Question

IB QUESTIONS:

Answer 1

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

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