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#### Question 1:

If in âˆ†ABC, ∠A = 45°, ∠B = 60° and ∠C = 75°, find the ratio of its sides.

Let $\frac{a}{\mathrm{sin}A}=\frac{b}{\mathrm{sin}B}=\frac{c}{\mathrm{sin}C}=k\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Then,

On multiplying by , we get:

Hence, the ratio of the sides is .

#### Question 2:

If in âˆ†ABC, ∠C = 105°, ∠B = 45° and a = 2, then find b.

#### Question 3:

In âˆ†ABC, if a = 18, b = 24 and c = 30 and ∠c = 90°, find sin A, sin B and sin C.

Given,∠C = 90°, a = 18, b = 24 and c = 30

According to sine rule, .

#### Question 4:

In triangle ABC, prove the following:

$\mathrm{Assume}\frac{a}{\mathrm{sin}A}=\frac{b}{\mathrm{sin}B}=\frac{c}{\mathrm{sin}C}=k\phantom{\rule{0ex}{0ex}}$

Consider the LHS of the equation .

.

#### Question 5:

In triangle ABC, prove the following:

Let $\frac{a}{\mathrm{sin}A}=\frac{b}{\mathrm{sin}B}=\frac{c}{\mathrm{sin}C}=k$    ...(1)

Consider the LHS of the equation

Hence proved.

#### Question 6:

In triangle ABC, prove the following:

Let $\frac{a}{\mathrm{sin}A}=\frac{b}{\mathrm{sin}B}=\frac{c}{\mathrm{sin}C}=k\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$     ...(1)

We need to prove:

Consider

#### Question 7:

In triangle ABC, prove the following:

Let $\frac{a}{\mathrm{sin}A}=\frac{b}{\mathrm{sin}B}=\frac{c}{\mathrm{sin}C}=k\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$    ...(1)

We need to prove:

Consider

#### Question 8:

In triangle ABC, prove the following:

Let $\frac{a}{\mathrm{sin}A}=\frac{b}{\mathrm{sin}B}=\frac{c}{\mathrm{sin}C}=k\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$    ...(1)

Then,
Consider the LHS of the equation .

#### Question 9:

In any triangle ABC, prove the following:

Let $\frac{a}{\mathrm{sin}A}=\frac{b}{\mathrm{sin}B}=\frac{c}{\mathrm{sin}C}=k\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Then,
Consider the RHS of the equation

#### Question 10:

In triangle ABC, prove the following:

Let
Then,
Consider the LHS of the equation .

#### Question 11:

In triangle ABC, prove the following:

Let
Then,

Consider the LHS of he equation .
LHS

#### Question 12:

In triangle ABC, prove the following:

Let
Then,

Consider the RHS of the equation .

#### Question 13:

In triangle ABC, prove the following:

Consider the LHS of the equation .

Let

Then,

#### Question 14:

In triangle ABC, prove the following:

Let $\frac{a}{\mathrm{sin}A}=\frac{b}{\mathrm{sin}B}=\frac{c}{\mathrm{sin}C}=k$

Then,

Consider the LHS of the equation .

#### Question 15:

In triangle ABC, prove the following:

Let $\frac{a}{\mathrm{sin}A}=\frac{b}{\mathrm{sin}B}=\frac{c}{\mathrm{sin}C}=k$

Then,

Consider the LHS of the equation .

#### Question 16:

In triangle ABC, prove the following:

Let $\frac{a}{\mathrm{sin}A}=\frac{b}{\mathrm{sin}B}=\frac{c}{\mathrm{sin}C}=k$

Then,

Consider the LHS of the equation .

#### Question 17:

In triangle ABC, prove the following:

Let $\frac{a}{\mathrm{sin}A}=\frac{b}{\mathrm{sin}B}=\frac{c}{\mathrm{sin}C}=k$

Then,

Consider the LHS of the equation .

#### Question 18:

In triangle ABC, prove the following:

Let $\frac{a}{\mathrm{sin}A}=\frac{b}{\mathrm{sin}B}=\frac{c}{\mathrm{sin}C}=k$

Then,

Consider the LHS of the equation .

