163^o347^o + 73^o167^o = - Vidyadip

MCQ

$\sin {163^o}\cos {347^o} + \sin {73^o}\sin {167^o} = $

A
$0$
✓
$1/2$
C
$1$
D
None of these

✓

Answer

Correct option: B.

$1/2$

b
(b) $\sin {163^o}\cos {347^o} + \sin {73^o}\sin {167^o}$

$ = \sin ({180^o} - {17^o})\cos ({360^o} - {13^o}) + \cos ({90^o} - {17^o}) \sin ({180^o} - {13^o})$

$ = \sin {17^o}\cos {13^o} + \cos {17^o}\sin {13^o}$

$ =\sin {30^o} = 1/2$.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "MCQ" from 3. trignometrical ratios,functions and identities→3. trignometrical ratios,functions and identities — all question types→MATHS STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

Number of natural solutions of the equation $x_1 + x_2 = 100$ , such that $x_1$ and $x_2$ are not multiple of $5$

The locus of the mid points of the chords of the circle $C_1:(x-4)^2+(y-5)^2=4$ which subtend an angle $\theta_i$ at the centre of the circle $C_1$, is a circle of radius $r_i$. If $\theta_1=\frac{\pi}{3}, \theta_3=\frac{2 \pi}{3}$ and $r_1^2=r_2^2+r_3^2$, then $\theta_2$ is equal to

In a triangle $A B C, \angle B A C=90^{\circ} ; A D$ is the altitude from $A$ on to $B C$. Draw $D E$ perpendicular to $A C$ and $D F$ perpendicular to $A B$. Suppose $A B=15$ and $B C=25$. Then the length of $E F$ is

Let $P$ be a moving point such that sum of its perpendicular distances from $2x + y = 3$ and $x - 2y + 1 = 0$ is always $2\, units$ then area bounded by locus of point $P$ is

The length of the minor axis (along $y-$axis) of an ellipse in the standard form is $\frac{4}{\sqrt{3}} .$ If this ellipse touches the line, $x+6 y=8 ;$ then its eccentricity is

Distance between the parallel lines $3x + 4y + 7 = 0$ and $3x + 4y - 5 = 0$ is

$2^{3 n}-7 n-1$ is divisible by:

The equation of the parabola whose vertex and focus lies on the $x$ - axis at distance $a$ and $a’$ from the origin, is

With one focus of the hyperbola $\frac{{{x^2}}}{9}\,\, - \,\,\frac{{{y^2}}}{{16}}\,\, = \,\,1$ as the centre , a circle is drawn which is tangent to the hyperbola with no part of the circle being outside the hyperbola. The radius of the circle is

$(p - q)x + (q - r)y + (r - p) = 0$

$(q - r)x + (r - p)y + (p - q) = 0$

$(r - p)x + (p - q)y + (q - r) = 0$ are