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3 years ago

Trig please help

Mathematics

2 answers:

jeyben [28]3 years ago

7 0

Answer:

one triangle

Step-by-step explanation:

The given angle is opposite the longest given side, so this case is not ambiguous. These numbers define one triangle.

_____

Comment on ambiguity

The law of sines only gives rise to ambiguity (two solutions) only if the given angle is opposite the shorter of the two given sides, and the ratio of side lengths (short/long) is greater than the sine of the given angle.

In the extremely rare case of the side ratio being exactly equal to the sine of the angle opposite the shorter side, the triangle is a right triangle and there is exactly one solution.

avanturin [10]3 years ago

3 0

Answer:

One

Step-by-step explanation:

19/sin(50) = 15/sinB

sinB = 0.6047719288

B = 37.2, 142.8

Since 142.8+50 = 192.8 > 180

It will not form a triangle.

So only one triangle possible with angle 37.2° at B

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ABC is translated 6 units up and 3 units left to create ∆A'B'C'.

aivan3 [116]

B is the right answer for this

6 0

4 years ago

5491 apples were harvested from one side of the McMillan farm. they harvested 512 from the other side of the farm.about how many

Andrews [41]

Answer:

6003

Step-by-step explanation:

Given : They harvested 5491 apples from one side of the McMillan farm.

They harvested 512 from the other side of the farm.

To Find : The total number of apples harvested at the McMillan.

Solution :

Since we know that

Number of apples harvested from one side of the McMillan farm = 5491

Number of apples harvested from other side of the McMillan farm = 512

Now we are supposed to calculate the total number of apples

So, we need to sum the number of apples of both the sides .

⇒5491+512

⇒6003

Hence , the number of apples harvested at the McMillan is 6003

3 0

3 years ago

Prove that: (Sec A- cosec A)(1+ tan A+cot A) = tan A× sec A - cot A × cosec A

mash [69]

Answer:

Step-by-step explanation:

$(sec A-cosecA)(1+tanA+cotA)=tanAsecA-cotAcosecA$

we take the LHS so here goes,

$(sec A-cosecA)(1+tanA+cotA)=tanAsecA-cotAcosecA\\(\frac{1}{cosA} -\frac{1}{sinA})(1+\frac{sinA}{cosA}+\frac{cosA}{sinA})\\\\(\frac{sinA-cosA}{sinAcosA})(\frac{sinAcosA+sin^2A+cos^2A}{sinAcosA})\\$

since , $sin^2A+cos^2A=1$

the identity becomes,

$(\frac{sinA-cosA}{sinAcosA})(\frac{1+sinAcosA}{sinAcosA})\\\\(\frac{sinA+sin^2AcosA-cosA-cos^2AsinA}{sin^2Acos^2A})\\\\$

now, we know,

$sin^2A=1-cos^2A$ and $cos^2A=1-sin^2A$

the identity becomes,

$(\frac{sinA+(1-cos^2A)cosA-cosA-(1-sin^2A)sinA}{sin^2Acos^2A} )\\\\$

$(\frac{sinA+cosA-cos^3A-cosA-sinA+sin^3A}{sin^2Acos^2A})$

sin A and cos A cancel out it becomes zero

$\frac{sin^3A-cos^3A}{sin^2Acos^2A} \\\\$

Splitting the denominator the identity becomes

$\frac{sin^3A}{sin^2Acos^2A}-\frac{cos^3A}{sin^2Acos^2A} \\\\\frac{sinA}{cos^2A} - \frac{cosA}{sin^2A} \\\\\frac{sinA}{cosA}(\frac{1}{cosA})-\frac{cosA}{sinA}(\frac{1}{sinA})\\\\$