#### Question 19:

In triangle ABC, prove the following:

Let $\frac{a}{\mathrm{sin}A}=\frac{b}{\mathrm{sin}B}=\frac{c}{\mathrm{sin}C}=k$

Then,

Consider the LHS of the equation .

Also,

Similarly,
$\frac{{\mathrm{cos}}^{2}A-{\mathrm{cos}}^{2}B}{a+b}=\frac{\mathrm{sin}B-\mathrm{sin}A}{k}$

Thus,

Hence, in any triangle ABC, .

#### Question 20:

In âˆ†ABC, prove that:

.

Consider
$a\mathrm{sin}\frac{A}{2}\mathrm{sin}\left(\frac{B-C}{2}\right)+b\mathrm{sin}\frac{B}{2}\mathrm{sin}\left(\frac{C-A}{2}\right)+c\mathrm{sin}\frac{C}{2}\mathrm{sin}\left(\frac{A-B}{2}\right)$

Hence proved.

#### Question 21:

In âˆ†ABC, prove that:

Let ABC be any triangle.

Suppose $\frac{a}{\mathrm{sin}A}=\frac{b}{\mathrm{sin}B}=\frac{c}{\mathrm{sin}C}=k$

Now,

Also,

From (1), (2) and (3), we get:

Hence proved.

#### Question 22:

In triangle ABC, prove the following:

So, from (1), we have
.

Hence proved.

#### Question 23:

Suppose $\frac{a}{\mathrm{sin}A}=\frac{b}{\mathrm{sin}B}=\frac{c}{\mathrm{sin}C}=k$

Consider:

From (1), (2) and (3), we get:

Hence proved.

#### Question 24:

In âˆ†ABC, prove that

#### Question 25:

In âˆ†ABC, prove that if θ be any angle, then b cosθ = c cos (A − θ) + a cos (C + θ).

Suppose  $\frac{a}{\mathrm{sin}A}=\frac{b}{\mathrm{sin}B}=\frac{c}{\mathrm{sin}C}=k$.     ...(1)

Consider the RHS of the equation b cosθ = c cos (A − θ) + a cos (C + θ).

#### Question 26:

In âˆ†ABC, if sin2A + sin2B = sin2C. show that the triangle is right-angled.

In âˆ† ABC,
Given,

Suppose $\frac{a}{\mathrm{sin}A}=\frac{b}{\mathrm{sin}B}=\frac{c}{\mathrm{sin}C}=k$ .

On putting these values in equation (1), we get:

Thus, âˆ† ABC is right-angled.

#### Question 27:

In âˆ†ABC, if a2, b2 and c2 are in A.P., prove that cot A, cot B and cot C are also in A.P.

Then,

a2, b2 and c2 are in A.P.

#### Question 28:

The upper part of a tree broken by the wind makes an angle of 30° with the ground and the distance from the root to the point where the top of the tree touches the ground is 15 m. Using sine rule, find the height of the tree.

Suppose BD be the tree and the upper part of the tree is broken over by the wind at point A.

#### Question 29:

At the foot of a mountain, the elevation of it summit is 45°; after ascending 1000 m towards the mountain up a slope of 30° inclination, the elevation is found to be 60°. Find the height of the mountain.

Suppose, AB is a mountain of height t + x.

Hence, height of the mountain = .

#### Question 30:

A person observes the angle of elevation of the peak of a hill from a station to be α. He walks c metres along a slope inclined at an angle β and finds the angle of elevation of the peak of the hill to be ϒ. Show that the height of the peak above the ground is .

Suppose, AB is a peak whose height above the ground is t+x.

#### Question 31:

If the sides a, b and c of âˆ†ABC are in H.P., prove that are in H.P.

#### Question 1:

In , show that its area is . units.

#### Question 2:

In , show that its area is units.

#### Question 3:

The sides of a triangle are a = 4, b = 6 and c = 8. Show that .

Given:
Then,

Hence proved.

#### Question 4:

In âˆ† ABC, if a = 18, b = 24 and c = 30, find cos A, cos B and cos C.

Hence,

#### Question 5:

In âˆ†ABC, prove the following:

Let ABC be any triangle.

Hence proved.

