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Zorluk: KolaySine and Cosine Rules

In ΔPQR\Delta PQR, side p=6 cmp = 6\text{ cm}, side q=62 cmq = 6\sqrt{2}\text{ cm}, and ∠P=30∘\angle P = 30^\circ. If ∠Q\angle Q is an acute angle, what is the measure of ∠Q\angle Q?

  1. 45∘45^\circCevap
  2. B
    135∘135^\circ
  3. C
    60∘60^\circ
  4. D
    30∘30^\circ

Cevap

45∘45^\circ
Using the Sine Rule sin⁡Qq=sin⁡Pp\frac{\sin Q}{q} = \frac{\sin P}{p}, we substitute the given values to find sin⁡Q=62⋅sin⁡30∘6=22\sin Q = \frac{6\sqrt{2} \cdot \sin 30^\circ}{6} = \frac{\sqrt{2}}{2}. Since ∠Q\angle Q is specified as an acute angle, ∠Q=45∘\angle Q = 45^\circ.

Adım Adım Çözüm

1
Set up the Sine Rule formula relating sides p,qp, q and their opposite angles P,QP, Q.
psin⁡P=qsin⁡Q\frac{p}{\sin P} = \frac{q}{\sin Q}
The Sine Rule allows finding an unknown angle when two side lengths and one opposite angle are known.
2
Substitute the given values p=6p = 6, q=62q = 6\sqrt{2}, and P=30∘P = 30^\circ into the formula.
6sin⁡30∘=62sin⁡Q\frac{6}{\sin 30^\circ} = \frac{6\sqrt{2}}{\sin Q}
Populating known quantities permits solving for sin⁡Q\sin Q.
3
Solve for sin⁡Q\sin Q.
sin⁡Q=62⋅sin⁡30∘6=2⋅12=22\sin Q = \frac{6\sqrt{2} \cdot \sin 30^\circ}{6} = \sqrt{2} \cdot \frac{1}{2} = \frac{\sqrt{2}}{2}
Simplifying the algebraic fraction yields the exact sine ratio for angle QQ.
4
Determine the acute angle QQ corresponding to sin⁡Q=22\sin Q = \frac{\sqrt{2}}{2}.
∠Q=45∘\angle Q = 45^\circ
The principal acute angle with sine equal to 22\frac{\sqrt{2}}{2} is 45∘45^\circ.

Anahtar Kavram

Applying the Sine Rule to calculate an unknown acute angle in a non-right-angled triangle
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Konular

Sine and Cosine Rules

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