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Zorluk: OrtaSine and Cosine Rules

In ΔKLM\Delta KLM, side k=5 cmk = 5\text{ cm}, side l=53 cml = 5\sqrt{3}\text{ cm}, and K=30\angle K = 30^\circ. If L\angle L is an obtuse angle, what is the measure of L\angle L?

  1. A
    6060^\circ
  2. 120120^\circCevap
  3. C
    135135^\circ
  4. D
    150150^\circ

Cevap

The measure of angle LL is 120120^\circ.
Applying the Sine Rule ksinK=lsinL\frac{k}{\sin K} = \frac{l}{\sin L} gives sinL=53sin305=32\sin L = \frac{5\sqrt{3} \cdot \sin 30^\circ}{5} = \frac{\sqrt{3}}{2}. The inverse sine gives an acute angle of 6060^\circ. Because the problem specifies that angle LL is obtuse, we find its supplementary angle in the second quadrant: 18060=120180^\circ - 60^\circ = 120^\circ.

Adım Adım Çözüm

1
Apply the Sine Rule relating sides k,lk, l and angles K,LK, L.
ksinK=lsinL\frac{k}{\sin K} = \frac{l}{\sin L}
The Sine Rule connects the ratio of side lengths to the sines of their opposite angles.
2
Substitute the given values into the Sine Rule equation.
5sin30=53sinL    50.5=53sinL    10=53sinL\frac{5}{\sin 30^\circ} = \frac{5\sqrt{3}}{\sin L} \implies \frac{5}{0.5} = \frac{5\sqrt{3}}{\sin L} \implies 10 = \frac{5\sqrt{3}}{\sin L}
Known values are k=5 cmk = 5\text{ cm}, l=53 cml = 5\sqrt{3}\text{ cm}, and sin30=12\sin 30^\circ = \frac{1}{2}.
3
Solve for sinL\sin L.
sinL=5310=32\sin L = \frac{5\sqrt{3}}{10} = \frac{\sqrt{3}}{2}
Isolating sinL\sin L yields the principal ratio.
4
Determine the obtuse angle solution for LL.
L=18060=120L = 180^\circ - 60^\circ = 120^\circ
Since sinL=32\sin L = \frac{\sqrt{3}}{2}, the acute solution is 6060^\circ, so the supplementary obtuse solution is 18060=120180^\circ - 60^\circ = 120^\circ.

Anahtar Kavram

Ambiguous Case of the Sine Rule
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