#### Question 6:

In âˆ†ABC, prove the following:

Consider

Hence proved.

#### Question 7:

In âˆ†ABC, prove the following:

LHS =

On using the cosine law, we get:
$\mathrm{LHS}=2\left[bc\left(\frac{{b}^{2}+{c}^{2}-{a}^{2}}{2bc}\right)+ca\left(\frac{{a}^{2}+{c}^{2}-{b}^{2}}{2ac}\right)+ab\left(\frac{{a}^{2}+{b}^{2}-{c}^{2}}{2ab}\right)\right]$
$={b}^{2}+{c}^{2}-{a}^{2}+{a}^{2}+{c}^{2}-{b}^{2}+{a}^{2}+{b}^{2}-{c}^{2}\phantom{\rule{0ex}{0ex}}={a}^{2}+{b}^{2}+{c}^{2}=\mathrm{RHS}$

Hence proved.

#### Question 8:

In âˆ†ABC, prove the following:

From (1), (2) and (3), we get:

#### Question 9:

In âˆ†ABC, prove the following:

Let ABC be any triangle.

Hence proved.

#### Question 10:

In âˆ†ABC, prove that:

#### Question 11:

a cos A + b cos B + c cos C = 2b sin A sin C

Hence, a cos A + b cos B + c cos C = 2b sin A sin C.

#### Question 12:

In âˆ†ABC, prove the following:

Hence proved.

#### Question 13:

In âˆ†ABC, prove the following:

$=2\left(ab+bc+ac\right)+{b}^{2}+{c}^{2}-{a}^{2}+{c}^{2}+{a}^{2}-{b}^{2}+{a}^{2}+{b}^{2}-{c}^{2}\phantom{\rule{0ex}{0ex}}={a}^{2}+{b}^{2}+{c}^{2}+2ab+2bc+2ac\phantom{\rule{0ex}{0ex}}={\left(a+b+c\right)}^{2}=\mathrm{RHS}\phantom{\rule{0ex}{0ex}}$

Hence proved.

#### Question 14:

In âˆ†ABC, prove the following:

#### Question 15:

In . Prove that .

Hence proved.

In prove that .

Hence proved.

#### Question 17:

If in , prove that the triangle is right-angled.

Let ABC be any triangle.
In $∆\mathrm{ABC}$,

Hence, $∆$ABC is right angled.

#### Question 18:

In , prove that the triangle is isosceles.

Let $∆ABC$ be any triangle.

Suppose $\frac{\mathrm{sin}A}{a}=\frac{\mathrm{sin}B}{b}=\frac{\mathrm{sin}C}{c}=k$
If  $\mathrm{cos}C=\frac{\mathrm{sin}A}{2\mathrm{sin}B}$, then

Thus, the lengths of two sides of the $∆ABC$ are equal.

Hence, $∆\mathrm{ABC}$ is an isosceles triangle.

#### Question 19:

Two ships leave a port at the same time. One goes 24 km/hr in the direction N 38° E and other travels 32 km/hr in the direction S 52° E. Find the distance between the ships at the end of 3 hrs.

#### Question 1:

Answer each of the following questions in one word or one sentence or as per exact requirement of the question.

Find the area of the triangle âˆ†ABC in which a = 1, b = 2 and $\angle C=60°$.

In âˆ†ABC, a = 1, b = 2 and $\angle C=60°$.

∴ Area of the âˆ†ABC

#### Question 2:

Answer each of the following questions in one word or one sentence or as per exact requirement of the question.

In a âˆ†ABC, if b = $\sqrt{3}$, c = 1 and $\angle A=30°$, find a.

In âˆ†ABC, b = $\sqrt{3}$, c = 1 and $\angle A=30°$.

Using cosine formula, we have

$\mathrm{cos}A=\frac{{b}^{2}+{c}^{2}-{a}^{2}}{2bc}\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}30°=\frac{{\left(\sqrt{3}\right)}^{2}+{\left(1\right)}^{2}-{a}^{2}}{2×\sqrt{3}×1}\phantom{\rule{0ex}{0ex}}⇒\frac{\sqrt{3}}{2}=\frac{4-{a}^{2}}{2\sqrt{3}}\phantom{\rule{0ex}{0ex}}⇒3=4-{a}^{2}\phantom{\rule{0ex}{0ex}}⇒{a}^{2}=4-3=1\phantom{\rule{0ex}{0ex}}⇒a=1$

#### Question 3:

Answer each of the following questions in one word or one sentence or as per exact requirement of the question.

