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vesna_86 [32]
3 years ago
9

In ΔWXY, if m∠W is five less than three times m∠Y and m∠X is 8 more than m∠W, find the measures of each angles?

Mathematics
1 answer:
yulyashka [42]3 years ago
5 0

Answer:

∡W = 73°

∡X = 81°

∡Y = 26°

Step-by-step explanation:

let 'y' = measure angle Y

let '3y - 5' = measure of angle W

let '3y + 3' = measure of angle X

add all together and set equal to 180

y + 3y + 3y - 2 = 180

7y = 182

y = 26

substitute 26 for y in '3y+3' to find measure of angle X

substitute 26 for y in '3y-5' to find measure of angle W

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\\\text{we need to calculate the height of two different side walls}\ h_1\ \text{and}\ h_2\\\text{use the Pythagorean theorem:}\\\\h_1^2=\left(\dfrac{l}{2}\right)^2+h^2\to h_1=\sqrt{\left(\dfrac{l}{2}\right)^2+h^2}=\sqrt{\dfrac{l^2}{4}+h^2}=\sqrt{\dfrac{l^2}{4}+\dfrac{4h^2}{4}}\\\\h_1=\sqrt{\dfrac{l^2+4h^2}{4}}=\dfrac{\sqrt{l^2+4h^2}}{\sqrt4}=\dfrac{\sqrt{l^2+4h^2}}{2}

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SA=lw+2\cdot\dfrac{lh_1}{2}+2\cdot\dfrac{wh_2}{2}\\\\SA=lw+2\!\!\!\!\diagup\cdot\dfrac{l\cdot\frac{\sqrt{l^2+4h^2}}{2}}{2\!\!\!\!\diagup}+2\!\!\!\!\diagup\cdot\dfrac{w\cdot\frac{\sqrt{w^2+4h^2}}{2}}{2\!\!\!\!\diagup}\\\\SA=lw+\dfrac{l\sqrt{l^2+4h^2}}{2}+\dfrac{w\sqrt{w^2+4h^2}}{2}\\\\SA=\dfrac{2lw}{2}+\dfrac{l\sqrt{l^2+4h^2}}{2}+\dfrac{w\sqrt{w^2+4h^2}}{2}\\\\SA=\dfrac{2lw+l\sqrt{l^2+4h^2}+w\sqrt{w^2+4h^2}}{2}

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Answer:

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Step-by-step explanation:

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