sarahl5
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what is the measure of the base angle,x, of the isosceles triangle shown below? round your answer to the nearest tenth of a degree.​

What Is The Measure Of The Base Anglex Of The Isosceles Triangle Shown Below Round Your Answer To The Nearest Tenth Of A Degree class=

Sagot :

Answer:

x ≈ 49.7°

Step-by-step explanation:

Since, the given triangle is an isosceles triangle,

Two sides having measure 17 units are equal.

Opposite angles of these equal sides will be equal.

Measure of the third angle = 180° - (x + x)°

By sine rule,

[tex]\frac{\text{sinx}}{17}=\frac{\text{sin}(180-2x)}{22}[/tex]

[tex]\frac{\text{sinx}}{17}=\frac{\text{sin}(2x)}{22}[/tex] [Since, sin(180 - θ) = sinθ]

[tex]\frac{\text{sinx}}{17}=\frac{2(\text{sin}x)(\text{cos}x)}{22}[/tex]

[tex]\frac{\text{sinx}}{17}=\frac{(\text{sin}x)(\text{cos}x)}{11}[/tex]

[tex]\frac{1}{17}=\frac{(\text{cos}x)}{11}[/tex]

cos(x) = [tex]\frac{11}{17}[/tex]

x = 49.68°

x ≈ 49.7°

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