Find the area of a triangle with a base length of 3 units and a height of 4 units

Please help me with this!!!

Mandarinka [93]

Answer:

p=3/8, q=1/2

Step-by-step explanation:

Please mark as brainliest

3 0

2 years ago

Not school work. y’all i need to know something.

LUCKY_DIMON [66]

Ask your teachers if you can still turn them in

4 0

3 years ago

Read 2 more answers

How is this solved using trig identities (sum/difference)?

GenaCL600 [577]

FIRST PART
We need to find sin α, cos α, and cos β, tan β
α and β is located on third quadrant, sin α, cos α, and sin β, cos β are negative

Determine ratio of ∠α
Use the help of right triangle figure to find the ratio
tan α = 5/12
side in front of the angle/ side adjacent to the angle = 5/12
Draw the figure, see image attached

Using pythagorean theorem, we find the length of the hypotenuse is 13
sin α = side in front of the angle / hypotenuse
sin α = -12/13

cos α = side adjacent to the angle / hypotenuse
cos α = -5/13

Determine ratio of ∠β
sin β = -1/2
sin β = sin 210° (third quadrant)
β = 210°

$cos \beta = -\frac{1}{2} \sqrt{3}$

$tan \beta= \frac{1}{3} \sqrt{3}$

SECOND PART
Solve the questions
Find sin (α + β)
sin (α + β) = sin α cos β + cos α sin β
$sin( \alpha + \beta )=(- \frac{12}{13} )( -\frac{1}{2} \sqrt{3})+( -\frac{5}{13} )( -\frac{1}{2} )$
$sin( \alpha + \beta )=(\frac{12}{26}\sqrt{3})+( \frac{5}{26} )$
$sin( \alpha + \beta )=(\frac{5+12\sqrt{3}}{26})$

Find cos (α - β)
cos (α - β) = cos α cos β + sin α sin β
$cos( \alpha + \beta )=(- \frac{5}{13} )( -\frac{1}{2} \sqrt{3})+( -\frac{12}{13} )( -\frac{1}{2} )$
$cos( \alpha + \beta )=(\frac{5}{26} \sqrt{3})+( \frac{12}{26} )$
$cos( \alpha + \beta )=(\frac{5\sqrt{3}+12}{26} )$

Find tan (α - β)
$tan( \alpha - \beta )= \frac{ tan \alpha-tan \beta }{1+tan \alpha tan \beta }$
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3} }{1+(\frac{5}{12}) ( \frac{1}{2} \sqrt{3})}$

Simplify the denominator
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3} }{1+(\frac{5\sqrt{3}}{24})}$
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3} }{ \frac{24+5\sqrt{3}}{24} }$

Simplify the numerator
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{6}{12} \sqrt{3} }{ \frac{24+5\sqrt{3}}{24} }$
$tan( \alpha - \beta )= \frac{ \frac{5-6\sqrt{3}}{12} }{ \frac{24+5\sqrt{3}}{24} }$

Simplify the fraction
$tan( \alpha - \beta )= (\frac{5-6\sqrt{3}}{12} })({ \frac{24}{24+5\sqrt{3}})$
$tan( \alpha - \beta )= \frac{10-12\sqrt{3} }{ 24+5\sqrt{3}}$

7 0

3 years ago

At a display booth at an amusement park, every visitor gets a gift bag. Some

kotegsom [21]

The bag will often contain all three items every 70 bags ⇒ answer C

Step-by-step explanation:

At a display booth at an amusement park, every visitor gets a gift bag

according to these information:

1. Hat Every 2nd visitor

2. T-shirt Every 7th visitor

3. Backpack Every 10th visitor

We need to know how often a bag will contain all three items

To solve this problem you must find a number divisible by 2, 7, and 10

To find that number start with the multiples of the largest one and

check them with the other two numbers

∵ The largest number is 10

∵ The multiples of 10 are 10 , 20 , 30 , 40 , 50 , 60 , 70 , 80 , .......

- Look for the first multiple of 10 is divisible by 7

∵ 70 is the first multiple of 10 divisible by 7

∵ 70 is divisible by 2 because its even number

∴ 70 is the smallest number divisible by 2, 7 and 10

The bag will often contain all three items every 70 bags

Learn more:

You can learn more about numbers in brainly.com/question/537998

#LearnwithBrainly

4 0

3 years ago

What’s the answer??????

NikAS [45]

Answer:

-3x - 5

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Find the area of a triangle with a base length of 3 units and a height of 4 units​

Find the area of a triangle with a base length of 3 units and a height of 4 units