In a âˆ†ABC, if $\mathrm{cos}A=\frac{\mathrm{sin}B}{2\mathrm{sin}C}$, then show that c = a.

Given: $\mathrm{cos}A=\frac{\mathrm{sin}B}{2\mathrm{sin}C}$

$⇒\frac{{b}^{2}+{c}^{2}-{a}^{2}}{2bc}=\frac{b}{2c}$      (Using sine rule and cosine rule)

$⇒{b}^{2}+{c}^{2}-{a}^{2}={b}^{2}$

$⇒{c}^{2}={a}^{2}$

$⇒c=a$

#### Question 4:

Answer each of the following questions in one word or one sentence or as per exact requirement of the question.

In a âˆ†ABC, if b = 20, c = 21 and $\mathrm{sin}A=\frac{3}{5}$, find a.

In âˆ†ABC, b = 20, c = 21 and $\mathrm{sin}A=\frac{3}{5}$.

Using cosine rule, we have

$⇒{a}^{2}=841-672=169\phantom{\rule{0ex}{0ex}}⇒a=13$

#### Question 5:

Answer each of the following questions in one word or one sentence or as per exact requirement of the question.

In a âˆ†ABC, if sinA and sinB are the roots of the equation ${c}^{2}{x}^{2}-c\left(a+b\right)x+ab=0$, then find $\angle C$.

It is given that sinA and sinB are the roots of the equation ${c}^{2}{x}^{2}-c\left(a+b\right)x+ab=0$.

$⇒\mathrm{sin}A+\mathrm{sin}B=\frac{\mathrm{sin}A+\mathrm{sin}B}{\mathrm{sin}C}\phantom{\rule{0ex}{0ex}}⇒\mathrm{sin}C=1=\mathrm{sin}90°\phantom{\rule{0ex}{0ex}}⇒C=90°$

#### Question 6:

Answer each of the following questions in one word or one sentence or as per exact requirement of the question.

In âˆ†ABC, if a = 8, b = 10, c = 12 and C = λA, find the value of λ.

Using cosine rule, we have

Now, using sine rule, we have

$\frac{a}{\mathrm{sin}A}=\frac{c}{\mathrm{sin}C}\phantom{\rule{0ex}{0ex}}⇒\frac{8}{\mathrm{sin}A}=\frac{12}{\mathrm{sin}\lambda A}\phantom{\rule{0ex}{0ex}}⇒\mathrm{sin}\lambda A=\frac{3}{2}\mathrm{sin}A\phantom{\rule{0ex}{0ex}}⇒\mathrm{sin}\lambda A=2×\frac{3}{4}\mathrm{sin}A$

#### Question 7:

Answer each of the following questions in one word or one sentence or as per exact requirement of the question.

If the sides of a triangle are proportional to 2, $\sqrt{6}$ and $\sqrt{3}-1$, find the measure of its greatest angle.

Let âˆ†ABC be the triangle such that a = 2, b = $\sqrt{6}$ and c = $\sqrt{3}-1$.

Clearly, b > a > c. Then,

$\angle$B is the greatest angle of âˆ†ABC.                  (Greatest side has greatest angle opposite to it)

Using cosine formula, we have

$\mathrm{cos}B=\frac{{c}^{2}+{a}^{2}-{b}^{2}}{2ca}\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}B=\frac{{\left(\sqrt{3}-1\right)}^{2}+{2}^{2}-{\left(\sqrt{6}\right)}^{2}}{2×\left(\sqrt{3}-1\right)×2}\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}B=\frac{3+1-2\sqrt{3}+4-6}{4\left(\sqrt{3}-1\right)}$
$⇒\mathrm{cos}B=\frac{2-2\sqrt{3}}{4\left(\sqrt{3}-1\right)}=\frac{-2\left(\sqrt{3}-1\right)}{4\left(\sqrt{3}-1\right)}\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}B=-\frac{1}{2}=\mathrm{cos}120°\phantom{\rule{0ex}{0ex}}⇒B=120°$

Hence, the measure of its greatest angle is 120º.

