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Draw a right triangle where tan B=3. Use the trigonometric ratio to label two side lengths and the Pythagorean Theorem to find the missing side.
sin B=3sqrt(10)/10, cos B=sqrt(10)/10, csc B=sqrt(10)/3, sec B=sqrt(10), cot B=1/3
In a right triangle, ∠A and ∠B are acute angles while ∠C is the right angle.
Tangent is defined as the ratio of the opposite side to the adjacent side of an angle. We are given that tan B=3.
Note that 3 can be written as 3 1.
tan = opp/adj ⇒ tan B = 3 1
Now, let's draw a right triangle satisfying these conditions.
Let's find the hypotenuse by substituting a= 1 and b= 3 into the Pythagorean Theorem.
a= 1, b= 3
Calculate power
Add terms
sqrt(LHS)=sqrt(RHS)
Rearrange equation
Note that when solving the equation we only kept the principal root. This is because c represents a side length, and therefore, must be positive.
Now that we know the length of all three sides of the triangle, we can find the values of the five remaining trigonometric functions. Remember to rationalize denominators if needed.
| Function | Substitute | Simplify |
|---|---|---|
| sin B=opp/hyp | sin B=3/sqrt(10) | sin B=3sqrt(10)/10 |
| cos B=adj/hyp | cos B=1/sqrt(10) | cos B=sqrt(10)/10 |
| csc B=hyp/opp | csc B=sqrt(10)/3 | csc B=sqrt(10)/3 |
| sec B=hyp/adj | sec B=sqrt(10)/1 | sec B=sqrt(10) |
| cot B=adj/opp | cot B=1/3 | cot B=1/3 |
a/b=a * sqrt(10)/b * sqrt(10)
a* a=a^2
( sqrt(a) )^2 = a
Let's now follow the same procedure to rationalize the denominator of 1sqrt(10).
a/b=a * sqrt(10)/b * sqrt(10)
a* a=a^2
( sqrt(a) )^2 = a