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Fofino [41]
3 years ago
10

Csc^2 xcos^2 x=csc^2 x-1 prove the identity

Mathematics
1 answer:
pochemuha3 years ago
6 0

Answer:

Step-by-step explanation:

Hello

csc^2(x)-1=\dfrac{1}{sin^2(x)}-1=\dfrac{1-sin^2(x)}{sin^2(x)}\\=\dfrac{cos^2(x)}{sin^2(x)}=cos^2(x)csc^2(x)

hope this helps

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One angle of the following triangle measures 2x. How many degrees does that angle
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<u>Given</u><u> info</u><u>:</u><u>-</u> In triangle (∆)ABC , in which ∠A = 2x, ∠B = x+15° and ∠C = 2x + 10°. Then find the value of x , also find the measure of each angles of a triangle.

<u>Explanation</u><u>:</u><u>-</u>

Let the angles be 2x, x+15 and 2x+10 respectively.

∵ Sum of the three angles of a triangle is 180°

∴ ∠A + ∠B + ∠C = 180° [Sum of ∠s of a ∆=180°]

→2x + x+15 + 2x+10 = 180°

→ 2x + x + 2x + 15 + 10 = 180°

→ 3x + 2x + 15 + 10 = 180°

→ 5x + 15 + 10 = 180°

→ 5x + 25 = 180°

→ 5x = 180°-25

→ 5x = 155°

→ x = 155°÷5 = 155/5 = 31.

Now, finding the measure of each angles of a ∆ABC by putting the original value of “x”.

∴ ∠A = 2x = 2(31) = 62°

∠B = x+15 = 31 + 15 = 46°

∠C = 2x + 10 = 2(31) + 10 = 62 + 10 = 72°.

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