#### Question 8:

Answer each of the following questions in one word or one sentence or as per exact requirement of the question.

If in a âˆ†ABC, $\frac{\mathrm{cos}A}{a}=\frac{\mathrm{cos}B}{b}=\frac{\mathrm{cos}C}{c}$, then find the measures of angles A, B, C.

In âˆ†ABC,

⇒ âˆ†ABC is an equilateral triangle.

A = B = C = 60º

#### Question 9:

Answer each of the following questions in one word or one sentence or as per exact requirement of the question.

In any triangle ABC, find the value of $a\mathrm{sin}\left(B-C\right)+b\mathrm{sin}\left(C-A\right)+c\mathrm{sin}\left(A-B\right)$.

Using sine rule, we have

$=k\left[\mathrm{sin}\left(B+C\right)\mathrm{sin}\left(B-C\right)+\mathrm{sin}\left(C+A\right)\mathrm{sin}\left(C-A\right)+\mathrm{sin}\left(A+B\right)\mathrm{sin}\left(A-B\right)\right]\phantom{\rule{0ex}{0ex}}=k\left({\mathrm{sin}}^{2}B-{\mathrm{sin}}^{2}C+{\mathrm{sin}}^{2}C-{\mathrm{sin}}^{2}A+{\mathrm{sin}}^{2}A-{\mathrm{sin}}^{2}B\right)\phantom{\rule{0ex}{0ex}}=k×0\phantom{\rule{0ex}{0ex}}=0$

Hence, the required value is 0.

#### Question 10:

Answer each of the following questions in one word or one sentence or as per exact requirement of the question.

In any âˆ†ABC, find the value of $\sum _{}^{}a\left(\mathrm{sin}B-\mathrm{sin}C\right)$.

Using sine rule, we have

$\frac{a}{\mathrm{sin}A}=\frac{b}{\mathrm{sin}B}=\frac{c}{\mathrm{sin}C}=k\phantom{\rule{0ex}{0ex}}⇒a=k\mathrm{sin}A,b=k\mathrm{sin}B,c=k\mathrm{sin}C$

$\therefore \sum _{}^{}a\left(\mathrm{sin}B-\mathrm{sin}C\right)\phantom{\rule{0ex}{0ex}}=\sum _{}^{}k\mathrm{sin}A\left(\mathrm{sin}B-\mathrm{sin}C\right)$

$=k\sum _{}^{}\mathrm{sin}A\left(\mathrm{sin}B-\mathrm{sin}C\right)$

$=k\left[\mathrm{sin}A\left(\mathrm{sin}B-\mathrm{sin}C\right)+\mathrm{sin}B\left(\mathrm{sin}C-\mathrm{sin}A\right)+\mathrm{sin}C\left(\mathrm{sin}A-\mathrm{sin}B\right)\right]$

$=k\left(\mathrm{sin}A\mathrm{sin}B-\mathrm{sin}A\mathrm{sin}C+\mathrm{sin}B\mathrm{sin}C-\mathrm{sin}B\mathrm{sin}A+\mathrm{sin}C\mathrm{sin}A-\mathrm{sin}C\mathrm{sin}B\right)$

$=k×0=0$

#### Question 1:

Mark the correct alternative in each of the following:

In any âˆ†ABC, $\sum _{}^{}{a}^{2}\left(\mathrm{sin}B-\mathrm{sin}C\right)=$

(a) ${a}^{2}+{b}^{2}+{c}^{2}$                            (b) ${a}^{2}$                                           (c) ${b}^{2}$                                          (d) 0

Using sine rule, we have

$\sum _{}^{}{a}^{2}\left(\mathrm{sin}B-\mathrm{sin}C\right)$

$={a}^{2}\left(\frac{b}{k}-\frac{c}{k}\right)+{b}^{2}\left(\frac{c}{k}-\frac{a}{k}\right)+{c}^{2}\left(\frac{a}{k}-\frac{b}{k}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{k}\left({a}^{2}b-{a}^{2}c+{b}^{2}c-{b}^{2}a+{c}^{2}a-{c}^{2}b\right)$

This expression cannot be simplified to match with any of the given options.

