204 \\ \text{divide both sides by 12} \\ \frac{12x}{12}=\frac{204}{12} \\ \text{Then} \\ x=17 \end{gathered}$

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$\begin{gathered} \angle NOP=(4x)^0 \\ \text{substitute the value of x, we have } \\ \angle NOP=4\times17=68^0 \\ \text{Hence } \\ \angle NOP=68^0 \end{gathered}$

Therefore

The measure of angle NOP is 68⁰

Answer; 68⁰ (The fourth Option )

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