Question

1) observe the figure and write the value of theta and alpha ​

Answer 1

$\large\underline{\bf{Solution-}}$

Given that

In right angle triangle PQR right-angled at P,

$\red{\rm :\longmapsto\: \bf \: PQ = 5}$

$\red{\rm :\longmapsto\: \bf \: PR = 5 \sqrt{3} }$

$\red{\rm :\longmapsto\: \bf \: QR = 10 \sqrt{3} }$

Now,

We know that,

In right angle triangle,

$\red{\rm :\longmapsto\: \bf \: sin \theta \: = \: \dfrac{Opposite \: Side}{Hypotenuse} }$

So,

$\red{\rm :\longmapsto\: \bf \: In \: \triangle \: PQR}$

${\rm :\longmapsto\: \bf \: sin \theta \: = \: \dfrac{5}{10} }$

${\rm :\longmapsto\: \bf \: sin \theta \: = \: \dfrac{1}{2}}$

${\rm :\longmapsto\: \bf \: sin \theta \: = sin \: 30 \degree}$

$\: \: \: \: \: \: \: \: \: \: \: \: \purple{\boxed{{\rm :\longmapsto\: \bf \: \theta \: = \: 30 \degree}}}$

Now,

We know, sum of interior angles of a triangle is 180°.

So,

$\red{\rm :\longmapsto\: \bf \: In \: \triangle \: PQR}$

$\rm :\longmapsto\: \theta \: + \alpha \: + \: 90 \degree \: = 180\degree$

$\rm :\longmapsto\: 30\degree \: + \alpha \: + \: 90 \degree \: = 180\degree$

$\rm :\longmapsto\: \alpha \: + \: 120 \degree \: = 180\degree$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \purple{ \boxed{\bf :\longmapsto\: \alpha \: \: = 60\degree}}$

Additional Information :-

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$

Relationship between sides and T ratios

sin θ = Opposite Side/Hypotenuse

cos θ = Adjacent Side/Hypotenuse

tan θ = Opposite Side/Adjacent Side

sec θ = Hypotenuse/Adjacent Side

cosec θ = Hypotenuse/Opposite Side

cot θ = Adjacent Side/Opposite Side

Reciprocal Identities

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

sin θ = 1/cosec θ

cos θ = 1/sec θ

tan θ = 1/cot θ

Co-function Identities

sin (90°−x) = cos x

cos (90°−x) = sin x

tan (90°−x) = cot x

cot (90°−x) = tan x

sec (90°−x) = cosec x

cosec (90°−x) = sec x

Fundamental Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1

Answer 2

Answer:

hello

Step-by-step explanation:

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