MHB Find cos theta and tan theta using sin theta

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Given that sin θ = 4/5, the adjacent side of the triangle is determined to be 3 using the Pythagorean theorem. Consequently, cos θ is calculated as 3/5 and tan θ as 4/3. However, it is noted that if θ is in the second quadrant, the cosine and tangent values would be negative. Therefore, both positive and negative values should be considered based on the quadrant in which the angle lies. This highlights the importance of quadrant consideration in trigonometric calculations.
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If sin $$\theta$$ =$$\frac{4}{5}$$ , find cos $$\theta$$ and tan $$\theta$$

Can you help me to solve. :)

Many thanks :)
 
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Re: Find cos theta and tan theta using sin thetha

mathlearn said:
If sin $$\theta$$ =$$\frac{4}{5}$$ , find cos $$\theta$$ and tan $$\theta$$

Can you help me to solve. :)

Many thanks :)

Hey mathlearn! ;)

The sine is the opposite side divided by the hypotenuse.
What would be the length of the adjacent side, knowing we have a right triangle?
And what would then be the cosine respectively the tangent?
 
Re: Find cos theta and tan theta using sin thetha

:)

mathlearn said:
If sin $$\theta$$ =$$\frac{4}{5}$$ , find cos $$\theta$$ and tan $$\theta$$

sin $$\theta$$ = $$\frac{opposite side}{hypotenuese}$$

$$\therefore sin $$ $$\theta$$ =$$\frac{4}{5}$$

So applying Pythagoras theorem

Hypotenuse2 = opposite side 2 + adjacent side2

$$5^{2}$$ = $$4^{2}$$ + $$ adjacent side^{2}$$

$$25$$ = $$16$$ + $$ adjacent side^{2}$$

$$25$$ - $$16$$= $$ adjacent side^{2}$$

$$9$$= $$ adjacent side^{2}$$

$$\sqrt{9}$$= $$ \sqrt{adjacent side^{2}}$$

$$3$$= $$adjacent side$$

$$\therefore cos \theta$$ = $$\frac{adjacent side}{hypotenuse }$$$$\therefore cos \theta$$ = $$\frac{3}{5}$$

and

$$\therefore tan \theta$$ = $$\frac{opposite side}{adjacent side}$$

$$\therefore tan \theta$$ = $$\frac{4}{3}$$

Correct I Guess?

Many Thanks :)
 
Yep. All correct. (Nod)
 
I like Serena said:
Yep. All correct. (Nod)

Actually, it's possible that the angle could be in the second quadrant, in which case the cosine and tangent values would be negative.

Without any information about which quadrant the angle lies, you would need to write both the positive and negative answers.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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