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Mathematics Question on Some Applications of Trigonometry

Consider a triangle PQRP Q R having sides of lengths p,qp, q and rr opposite to the angles P,QP, Q and RR, respectively. Then which of the following statements is(are) TRUE?

A

cos⁡P≥1−p22qr\cos P \geq 1-\frac{ p ^{2}}{2 qr }

B

cos⁡R≥(q−rp+q)cos⁡P+(p−rp+q)cos⁡Q\cos R \geq\left(\frac{q-r}{p+q}\right) \cos P+\left(\frac{p-r}{p+q}\right) \cos Q

C

q+rp<2sin⁡Qsin⁡Rsin⁡P\frac{q+r}{p} < 2 \frac{\sqrt{\sin Q \sin R}}{\sin P}

D

If p<qp < q and p<rp < r, then cos⁡Q>pr\cos Q >\frac{ p }{ r } and cos⁡R>pq\cos R >\frac{ p }{ q }

Answer

cos⁡P≥1−p22qr\cos P \geq 1-\frac{ p ^{2}}{2 qr }

Explanation

Solution

(A) cos⁡P=q2+r2−p22qr\cos P = \frac{{q^2 + r^2 - p^2}}{{2qr}}

And q2+r22≥q2⋅r2(AM≥GM)\frac{{q^2 + r^2}}{2} \geq \sqrt{q^2 \cdot r^2} \quad \text{(AM} \geq \text{GM)}

⇒⇒ q2+r2≥2qrq^2 + r^2 \geq 2qr

So cos⁡P≥2qr−p22qr\cos P \geq \frac{{2qr - p^2}}{{2qr}}

cos⁡P≥1−p22qr\cos P \geq 1 - \frac{{p^2}}{{2qr}}

(B) (q−r)cos⁡P+(p−r)cos⁡Qp+q=(qcos⁡P+pcos⁡Q)−r(cos⁡P+cos⁡Q)p+q\frac{{(q-r) \cos P + (p-r) \cos Q}}{{p+q}} = \frac{{(q \cos P + p \cos Q) - r(\cos P + \cos Q)}}{{p+q}}

=r(1−cos⁡P−cos⁡Q)p+q\frac{r(1 - \cos P - \cos Q)}{p+q}

=r(q−pcos⁡R)−(p−qcos⁡R)p+q\frac{{r(q-p \cos R) - (p-q \cos R)}}{{p+q}}

= (r−p−q)+(p+q)cos⁡Rp+q\frac{{(r-p-q) + (p+q)\cos R}}{{p+q}}

= cos⁡R+r−q−pp+q≤cos⁡R\cos R + \frac{{r-q-p}}{{p+q}} \leq \cos R

(C) q+rp=sin⁡Q+sin⁡Rsin⁡P≥2sin⁡Qsin⁡Rsin⁡P\frac{{q+r}}{p} = \frac{{\sin Q + \sin R}}{{\sin P}} \geq 2\sqrt{\frac{{\sin Q \sin R}}{{\sin P}}}

(D) If p<qandq<rp < q \quad \text{and} \quad q < r

So, p is the smallest side, therefore one of Q or R can be obtuse

So, one of cos Q or cos R can be negative

Therefore, cos⁡Q>prandcos⁡R>pq cannot hold always.\cos Q > \frac{p}{r} \quad \text{and} \quad \cos R > \frac{p}{q} \text{ cannot hold always.}