We know that
cos A=adjacent side angle A/hypotenuse
adjacent side angle A=24 units
hypotenuse=26 units
cos A=24/26-----> 12/13
cos B=adjacent side angle B/hypotenuse
adjacent side angle B=10 units
hypotenuse=26 units
cos B=10/26------> 5/13
the answers are
cos A=12/13
cos B=5/13
cot A=adjacent side angle A/opposite side angle A
adjacent side angle A=24 units
opposite side angle A=10 units
cot A=24/10------> cot A=12/5
cot B=adjacent side angle B/opposite side angle B
adjacent side angle B=10 units
opposite side angle B=24 units
cot B=10/24------> cot B=5/12
<span>f(-10)=12, x = -10, y = 12
f(16)=-1, x = 16, y = -1.
so, we have two points, let's check with that,
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
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![\bf \stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)} y-12=-\cfrac{1}{2}[x-(-10)] \\\\\\ y-12=-\cfrac{1}{2}(x+10)\implies y-12=-\cfrac{1}{2}x-5\implies y=-\cfrac{1}{2}x+7](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7Bpoint-slope%20form%7D%7D%7By-%20y_1%3D%20m%28x-%20x_1%29%7D%20y-12%3D-%5Ccfrac%7B1%7D%7B2%7D%5Bx-%28-10%29%5D%0A%5C%5C%5C%5C%5C%5C%0Ay-12%3D-%5Ccfrac%7B1%7D%7B2%7D%28x%2B10%29%5Cimplies%20y-12%3D-%5Ccfrac%7B1%7D%7B2%7Dx-5%5Cimplies%20y%3D-%5Ccfrac%7B1%7D%7B2%7Dx%2B7)
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