However, if the quesion is "In any âˆ†ABC, $\sum _{}^{}{a}^{2}\left({\mathrm{sin}}^{2}B-{\mathrm{sin}}^{2}C\right)=$", then the solution is as follows.

Using sine rule, we have

$\sum _{}^{}{a}^{2}\left({\mathrm{sin}}^{2}B-{\mathrm{sin}}^{2}C\right)$

$={a}^{2}\left(\frac{{b}^{2}}{{k}^{2}}-\frac{{c}^{2}}{{k}^{2}}\right)+{b}^{2}\left(\frac{{c}^{2}}{{k}^{2}}-\frac{{a}^{2}}{{k}^{2}}\right)+{c}^{2}\left(\frac{{a}^{2}}{{k}^{2}}-\frac{{b}^{2}}{{k}^{2}}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{{k}^{2}}\left({a}^{2}{b}^{2}-{a}^{2}{c}^{2}+{b}^{2}{c}^{2}-{b}^{2}{a}^{2}+{c}^{2}{a}^{2}-{c}^{2}{b}^{2}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{{k}^{2}}×0\phantom{\rule{0ex}{0ex}}=0$

Hence, the correct answer is option (d).

Disclaimer: The question given in the book in incorrect or there is some printing mistake in the question.

#### Question 2:

Mark the correct alternative in each of the following:

In a âˆ†ABC, if a = 2, $\angle B=60°$ and $\angle C=75°$, then b =

(a) $\sqrt{3}$                            (b) $\sqrt{6}$                           (c) $\sqrt{9}$                           (d) $1+\sqrt{2}$

It is given that a = 2, $\angle B=60°$ and $\angle C=75°$.

In âˆ†ABC,

Using sine rule, we get

Hence, the correct answer is option (b).

#### Question 3:

Mark the correct alternative in each of the following:

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

(a) $\frac{\mathrm{\pi }}{6}$                            (b) $\frac{\mathrm{\pi }}{3}$                            (c) $\frac{\mathrm{\pi }}{2}$                            (d) $\frac{2\mathrm{\pi }}{3}$

Let âˆ†ABC be the given triangle such that its sides are in the ratio $1:\sqrt{3}:2$.

$\therefore a=k,b=\sqrt{3}k,c=2k$

Now, ${a}^{2}+{b}^{2}={k}^{2}+3{k}^{2}=4{k}^{2}={c}^{2}$

So, âˆ†ABC is a right triangle right angled at C.

$\therefore C=90°$

Using sine rule, we have

Thus, the measure of its greatest angle is $\frac{\mathrm{\pi }}{2}$.

Hence, the correct answer is option (c).

#### Question 4:

Mark the correct alternative in each of the following:

In any âˆ†ABC, 2(bc cosA + ca cosB + ab cosC) =

(a) $abc$                            (b) $a+b+c$                           (c) ${a}^{2}+{b}^{2}+{c}^{2}$                           (d) $\frac{1}{{a}^{2}}+\frac{1}{{b}^{2}}+\frac{1}{{c}^{2}}$

Using cosine rule, we have

$2\left(bc\mathrm{cos}A+ca\mathrm{cos}B+ab\mathrm{cos}C\right)\phantom{\rule{0ex}{0ex}}=2bc\left(\frac{{b}^{2}+{c}^{2}-{a}^{2}}{2bc}\right)+2ca\left(\frac{{c}^{2}+{a}^{2}-{b}^{2}}{2ca}\right)+2ab\left(\frac{{a}^{2}+{b}^{2}-{c}^{2}}{2ab}\right)\phantom{\rule{0ex}{0ex}}={b}^{2}+{c}^{2}-{a}^{2}+{c}^{2}+{a}^{2}-{b}^{2}+{a}^{2}+{b}^{2}-{c}^{2}\phantom{\rule{0ex}{0ex}}={a}^{2}+{b}^{2}+{c}^{2}$

Hence, the correct answer is option (c).

