a) Solve the triangle ABC if a = 1 , b = 3 and C = 30^ . [2] b) Prove that the equation 9x^2 16y^2 + 18x + 32y 151 = 0 represents equation of hyperbola. Also find eccentricity and foci of the given…

a) Solve the triangle ABC if $a=1$, $b=\sqrt{3}$ and $C=30^{\circ }$.

Step 1: Identify given data

• Side $a=1$ (opposite to angle $A$)
• Side $b=\sqrt{3}$ (opposite to angle $B$)
• Angle $C=30^{\circ }$

Step 2: Use the Law of Sines to find angle $B$ $$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$$ First, find $\sin A$ using the Law of Sines: $$\frac{a}{\sin A}=\frac{c}{\sin C}⟹\frac{1}{\sin A}=\frac{c}{\sin 30^{\circ }}=\frac{c}{0.5}=2c$$ But we need another approach. Instead, use the Law of Sines directly to find angle $B$: $$\frac{a}{\sin A}=\frac{b}{\sin B}⟹\frac{1}{\sin A}=\frac{\sqrt{3}}{\sin B}$$ We can use the Law of Cosines to find side $c$ first: $$c^2=a^2+b^2-2ab\cos C=1^2+(\sqrt{3}{)}^2-2(1)(\sqrt{3})\cos 30^{\circ }$$ $$c^2=1+3-2\sqrt{3}\cdot \frac{\sqrt{3}}{2}=4-3=1⟹c=1$$

Now, use the Law of Sines to find angle $A$: $$\frac{a}{\sin A}=\frac{c}{\sin C}⟹\frac{1}{\sin A}=\frac{1}{\sin 30^{\circ }}=\frac{1}{0.5}=2$$ $$\sin A=\frac{1}{2}⟹A=30^{\circ }(\text{since }A\text{ must be acute})$$

Now, find angle $B$: $$A+B+C=180^{\circ }⟹30^{\circ }+B+30^{\circ }=180^{\circ }⟹B=120^{\circ }$$

Final Answer:

• $A=\boxed{30^{\circ }}$
• $B=\boxed{120^{\circ }}$
• $C=30^{\circ }$ (given)
• $a=1$ (given)
• $b=\sqrt{3}$ (given)
• $c=\boxed{1}$

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b) Prove that the equation $9x^2-16y^2+18x+32y-151=0$ represents the equation of a hyperbola. Also find eccentricity and foci of the given equation of hyperbola.

Step 1: Rewrite the equation in standard form $$9x^2-16y^2+18x+32y-151=0$$ Group $x$ and $y$ terms: $$9x^2+18x-16y^2+32y=151$$

Step 2: Complete the square for $x$ and $y$ For $x$: $$9(x^2+2x)=9\left[(x^2+2x+1)-1\right]=9(x+1{)}^2-9$$

For $y$: $$-16(y^2-2y)=-16\left[(y^2-2y+1)-1\right]=-16(y-1{)}^2+16$$

Substitute back: $$9(x+1{)}^2-9-16(y-1{)}^2+16=151$$ $$9(x+1{)}^2-16(y-1{)}^2+7=151$$ $$9(x+1{)}^2-16(y-1{)}^2=144$$

Divide by 144 to get the standard form: $$\frac{9(x+1{)}^2}{144}-\frac{16(y-1{)}^2}{144}=1$$ $$\frac{(x+1{)}^2}{16}-\frac{(y-1{)}^2}{9}=1$$

Step 3: Identify the conic section The equation is of the form: $$\frac{(x-h{)}^2}{a^2}-\frac{(y-k{)}^2}{b^2}=1$$ This represents a hyperbola centered at $(-1,1)$.

Step 4: Find eccentricity $e$ and foci For hyperbolas, $e=\sqrt{1+\frac{b^2}{a^2}}$. Here, $a^2=16⟹a=4$, $b^2=9⟹b=3$.

$$e=\sqrt{1+\frac{9}{16}}=\sqrt{\frac{25}{16}}=\frac{5}{4}$$

The distance of each focus from the center is $c=ae$: $$c=4\cdot \frac{5}{4}=5$$

Thus, the foci are located at: $$(-1\pm 5,1)⟹(-6,1)\text{ and }(4,1)$$

Final Answer:

• The equation represents a hyperbola.
• Eccentricity: $e=\boxed{\frac{5}{4}}$
• Foci: $\boxed{(-6,1)}$ and $\boxed{(4,1)}$