Question

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If in a right angled ∆ABC, tan B = 12/5, then find sin B.
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Answer 1

✰ Question ༄ :

If in a right angled ∆ABC, tan B = 12/5, then find sin B.

✰ Given ༄ :

➱ tan B = 12/5

✰ To Find ༄ :

➱ sin B

✰ Concept ༄ :

In this question we have to firstly find the third side of rectangle using Pythagoras Theorem and then find the sin B using Trigonometric Ratios .

✰ Solution ༄ :

As we know that in right angled ∆ ABC ,

$: \implies \: tan \: B = \boxed{ \pink{ \frac{perpendicular}{base} }}$

So ,

$: \implies \: tan \: B = \frac{BC }{BA}$

Therefore values of :

• BC = 12 cm

• BA = 5 cm

Now finding the third side of right angled triangle using Pythagoras Theorem :

• H² = P² + B²

Where ,

• H = Hypotenuse = AC

• P = Perpendicular = BC = 12 cm

• B = Base = BA = 5 cm

So ,

• AC² = 12² + 5²

• AC² = 144 + 25

• AC² = 169

• AC = √169

• AC = 13 cm

Now ,

$: \implies \: sin \:B = \boxed{ \pink{\frac{perpendicular}{hypotenuse} }}$

So ,

$: \implies \: sin \:B = \frac{BC }{AC}$

Therefore ,

$: \implies \: sin \:B = \boxed{ \bold \blue{ \frac{12}{13} }}$

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

Trigonometric Identities :

➱ Cos² θ + Sin² θ = 1.

➱ 1 + Tan² θ = Sec² θ

➱ 1 + Cot² θ = Cosec² θ

Answer 2

Step-by-step explanation:

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