#### Question 5:

Mark the correct alternative in each of the following:

In a triangle ABC, a = 4, b = 3, $\angle A=60°$ then c is a root of the equation

(a) ${c}^{2}-3c-7=0$                 (b) ${c}^{2}+3c+7=0$                (c) ${c}^{2}-3c+7=0$                (d) ${c}^{2}+3c-7=0$

It is given that a = 4, b = 3 and $\angle A=60°$.

Using cosine rule, we have

$\mathrm{cos}A=\frac{{b}^{2}+{c}^{2}-{a}^{2}}{2bc}\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}60°=\frac{9+{c}^{2}-16}{2×3×c}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{2}=\frac{{c}^{2}-7}{6c}\phantom{\rule{0ex}{0ex}}⇒{c}^{2}-7=3c\phantom{\rule{0ex}{0ex}}⇒{c}^{2}-3c-7=0$

Thus, c is the root of ${c}^{2}-3c-7=0$.

Hence, the correct answer is option (a).

#### Question 6:

Mark the correct alternative in each of the following:

In a âˆ†ABC, if $\left(c+a+b\right)\left(a+b-c\right)=ab$, then the measure of angle C is

(a) $\frac{\mathrm{\pi }}{3}$                            (b) $\frac{\mathrm{\pi }}{6}$                            (c) $\frac{2\mathrm{\pi }}{3}$                            (d) $\frac{\mathrm{\pi }}{2}$

Given: $\left(c+a+b\right)\left(a+b-c\right)=ab$

$⇒{\left(a+b\right)}^{2}-{c}^{2}=ab\phantom{\rule{0ex}{0ex}}⇒{a}^{2}+{b}^{2}+2ab-{c}^{2}=ab\phantom{\rule{0ex}{0ex}}⇒{a}^{2}+{b}^{2}-{c}^{2}=-ab\phantom{\rule{0ex}{0ex}}⇒\frac{{a}^{2}+{b}^{2}-{c}^{2}}{2ab}=-\frac{1}{2}$

Thus, the measure of angle C is $\frac{2\mathrm{\pi }}{3}$.

Hence, the correct answer is option (c).

#### Question 7:

Mark the correct alternative in each of the following:

In any âˆ†ABC, the value of  $2ac\mathrm{sin}\left(\frac{A-B+C}{2}\right)$ is

(a) ${a}^{2}+{b}^{2}-{c}^{2}$                            (b) ${c}^{2}+{a}^{2}-{b}^{2}$                            (c) ${b}^{2}-{c}^{2}-{a}^{2}$                            (d) ${c}^{2}-{a}^{2}-{b}^{2}$

In âˆ†ABC,

$\therefore 2ac\mathrm{sin}\left(\frac{A-B+C}{2}\right)\phantom{\rule{0ex}{0ex}}=2ac\mathrm{sin}\left(\frac{\mathrm{\pi }-2B}{2}\right)\phantom{\rule{0ex}{0ex}}=2ac\mathrm{sin}\left(\frac{\mathrm{\pi }}{2}-B\right)\phantom{\rule{0ex}{0ex}}=2ac\mathrm{cos}B$

Hence, the correct answer is option (b).

#### Question 8:

Mark the correct alternative in each of the following:

In any âˆ†ABC, $a\left(b\mathrm{cos}C-c\mathrm{cos}B\right)=$

(a) ${a}^{2}$                            (b) ${b}^{2}-{c}^{2}$                              (c) 0                               (d) ${b}^{2}+{c}^{2}$

$a\left(b\mathrm{cos}C-c\mathrm{cos}B\right)\phantom{\rule{0ex}{0ex}}=ab\left(\frac{{a}^{2}+{b}^{2}-{c}^{2}}{2ab}\right)-ca\left(\frac{{c}^{2}+{a}^{2}-{b}^{2}}{2ca}\right)\phantom{\rule{0ex}{0ex}}=\frac{{a}^{2}+{b}^{2}-{c}^{2}-{c}^{2}-{a}^{2}+{b}^{2}}{2}\phantom{\rule{0ex}{0ex}}=\frac{2{b}^{2}-2{c}^{2}}{2}\phantom{\rule{0ex}{0ex}}={b}^{2}-{c}^{2}$