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Calculation:

Given,

Let$p = \sin 35^\circ$,$q = \sin 25^\circ$ and $r = \sin(-95^\circ)$

We need to find the value of p + q + r .

First, we know that:

$\sin(-95^\circ) = -\sin(95^\circ)$.

Thus, the expression p + q + r becomes:

$p + q + r = \sin 35^\circ + \sin 25^\circ - \sin 95^\circ$.

Using the identity for the sum of two sine functions:

$\sin A + \sin B = 2 \sin \left( \frac{A + B}{2} \right) \cos \left( \frac{A - B}{2} \right)$.

Substitute $A = 35^\circ$ and $B = 25^\circ$

$\sin 35^\circ + \sin 25^\circ = 2 \sin 30^\circ \cos 5^\circ$.

Since $\sin 30^\circ = \frac{1}{2}$, we get:

$\sin 35^\circ + \sin 25^\circ = \cos 5^\circ$.

Now, we know that:

$\sin 95^\circ = \cos 5^\circ$.

Therefore:

$p + q + r = \cos 5^\circ - \cos 5^\circ = 0$.

∴ The value of p + q + r  is 0.

Hence, the correct answer is Option 0.

Calculation:

Given,

The expression to simplify is pq + qr + rp, where:

$p = \sin(35^\circ)$

$q = \sin(25^\circ)$

$r = \sin(-95^\circ) = -\cos(5^\circ)$

Thus, we need to evaluate:

$pq + qr + rp = \sin(35^\circ) \sin(25^\circ) + \sin(25^\circ) (-\cos(5^\circ)) + (-\cos(5^\circ)) \sin(35^\circ)$

Using trigonometric identities, we simplify the expression:

$pq + qr + rp = -\sin^2(25^\circ) - \sin(95^\circ) \sin(35^\circ)$

Using the sum-to-product identity and known values for sine and cosine:

$= \frac{1}{2} [-2 \sin(25^\circ) - 2 \sin(95^\circ) \cdot \sin(35^\circ)]$

$= \frac{1}{2} [-1 + \cos(50^\circ) + \cos(130^\circ) - \cos(60^\circ)]$

$= \frac{1}{2} [-1 + \cos(50^\circ) - \cos(50^\circ) - \frac{1}{2}]$

$= \frac{1}{2} \left[ -1 - \frac{1}{2} \right] = -\frac{3}{4}$

∴ The value of pq + qr + rp  is $-\frac{3}{4} .$

Hence, the correct answer is Option 1.

Calculation:

Given,

The expression to simplify is $p^2 + q^2 + r^2$, where:

$p^2 + q^2 + r^2$

$q = \sin(25^\circ)$

$r = \sin(-95^\circ) = -\cos(5^\circ)$

We are given that p + q + r = 0 . Using the identity for squaring a sum:

$(p + q + r)^2 = p^2 + q^2 + r^2 + 2(pq + qr + rp)$

Since p + q + r = 0, we substitute this into the identity:

$0 = p^2 + q^2 + r^2 + 2(pq + qr + rp)$

From previous calculations, we know that:

$pq + qr + rp = -\frac{3}{4}$.

Substitute this value into the equation:

$0 = p^2 + q^2 + r^2 + 2\left( -\frac{3}{4} \right)$

Simplifying:

$0 = p^2 + q^2 + r^2 - \frac{3}{2}$

$p^2 + q^2 + r^2 = \frac{3}{2}$.

Hence, the correct answer is Option 3.

Calculation:

Given,

We are given the expression:

$p = | \sin \alpha - \sin(\alpha - 90^\circ) |$

Using the identity for sine of a difference:

$\sin(\alpha - 90^\circ) = -\cos \alpha$

Substituting this identity into the expression for p :

$p = | \sin \alpha + \cos \alpha |$

The minimum value of $| \sin \alpha + \cos \alpha |$ occurs when $\sin \alpha = -\cos \alpha$, which happens when$\alpha = 135^\circ$. At this point:

$\sin 135^\circ = \frac{1}{\sqrt{2}}, \cos 135^\circ = -\frac{1}{\sqrt{2}}$

Therefore, at $\alpha = 135^\circ$, the sum of $\sin \alpha$ and $\cos \alpha$ is 0, so:

$p = 0$

∴ The minimum value of p  is 0.

Hence, the correct answer is option (a) 0.

Calculation:

Given,

The expression for p  is:

$p = |\sin \alpha - \sin(\alpha - 90^\circ)|$

Using the identity for $\sin(\alpha - 90^\circ)$

$\sin(\alpha - 90^\circ) = \cos \alpha$

Substituting this into the expression for p :

$p = |\sin \alpha - \cos \alpha|$

The expression  $|\sin \alpha - \cos \alpha|$  achieves its maximum value when the difference between $\sin \alpha$ and $\cos \alpha$ is largest.

