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Trigonometric Functions

Multiple Choice Questions

191.

$\cos^{2} (76 °) + \cos^{2} (16 °) - \cos (76 °) \cos (16 °) is equal to$

0
$\frac{1}{2}$
$- \frac{1}{2}$
$\frac{3}{4}$

192.

$\sinh (ix) is equal to$

$isin (x)$
$\sin (ix)$
$- isin (x)$
$isin (ix)$

193.

The perimeter of a triangle is 16cm, one of the sides is of length 6cm. If the area of the triangle is 12 sq cm. Then the triangle is

right angled
isosceles
equilateral
scalene

194.
If $∆$ ABC is right angled at A, then r₂ + r₃ is equal to

r₁ - r

r₁ + r

r - r₁

R

r₁ - r

$Given that, ∠ A = 90 ° Now, r_{2} + r_{3} = 4 Rcos (\frac{A}{2}) \sin (\frac{B}{2}) \cos (\frac{C}{2}) + 4 Rcos (\frac{A}{2}) \cos (\frac{B}{2}) \sin (\frac{C}{2}) = 4 Rcos (\frac{A}{2}) [\sin (\frac{B}{2}) \cos (\frac{C}{2}) + \cos (\frac{B}{2}) \sin (\frac{C}{2})] = 4 Rcos (45 °) . \sin (\frac{B + C}{2}) = 4 R \frac{1}{\sqrt{2}} . \sin (\frac{π}{2} - \frac{A}{2}) = 4 R \frac{1}{\sqrt{2}} \cos (\frac{A}{2}) = 4 R \frac{1}{\sqrt{2}} \cos (45 °) = 4 R \frac{1}{\sqrt{2}} . \frac{1}{\sqrt{2}} = 2 R Also, r_{1} + r_{2} + r_{3} - r = 4 R = 2 (2 R) r_{1} + r_{2} + r_{3} - r = 2 (r_{2} + r_{3}) r_{2} + r_{3} = r_{1} - r$

195.

If in $∆ ABC$ , r₁ < r₂ < r₃, then :

a < b < c
a > b > c
b > a < c
a < c < b

196.

In a $∆ ABC$ , if 3a = b + c, then $\cot (\frac{B}{2}) \cot (\frac{C}{2})$ is equal to :

197.

If in a triangle, if b = 20, c = 21 and $\sin (A) = \frac{3}{5}$ , then a is equal to

198.

A tower subtends angles $α, 2 α and 3 α$ respectively at points A, B and C, all lying on a horizontal line through the foot of the tower,then $\frac{AB}{BC}$ is equal to :

$\frac{\sin (3 α)}{\sin (2 α)}$
$1 + 2 \cos (2 α)$
$2 \cos (2 α)$
$\frac{\sin (2 α)}{\sin (α)}$

199.

The co-ordinate axes are rotated through an angle 135°. If the co-ordinates of a point P inthe new system are known to be (4, - 3), then the co-ordinates of P in the original system are :