The sum of three and a number is 3+x
You first wanna find <BAD, because if AB is perpendicular to AC, then it has to form a 90 degree angle. So 90-56=34 degrees. So now you have a 34 & 63 degrees in the ABD triangle. In a triangle, all angles add up to equal 180 degrees. So 34+63+x=180...and x=83. So <ADB= 83 degrees. Now you want to find angle ADC...which you can just subtract 83 from 180 (because <ADB & <ADC forms 180 degree angle). You will then get 97 as angle ADC. So, the same thing as before, add up 56+97+x=180, because all angles (in the triangle ADC) add up to be 180 degrees. X will then equal 27 degrees. Angle ACB= 27 degrees.
Answer:
A Pyramid
Step-by-step explanation:
We can use the Сosine formula to solve this problem.
<span>First we must find the third side (АС) of the triangle:
</span>
<span>
The smallest angle of the triangle lies opposite the smallest side, so we need to find m</span>∠C.
Now we can use Bradis's Table (I don't know the name in English, maybe Trigonometric Table?) to find m∠C:
m∠С = 38°42' = 38.7°
Answer: 38.7°
<span>I hope this helps</span>
(a) Yes all six trig functions exist for this point in quadrant III. The only time you'll run into problems is when either x = 0 or y = 0, due to division by zero errors. For instance, if x = 0, then tan(t) = sin(t)/cos(t) will have cos(t) = 0, as x = cos(t). you cannot have zero in the denominator. Since neither coordinate is zero, we don't have such problems.
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(b) The following functions are positive in quadrant III:
tangent, cotangent
The following functions are negative in quadrant III
cosine, sine, secant, cosecant
A short explanation is that x = cos(t) and y = sin(t). The x and y coordinates are negative in quadrant III, so both sine and cosine are negative. Their reciprocal functions secant and cosecant are negative here as well. Combining sine and cosine to get tan = sin/cos, we see that the negatives cancel which is why tangent is positive here. Cotangent is also positive for similar reasons.
