Question

hii can you help me out with this trigo qn !

Answer 1

The Solution:

Given:

Part(a)

(i) The value of the ratio of

$\frac{\sin\angle PRQ}{\sin\angle RPQ}$

Applying the Law of sine:

$\frac{\sin P}{p}=\frac{\sin(Q)}{q}=\frac{\sin(R)}{r}$

In this case,

$\begin{gathered} \sin\angle PQR=\sin\angle Q \\ \sin\operatorname{\angle}PRQ=\sin\operatorname{\angle}R \\ \sin\operatorname{\angle}RPQ=\sin\operatorname{\angle}P \end{gathered}$

So, the ratio

$\frac{\sin\operatorname{\angle}PRQ}{\sin\operatorname{\angle}RPQ}=\frac{\sin\operatorname{\angle}R}{\sin\operatorname{\angle}P}$

Applying the Law of sine to get the above ratio, we get

$\begin{gathered} \frac{\sin\angle R}{r}=\frac{\sin P}{p} \\ \\ Where \\ r=2x \\ p=3x \end{gathered}$

Substituting, we get

$\frac{\sin\operatorname{\angle}R}{2x}=\frac{\sin\angle P}{3x}$

Cross multiplying, we get

$\begin{gathered} \frac{\sin\angle R}{\sin\angle P}=\frac{2x}{3x}=\frac{2}{3} \\ Thus, \\ \frac{\sin\operatorname{\angle}PRQ}{\sin\operatorname{\angle}RPQ}=\frac{\sin\operatorname{\angle}R}{\sin\operatorname{\angle}P}=\frac{2}{3} \end{gathered}$

(ii) To find:

$\cos\angle PQR=\cos\angle Q$

Applying the Law of Cosine:

$\cos\angle PQR=\cos\angle Q=\frac{p^2+r^2-q^2}{2pr}$

Where,

$p=3x,r=2x,q=4x$

Substituting, we get

$\begin{gathered} \cos\angle Q=\frac{(3x)^2+(2x)^2-(4x)^2}{2(3x)(2x)}=\frac{9x^2+4x^2-16x^2}{12x^2}=\frac{13x^2-16x^2}{12x^2} \\ \\ \cos\angle Q=\frac{-3x^2}{12x^2}=\frac{-1}{4}=-0.25 \\ \\ \angle Q=\cos^{-1}(-0.25)=104.4775\approx104.48^o \end{gathered}$

Thus,

$\cos\angle PQR=104.48^o$

Part (b)

To find the value of x if the area of the triangle PQR is 12 square centimeters.

By area of triangle formula:

$\begin{gathered} Area=\frac{1}{2}pr\sin Q \\ Where \\ Area=12cm^2 \\ p=3x \\ r=2x \\ \angle Q=104.48^o \end{gathered}$

Substituting these values, we get

$\begin{gathered} 12=\frac{1}{2}\times3x\times2x\times\sin104.48 \\ cross\text{ multiplying, we get} \\ 24=6x^2(0.96823) \\ 24=5.8094x^2 \end{gathered}$

Dividing both sides by 5.8094, we get

$\begin{gathered} x^2=\frac{24}{5.8094}=4.131235 \\ \text{ Taking the square root of both sides, we get} \\ x=\sqrt{4.131235}=\pm2.0325 \\ x=2.0325\text{ \lparen since x cannot be negative\rparen} \end{gathered}$

Thus, the value of x is 2.0325