Question

If angle b and angle q are acute angles such that sin b = sin q . Then prove that angle b = angle q
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Answer 1

Answer:

Given that ∠B and ∠Q are acute angle and

sinB=sinQ__ (A)

From ΔACB and ΔPRQ

sinB=

AB

AC

__(1)

sinQ=

PQ

PR

___(2)

From equation (A)

sinB=sinQ

AB

AC

=

PQ

PR

let

AB

AC

=

PQ

PR

=k

∴

PR

AC

=

PQ

AB

=k __(3)

Now,

AC=k×PR

AB=k×PQ

From ΔACB

By Pythagoras theorem

AB

2

=AC

2

+BC

2

(k×PR)

2

=(k×PQ)

2

+BC

2

⇒k

2

×PR

2

=k

2

×PQ

2

−BC

2

⇒BC

2

=k

2

×PR

2

−k

2

PQ

2

=k

2

[PR

2

−PQ

2

]

∴BC=

k

2

[PR

2

−PQ

2

]

From ΔPRQ

By Pythagoras theorem

PQ

2

=PR

2

+QR

2

⇒QR

2

=PQ

2

−PR

2

∴QR=

PQ

2

−PR

2

Consider that

QR

BC

=k __(4)

From equation (3) and (4) to,

PR

AC

=

PQ

AB

=

QR

BC

Hence, ΔACB∼ΔPRQ (sss similarity)

∠B=∠Q

Hence, this is the answer

Step-by-step explanation:

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Answer 2

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

Let us consider two right angle triangles abc and pqr right angled at c and r respectively and b and q are acute angles such that

$\rm \: sinb = sinq$

$\rm\implies \:\dfrac{ac}{ab} = \dfrac{pr}{pq}$

$\rm\implies \:\dfrac{ac}{pr} = \dfrac{ab}{pq} = k \: (say) - - - (1)$

$\rm\implies \:ac = k \: (pr) \: \: and \: \: ab = k \: (pq)$

Now, In right triangle abc

$\rm \: {ab}^{2} = {bc}^{2} + {ac}^{2}$

$\rm \: {bc}^{2} = {ab}^{2} - {ac}^{2}$

On substituting the values of ab and ac from above, we get

$\rm \: {bc}^{2} = {(k \: pq)}^{2} - {(k \: pr)}^{2}$

$\rm \: {bc}^{2} = { {k}^{2} \: pq}^{2} - { {k}^{2} \: pr}^{2}$

$\rm \: {bc}^{2} = { {k}^{2} ( \: pq}^{2} - {\: pr}^{2} )$

$\rm \: {bc}^{2} = {k}^{2} \: {qr}^{2}$

$\rm\implies \:bc \: = \: k \: qr$

$\rm\implies \:\dfrac{bc}{qr} = k - - - - (2)$

So, from equation (1) and (2), we get

$\rm\implies \:\dfrac{ac}{pr} = \dfrac{ab}{pq} = \dfrac{bc}{qr}$

$\rm\implies \: \triangle \: acb \: \sim \: \triangle \: prq \: \: \: \: \: \: \: \: \{By \: SSS \}$

9

$\rm\implies \:b = q \: \: \: \: \: \: \{By \: CPCT \}$

Hence, Proved

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More To Know :-

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

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