M.C.Q (1 Marks)

MCQ 11 Mark

In a triangle ABC, a = 4, b = 3, $\angle\text{A}=60^{\circ}$ then c is a root of the equation:

✓
$\text{c}^2-3\text{c}-7=0$
B
$\text{c}^2+3\text{c}+7=0$
C
$\text{c}^2-3\text{c}+7=0$
D
$\text{c}^2+3\text{c}-7=0$

Answer

Correct option: A.

$\text{c}^2-3\text{c}-7=0$

It is given that a = 4, b = 3 and $\angle\text{A}=60^{\circ}$
Using cosine rule, we have
$\cos\text{A}=\frac{\text{b}^2+\text{c}^2-\text{a}^2}{2\text{bc}}$
$\Rightarrow\cos60^{\circ}=\frac{9+\text{c}^2-16}{2\times3\times\text{c}}$
$\Rightarrow\frac{1}{2}=\frac{\text{c}^2-7}{6\text{c}}$
$\Rightarrow\text{c}^2-7=3\text{c}$
$\Rightarrow\text{c}^2-3\text{c}-7=0$
Thus, c is the root of $\text{c}^2-3\text{c}-7=0.$
Hence, the correct answer is option (a).

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MCQ 21 Mark

In any $\triangle\text{ABC},2(\text{bc}\cos\text{A + ca}\cos\text{B + ab}\cos\text{C})=$

A
$\text{abc}$
B
$\text{a + b + c}$
✓
$\text{a}^2+\text{b}^2+\text{c}^2$
D
$\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}+\frac{1}{\text{c}^2}$

Answer

Correct option: C.

$\text{a}^2+\text{b}^2+\text{c}^2$

Using cosine rule, we have
$2(\text{bc}\cos\text{A}+\text{ca}\cos\text{B}+\text{ab}\cos\text{C})$
$=2\text{bc}\Big(\frac{\text{b}^2+\text{c}^2-\text{a}^2}{2\text{bc}}\Big)+2\text{ca}\Big(\frac{\text{c}^2+\text{a}^2-\text{b}^2}{2\text{ca}}\Big)+2\text{ab}\Big(\frac{\text{a}^2+\text{b}^2-\text{c}^2}{2\text{ab}}\Big)$
$=\text{b}^2+\text{c}^2-\text{a}^2+\text{c}^2+\text{a}^2-\text{b}^2+\text{a}^2+\text{b}^2-\text{c}^2$
$=\text{a}^2+\text{b}^2+\text{c}^2$
Hence, the correct answer is option (c).

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MCQ 31 Mark

In any $\triangle\text{ABC},\sum\text{a}^2(\sin\text{B}-\sin\text{C})=$

A
$\text{a}^2+\text{b}^2+\text{c}^2$
B
$\text{a}^2$
C
$\text{b}^2$
✓
$0$

Answer

Correct option: D.

$0$

Using sine rule, we have
$\sum\text{a}^2(\sin\text{B}-\sin\text{C})$
$\text{a}^2\Big(\frac{\text{b}}{\text{k}}-\frac{\text{c}}{\text{k}}\Big)+\text{b}^2\Big(\frac{\text{c}}{\text{k}}-\frac{\text{a}}{\text{k}}\Big)+\text{c}^2\Big(\frac{\text{a}}{\text{k}}-\frac{\text{b}}{\text{k}}\Big)$
$=\frac{1}{\text{k}}(\text{a}^2\text{b}-\text{a}^2\text{c}+\text{b}^2\text{c}-\text{b}^2\text{a}+\text{c}^2\text{a}-\text{c}^2\text{b})$
This expression cannot be simplified to match with any of the given options.
However, if the quesion is "In any $\triangle\text{ABC},\sum\text{a}^2(\sin\text{B}-\sin\text{C})=",$ then the solution is as follows.
Using sine rule, we have
$\sum\text{a}^2(\sin^2\text{B}-\sin^2\text{C})$
$\text{a}^2\Big(\frac{\text{b}^2}{\text{k}^2}-\frac{\text{c}^2}{\text{k}^2}\Big)+\text{b}^2\Big(\frac{\text{c}^2}{\text{k}^2}-\frac{\text{a}^2}{\text{k}^2}\Big)+\text{c}^2\Big(\frac{\text{a}^2}{\text{k}^2}-\frac{\text{b}^2}{\text{k}^2}\Big)$
$\frac{1}{\text{k}^2}(\text{a}^2\text{b}^2-\text{a}^2\text{c}^2+\text{b}^2\text{c}^2-\text{b}^2\text{a}^2+\text{c}^2\text{a}^2-\text{c}^2\text{b}^2)$
$=\frac{1}{\text{k}^2}\times0$
$=0$
Hence, the correct answer is option (d).

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MCQ 41 Mark

In a $\triangle\text{ABC},$ if (c + a + b)(a + b − c) = ab, then the measure of angle C is:

A
$\frac{\pi}{3}$
B
$\frac{\pi}{6}$
✓
$\frac{2\pi}{3}$
D
$\frac{\pi}{2}$

Answer

Correct option: C.

$\frac{2\pi}{3}$

Given: $\text{(c + a + b) (a + b − c) = ab}$
$\Rightarrow(\text{a + b})^2-\text{c}^2=\text{ab}$
$\Rightarrow\text{a}^2+\text{b}^2+2\text{ab}-\text{c}^2=\text{ab}$
$\Rightarrow\text{a}^2+\text{b}^2-\text{c}^2=-\text{ab}$
$\Rightarrow\frac{\text{a}^2+\text{b}^2-\text{c}^2}{2\text{ab}}=-\frac{1}{2}$
$\Rightarrow\cos\text{C}=-\frac{1}{2}=\cos\frac{2\pi}{3}$ (Using cosine rule)
$\Rightarrow\text{C}=\frac{2\pi}{3}$
Thus, the measure of angle C is $\frac{2\pi}{3}.$
Hence, the correct answer is option (c).

