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kari74 [83]
3 years ago
13

IB QUESTIONS:

Mathematics
1 answer:
enyata [817]3 years ago
4 0

Step-by-step explanation:

Using SOHCAHTOA

\tan =  \frac{opposite}{adjacent}  \\  \tan {31}^{0} =  \frac{450}{x}  \\  \tan {31}^{0} = 0.6009 \\ 0.6009 = \frac{450}{x} \\ cross \: multiply \\ 0.6009x = 450 \\ divide \: both \: sides \: by \: 0.6009 \\ x =  \frac{450}{0.6009}x \\  = 748.9

Second Question Explanation:

a. The base angles of an isosceles triangle are equal so angle ABC is also 32°

but the sum of angles in a triangle = 180° therefore

let angle CAB be x

{32}^{0}  +  {32}^{0} +  {x}^{0} =  {180}^{0}  \\ 64 + x =  180 \\  {x}^{0}  = 180 - 64 \\  {x}^{0}  =  {116}^{0}

b. If you share a triangle in half you get 2 right angles so I'll use SOHCAHTOA to find the length AB

First the base of the right angled triangle will be 20/2 = 10

\cos =  \frac{adjacent}{hypotenuse} \\  \cos {32}^{0} =  \frac{10}{ab} \\  \cos32 = 0.8480 \\ 0.8480 =  \frac{10}{ab}  \\ ab =  \frac{10}{0.8480}  \\ line \: ab = 11.8

c. To find the are we'll need to get the height first

using the right angled triangle With base 10cm and hypotenuse 11. 8 to find the height we'll use the Pythagorean theorem

{(11.8)}^{2}  =  {x}^{2} +  {10}^{2}  \\ 1324.24 =  {x}^{2}  + 100 \\  {x}^{2}  = 139.24 - 100 \\  {x}^{2}  = 39.24  \\  \sqrt{ {x}^{2} }  =  \sqrt{39.24}  \\ x =  \sqrt{39.24 }  \\ x = 6.3

now to find the area

area \: of \: triangle =  \frac{1}{2}  \times base \times height \\   =  \frac{1}{2}  \times 20 \times 6.3 \\  \frac{20 \times 6.3}{2}  \\  = \frac{126}{2}  \\ 63 {cm}^{2}

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