In $\triangle PQR$, if $3\sin P+4\cos Q=6$ and $4\sin Q+3\cos P=1$, then the angle $R$ is equal to

Question

In $\triangle PQR$, if $3\sin P+4\cos Q=6$ and $4\sin Q+3\cos P=1$, then the angle $R$ is equal to

My attempt is as follows:-

Squaring both equations and adding

$$9+16+24\sin(P+Q)=37$$ $$\sin(P+Q)=\dfrac{1}{2}$$

either $P+Q=\dfrac{\pi}{6}$ or $P+Q=\dfrac{5\pi}{6}$

If $P+Q=\dfrac{\pi}{6}$, then $R=\dfrac{5\pi}{6}$ otherwise $R=\dfrac{\pi}{6}$

Let's see case $1$: $P+Q=\dfrac{\pi}{6}$

$$3\sin P+4\cos\left(\dfrac{\pi}{6}-P\right)=6$$ $$3\sin P+4\left(\dfrac{\sqrt{3}}{2}\cos P+\dfrac{1}{2}\cdot\sin P\right)=6$$ $$3\sin P+2\sqrt{3}\cos P+2\sin P=6$$ $$5\sin P+2\sqrt{3}\cos P=6\tag{1}$$

$$4\left(\dfrac{1}{2}\cdot\cos P-\sin P\cdot\dfrac{\sqrt{3}}{2}\right)+3\cos P=1$$ $$-2\sqrt{3}\sin P+5\cos P=1\tag{2}$$

$$\cos P=\dfrac{12\sqrt{3}+5}{37}$$ $$\sin P=\dfrac{30-2\sqrt{3}}{37}$$

Using calculator I found $\cos P=0.69$, this means $P>\dfrac{\pi}{6}$ because $\cos \dfrac{\pi}{6}=0.866$, this mean $Q$ will be negative because $Q=\dfrac{\pi}{6}-P$. So this cannot be the case hence $P+Q$ would be $\dfrac{5\pi}{6}$ and $R$ will be $\dfrac{\pi}{6}$

This is the correct answer also , but I want to know does there exist any better way to decide on the value of $P+Q$. I am asking this because I had to use the calculator for finding the value of $\cos P$.

Answer 1

A solution using complex numbers geometry :

The 2 relationships can be grouped into a single one by adding the second one to $i$ times the first one, giving :

$$3e^{iP}+4ie^{-iQ}=1+6i \ \ \iff \ \ \underbrace{3e^{iP}}_A+\underbrace{4e^{i(\pi/2-Q)}}_B=\underbrace{1+6i}_C \tag{1}$$

This defining relationship between points ("affixes") of these complex numbers can be written under a vectorial form :

$$\vec{OA}+\vec{OB}=\vec{OC}$$

meaning that $OBCA$ is a parallelogram with prescribed lengths $OA, OB, OC$ which will not leave much degrees of freedom as we are going to see it.

Remark: the polar angles of $\vec{OA}$ and $\vec{OB}$ are $P$ and $\pi/2-Q$ resp. (the latter being in $(-\pi/2,\pi/2)$).

Therefore

$$\alpha := angle(OB,OA)=P-(\pi/2-Q)\tag{2}$$

In parallelogram $OBCA$, we have the following classical relationship between the sides and the diagonals (see here).

$$p^2+q^2=2(a^2+b^2)\tag{3}$$

With $a=OA=3, b=OB=4, p=OC=\sqrt{1^2+6^2}=\sqrt{37}$, we deduce from (3) that the second diagonal has its length $q$ given by :

$$37+q^2=2(3^2+4^2) \ \ \implies \ \ AB=q=\sqrt{13}$$

Let us apply the cosine formula to triangle $OAB$ :

$$AB^2=OA^2+OB^2-2OA.OB \cos \alpha \ \ \iff \ \ 13=3^2+4^2-2.3.4 \cos \alpha$$

giving

$$\cos(OB,OA)=\cos \alpha = \dfrac12 \ \ \implies \ \ \alpha := angle(OB,OA)=\dfrac{\pi}{3}\tag{4}$$

Identifying (4) and (2), we get :

$$P+Q=\dfrac{\pi}{2}+\dfrac{\pi}{3}=\dfrac{5\pi}{6} \ \ \implies \ \ R=\pi-(P+Q)=\dfrac{\pi}{6}$$

as awaited.

But there is a case we haven't yet considered :

It has been assumed implicitly that the polar angle of $\vec{OB}$ is less than the polar angle of $\vec{OA}$. We could have had the inverse situation, which geometrically corresponds to a symmetry of parallelogram $OBCA$ with respect to its diagonal $OC$. Fortunately, this cannot arise, because the polar angle for $\vec{OB}$ would have been outside $(-\pi/2,\pi/2)$, contradicting the remark done upwards.

Answer 2

If $R=\frac{5\pi}6$, we have

$$3\sin P + 4\cos(\frac\pi6-P)=6$$

Note that the RHS is an increasing function of $P$ for $P\in(0,\frac\pi6]$, whose maximal value is at $P=\frac\pi6$, i.e.

$$RHS_{max}=3\cdot \frac12+4\cdot 1 = 5.5 <6$$

Thus, $R\ne \frac{5\pi}6$.