We can rewrite $\sin \alpha - \cos \alpha$ as:

$\sin \alpha - \cos \alpha = \sqrt{2} \left( \sin \left( \alpha - 45^\circ \right) \right)$

Since $\sin(\alpha - 45^\circ)$ has a maximum value of 1, the maximum value of $\sqrt{2} \left( \sin(\alpha - 45^\circ) \right)$ is:

$\sqrt{2}$

Hence, the correct answer is Option 2

Calculation:

Given,

The sides of the triangle are:

a = 5 cm

b = 7 cm

c = 3 cm

Statement I: The triangle is obtuse-angled.

Using the Cosine Rule:

$\cos B = \frac{a^2 + c^2 - b^2}{2ac}$

Substitute the values for a, b = 7 and c

$\cos B = \frac{5^2 + 3^2 - 7^2}{2 \times 5 \times 3} = \frac{25 + 9 - 49}{30} = \frac{-15}{30} = -\frac{1}{2}$

Now, calculate angle B:

$B = \cos^{-1}\left(-\frac{1}{2}\right) = 120^\circ$

Thus, angle B  is $120^\circ$, which confirms that the triangle is an obtuse-angled triangle, 

Statement II: The sum of the acute angles of the triangle is also acute.

Since the sum of the angles in a triangle is $180^\circ$, the sum of the acute angles is:

$A + C = 180^\circ - 120^\circ = 60^\circ$

This confirms that the sum of the acute angles is acute,

∴ Both statements are correct.

Hence, the correct answer is Option 3.

Calculation:

Given,

The sides of the triangle are: a = 5 Cm,  b = 7 cm,  c = 3 cm.

Apply the Cosine Rule to find angle B:

Using the Cosine Rule:

$\cos B = \frac{a^2 + c^2 - b^2}{2ac}$

Substitute the values of a, b and c

$\cos B = \frac{5^2 + 3^2 - 7^2}{2 \times 5 \times 3} = \frac{25 + 9 - 49}{30} = \frac{-15}{30} = -\frac{1}{2}$

Now, calculate angle B :

$B = \cos^{-1}\left(-\frac{1}{2}\right) = 120^\circ$

Thus, angle B  is $120^\circ$

Hence, the correct answer is Option 3.

Calculation:

Given,

The sides of the triangle are: a = 5 cm, c = 3 cm, and the angle B = $120^\circ$ between sides a and c.

The area of the triangle is given by the formula:

$\text{Area} = \frac{1}{2} \times a \times c \times \sin B$

$\text{Area} = \frac{1}{2} \times 5 \times 3 \times \sin(120^\circ)$

We know that$\sin(120^\circ) = \frac{\sqrt{3}}{2}$, so:

$\text{Area} = \frac{1}{2} \times 5 \times 3 \times \frac{\sqrt{3}}{2}$

$\text{Area} = \frac{1}{2} \times 15 \times \frac{\sqrt{3}}{2} = \frac{15 \times \sqrt{3}}{4}$

∴ The area of the triangle is $\frac{15\sqrt{3}}{4}$ square cm.

Hence, the Correct answer is Option 1.

Calculation:

The angles of elevation are:

From point P,$\theta = 30^\circ$; from point Q, $\theta = 45^\circ$; and from point R, $\theta = 60^\circ$

Using the tangent formula for each angle:

$\tan(60^\circ) = \frac{h}{x} \implies h = \sqrt{3}x$

$\tan(45^\circ) = \frac{h}{QN} \implies h = QN$

$\tan(30^\circ) = \frac{h}{PN} \implies PN = h{\sqrt{3}}$

from the figure PN = h + a 

$h+a = h\sqrt3$

$h = \frac{a}{\sqrt3 -1} \implies \frac{a \sqrt3 +1}{2}$

= $(\frac{3 + \sqrt3}{2}) a$

Hence, the correct answer is Option 2.

Calculation

$\tan(60^\circ) = \frac{h}{x} \Rightarrow h = \sqrt{3}x$

$\tan(45^\circ) = \frac{h}{b + x} \Rightarrow h = b + x$

$h = b + x \Rightarrow \sqrt{3}x = b + x$

$x = \frac{b(\sqrt{3} + 1)}{2}$

$h = MN = \sqrt{3}x = \left( \frac{3 + \sqrt{3}}{2} \right) b$

Hence, the correct answer is Option 1.