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MCQ 51 Mark

In the sides of a triangle are in the ratio $1:\sqrt{3}:2,$ then the measure of its greatest angle is:

A
$\frac{\pi}{6}$
B
$\frac{\pi}{3}$
✓
$\frac{\pi}{2}$
D
$\frac{2\pi}{3}$

Answer

Correct option: C.

$\frac{\pi}{2}$

Let $\triangle\text{ABC}$ be the given triangle such that its sides are in the ratio $1:\sqrt{3}:2.$
$\therefore\text{a = k,b}=\sqrt{3}\text{k,c}=2\text{k}$
Now, $\text{a}^2+\text{b}^2=\text{k}^2+3\text{k}^2=4\text{k}^2=\text{c}^2$
So, $\triangle\text{ABC}$ is a right triangle right angled at C.
$\therefore\text{C}=90^{\circ}$
Using sine rule, we have
$\frac{\text{a}}{\sin\text{A}}=\frac{\text{b}}{\sin\text{B}}=\frac{\text{c}}{\sin\text{C}}$
$\Rightarrow\frac{\text{k}}{\sin\text{A}}=\frac{\sqrt{3}\text{k}}{\sin\text{B}}=\frac{2\text{k}}{\sin90^{\circ}}$
$\Rightarrow\sin\text{A}=\frac{1}{2}$ and $\sin\text{B}=\frac{\sqrt{3}}{2}$
$\Rightarrow\text{A}=30^{\circ}$ and $\text{B}=60^{\circ}$
Thus, the measure of its greatest angle is $\frac{\pi}{2}$
Hence, the correct answer is option (c).

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MCQ 61 Mark

In any $\triangle\text{ABC},\text{a}(\text{b}\cos\text{C}-\text{c}\cos\text{B})=$

A
$a^2$
✓
$b^2- c^2$
C
$0$
D
$b^2+ c^2$

Answer

Correct option: B.

$b^2- c^2$

Using cosine rule, we have
$\text{a(b}\cos\text{C}-\text{c}\cos\text{B})$
$=\text{ab}\Big(\frac{\text{a}^2+\text{b}^2-\text{c}^2}{2\text{ab}}\Big)-\text{ca}\Big(\frac{\text{c}^2+\text{a}^2-\text{b}^2}{2\text{ca}}\Big)$
$=\frac{\text{a}^2+\text{b}^2-\text{c}^2-\text{c}^2-\text{a}^2+\text{b}^2}{2}$
$=\frac{2\text{b}^2-2\text{c}^2}{2}$
$=\text{b}^2-\text{c}^2$
Hence, the correct answer is option $(b).$

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MCQ 71 Mark

In a $\triangle\text{ABC},$ if a = 2, $\angle\text{B}=60^{\circ}$ and $\angle\text{C}=75^{\circ},$ then b =

A
$\sqrt{3}$
✓
$\sqrt{6}$
C
$\sqrt{9}$
D
$1+\sqrt{2}$

Answer

Correct option: B.

$\sqrt{6}$

It is given that $\text{a}=2,\angle\text{B}=60^{\circ}$ and $\angle\text{C}=75^{\circ}.$
In $\triangle\text{ABC},$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^{\circ}$ (Angle sum property)
$\Rightarrow\angle\text{A}+60^{\circ}+75^{\circ}=180^{\circ}$
$\Rightarrow\angle\text{A}=180^{\circ}-135^{\circ}=45^{\circ}$
Using sine rule, we get
$\frac{2}{\sin45^{\circ}}=\frac{\text{b}}{\sin60^{\circ}}$ $\Big(\frac{\text{a}}{\sin\text{A}}=\frac{\text{b}}{\sin\text{B}}=\frac{\text{c}}{\sin\text{C}}\Big)$
$\Rightarrow\text{b}=\frac{2\times\frac{\sqrt{3}}{2}}{\frac{1}{\sqrt{2}}}=\sqrt{6}$
Hence, the correct answer is option (b).

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MCQ 81 Mark

In any $\triangle\text{ABC},$ the value of $2\text{ac}\sin\Big(\frac{\text{A}-\text{B + C}}{2}\Big)$ is:

A
$\text{a}^2+\text{b}^2-\text{c}^2$
✓
$\text{c}^2+\text{a}^2-\text{b}^2$
C
$\text{b}^2-\text{c}^2-\text{a}^2$
D
$\text{c}^2-\text{a}^2-\text{b}^2$

Answer

Correct option: B.

$\text{c}^2+\text{a}^2-\text{b}^2$

In $\triangle\text{ABC},$
$\text{A + B + C}=\pi$ (Angle sum property)
$\Rightarrow\text{A + C}=\pi-\text{B}$
$\therefore2\text{ac}\sin\Big(\frac{\text{A}-\text{B + C}}{2}\Big)$
$=2\text{ac}\sin\Big(\frac{\pi-2\text{B}}{2}\Big)$
$=2\text{ac}\sin\Big(\frac{\pi}{2}-\text{B}\Big)$
$=2\text{ac}\cos\text{B}$
$=2\text{ac}\Big(\frac{\text{c}^2+\text{a}^2-\text{b}^2}{2\text{ca}}\Big)$ (Using cosine rule)
$=\text{c}^2+\text{a}^2-\text{b}^2$
Hence, the correct answer is option (b).

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MATHS M.C.Q (1 Marks)

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