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3 years ago

7

How many different triangles can you make with 14cm, 10cm and 50cm sides?

2 answers:

Dominik [7]3 years ago

7 0

Answer:

0

Step-by-step explanation:

ANEK [815]3 years ago

5 0

Hello from MrBillDoesMath!

Answer:

0 (zero)

Discussion:

From the "triangle inequality", the sum of any two sides of a traingle must be greater than the length of the third side. As 14 + 10 is not greater than (is less than) 50, o no such triangle exists

Thank you,

MrB

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For positive acute angles A and B, it is known that tan A = 35/12 and sin B = 20/29. Find the value of sin(A - B ) in the simple

almond37 [142]

Answer:

$\displaystyle \sin(A-B)=\frac{495}{1073}$

Step-by-step explanation:

We are given that:

$\displaystyle \tan(A)=\frac{35}{12}\text{ and } \sin(B)=\frac{20}{29}$

Where both A and B are positive acute angles.

And we want to find he value of sin(A-B).

Using the first ratio, we can conclude that the opposite side is 35 and the adjacent side is 12.

Then by the Pythagorean Theorem, the hypotenuse is:

$h = \sqrt{35^2 + 12^2} =37$

Using the second ratio, we can likewise conclude that the opposite side is 20 and the hypotenuse is 29.

Then by the Pythagorean Theorem, the adjacent is:

$a=\sqrt{29^2-20^2}=21$

Therefore, we can conclude that:

So, for A, the adjacent is 12, opposite is 35, and the hypotenuse is 37.

For B, the adjacent is 21, opposite is 20, and the hypotenuse is 29.

We can rewrite sin(A-B) as:

$\sin(A-B)=\sin(A)\cos(B)-\cos(A)\sin(B)$

Using the above conclusions, this yields: (Note that since A and B are positive acute angles, all resulting ratios will be positive.)

$\displaystyle \sin(A-B)=\Big(\frac{35}{37}\Big)\Big(\frac{21}{29}\Big)-\Big(\frac{12}{37}\Big)\Big(\frac{20}{29}\Big)$

Evaluate:

$\displaystyle \sin(A-B)=\frac{735-240}{1073}=\frac{495}{1073}$

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