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Trigonometric Ratios | Trigonometric Ratios Exercise 17

Question 7

. If tan = 4/3

find the value of sin .. cos … (both sin and cos are Positive)

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Let ∆ABC be a right-angled

∠ACB = θ

Given that, tan θ = 4/3

(AB/BC = 4/3)

Give that, tan θ = 4/3

(AB/BC = 4/3)

Let AB = 4x,

then BC = 3x

In right-angled ∆ABC

By Pythagoras theorem, we get

\begin{array}{l} A C^{2}=A B^{2}+B C^{2} \\ A C^{2}=A B^{2}+B C^{2} \\ A C^{2}=A B^{2}+B C^{2} \\ A C^{2}=A B^{2}+B C^{2} \\ \left(A C^{2}=(4 x)^{2}+(3 x)\right. \\ A C^{2}=16 x^{2}+9 x^{2} \\ A C^{2}=25 x^{2} \\ A C^{2}=(5 x)^{2} \\ A C=5 x \end{array}

AC = 5x

Sin θ = perpendicular/Hypotenuse

= AB/AC

= 4x/5x

= 4/5

Cos θ = Base/Hypotenuse

= BC/AC

= 3x/5x

= 3/5

Sin θ + cos θ

= 4/5 + 3/5

= (4 + 3)/5

= 7/5

Hence, Sin θ + cos θ = 7/5 = 1 (2/5)

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Our top 5% students will be awarded a special scholarship to Lido.

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Question 7

. If tan = 4/3

find the value of sin .. cos … (both sin and cos are Positive)

Looking to do well in your science exam ? Learn from an expert tutor. Book a free class!

Let ∆ABC be a right-angled

∠ACB = θ

Given that, tan θ = 4/3

(AB/BC = 4/3)

Give that, tan θ = 4/3

(AB/BC = 4/3)

Let AB = 4x,

then BC = 3x

In right-angled ∆ABC

By Pythagoras theorem, we get

\begin{array}{l} A C^{2}=A B^{2}+B C^{2} \\ A C^{2}=A B^{2}+B C^{2} \\ A C^{2}=A B^{2}+B C^{2} \\ A C^{2}=A B^{2}+B C^{2} \\ \left(A C^{2}=(4 x)^{2}+(3 x)\right. \\ A C^{2}=16 x^{2}+9 x^{2} \\ A C^{2}=25 x^{2} \\ A C^{2}=(5 x)^{2} \\ A C=5 x \end{array}

AC = 5x

Sin θ = perpendicular/Hypotenuse

= AB/AC

= 4x/5x

= 4/5

Cos θ = Base/Hypotenuse

= BC/AC

= 3x/5x

= 3/5

Sin θ + cos θ

= 4/5 + 3/5

= (4 + 3)/5

= 7/5

Hence, Sin θ + cos θ = 7/5 = 1 (2/5)

Our top 5% students will be awarded a special scholarship to Lido.

subject-cta
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