Naming the sides in a right angled triangle - class-XI

$ \displaystyle \sin 45^{\circ} \cos 45 \left ( \tan 45^{\circ}+\cot 45^{\circ} \right )^{2}   $

$=\dfrac{1}{\sqrt2} \times \dfrac{1}{\sqrt2} (1+1)^2 $

Two angles are called adjacent if
they lie in same plane and have a common vertex,they have a ray in common and the intersection of their interiors is empty.
do (D) all the above is correct

In a cyclic quadrilateral, the angle subtended by a chord at the center is twice the angle at the circumference. If angle ABC = 60, the central angle AOC = 120. In triangle OAC, OA = OC (radii), so angle OAC = (180 - 120)/2 = 30 degrees.

In the early 9th century AD, Muhammad ibn Mūsā al-Khwārizmī produced accurate sine and cosine tables, and the first table of tangents. He was also a pioneer in spherical trigonometry.

Hipparchus the founder of trigonometry. 

The history of trigonometry goes back to the earliest recorded mathematics in Egypt and Babylon.

Trigonometry is used mainly due to the purpose of time keeping and astronomy.

The first recorded use of trigonometry came from the Hellenistic mathematician Hipparchus.

Trigonometry is a branch of mathematics that studies relationships involving lengths and angles of triangles.

The term trigonometry was first invented by the German mathematician Bartholomaeus Pitiscus.

Trigonometry was first published by Bartholomaeus Pitiscus in 1595.

In $\triangle ABC,a=8,b=9,c=10$

Given, $\sin \theta + \cos \theta = 1$

Since $0 < x < \dfrac{\pi}{4}$

The angle of elevation and angle of depression are measured with Line of Sight which is always Horizontal.

$f _n(\theta)=(\tan \frac{\theta}{2})(1+\text{sec}\theta)(1+\text{sec} 2\theta)(1+\text{sec}2^2 \theta)\cdots(1+\text{sec}2^n \theta)$

$8\sin\theta \cos\theta\cos 2\theta \cos 4\theta =\sin x$

Given $11\sin^2 x+7\cos^2 x=8$

$\cos x+\cos y+\cos \theta=0$

Using the identity tan(theta) * tan(60-theta) * tan(60+theta) = tan(3theta), we have tan(3theta) = 1/sqrt(3). Thus, 3theta = npi + pi/6, which simplifies to theta = n*pi/3 + pi/18.

Using the formulas r1 = stan(A/2), r2 = stan(B/2), r3 = stan(C/2), and r = stan(A/2)*tan(B/2)*tan(C/2), the expression simplifies to 0 based on trigonometric identities in a triangle.

Using the Law of Sines and Mollweide's formulas, the correct relation is (b-c)cos(A/2) = a*sin((B-C)/2).

Ptolemy used length of chords to define his trigonometric functions

Let $f(x)=\tan {x}$
The points of discontinuity of $f(x)$ are those points where $\tan {x}$ is infinite. This gives
$\tan { x } =\infty $
ie $\tan { x } =\tan { \cfrac { \pi  }{ 2 }  } $
$x=\left( 2n+1 \right) \cfrac { \pi  }{ 2 } ,n\in I$

Trigonometry in the Greek language is triangle measure.
(Trigono - triangle + metron- measure)

$\text {Hipparchus}$ gave the first table of chords analogus to modern table of sine values, and used them to solve trigonometric problems

$\sqrt {2}\sec 45^{\circ} - \tan 30^{\circ} = \sqrt {2}\times \sqrt {2} - \dfrac {1}{\sqrt {3}} = \dfrac {2\sqrt {3} - 1}{\sqrt {3}}$.

Given, $XZ=YZ$ which implies angles $ZXY$ and $ZYX$ are equal and let it be $\theta$.
We have $ZXY+ZYX+XZY = 180$ , which implies $\theta+\theta+a=180$
Which implies $\theta =\dfrac { (180-a)}{2}$

Given, $\tan A =  \dfrac {1 - \cos B}{\sin B} $

$\sin15^o$

$Sin \theta=\dfrac{P}{H}=\dfrac{3}{5}$

The side opposite to the right angle in a right angled triangle is called Hypotenuse.

The Pythagorean theorem states that a^2 + b^2 = c^2. The area of a semicircle is (pi * r^2) / 2. For a triangle with sides a, b, and hypotenuse c, the areas are (pi * (a/2)^2) / 2, (pi * (b/2)^2) / 2, and (pi * (c/2)^2) / 2. Since a^2 + b^2 = c^2, the sum of the areas of the semicircles on the legs equals the area of the semicircle on the hypotenuse, not the difference.

Let AB be the tower of height h meter and BC be the height of flag staff surmounted on the tower. Let the point of the plane be D at a distance m meter from the foot of the tower.

In $\bigtriangleup$ ABD,

$\displaystyle tan 30^{\circ} = \frac{AB}{BD}$

$\displaystyle \Rightarrow \frac{1}{\sqrt3} = \frac{h}{x} \Rightarrow x = \sqrt 3 h$.......(1)

In $\displaystyle \bigtriangleup ADC, tan 60^{\circ} = \frac{AC}{AD}$

$\displaystyle \Rightarrow \sqrt3 = \frac{5+h}{x} \Rightarrow x = \frac{5 +h}{\sqrt3}$ .....(2)

From (1) and (2), $\displaystyle \sqrt 3h = \frac{5+h}{\sqrt 3}$

$\Rightarrow 3h = 5 + h \Rightarrow 2h = 5$

$\Rightarrow \displaystyle h = \frac{5}{2} = 2.5 m$

So, the height of tower = 2.5m

$\tan { 60°=\sqrt { 3 }  } =\dfrac { p }{ b } $

Length of Ladder = 13 ft