Question

PLEASE SOLVE THE PROBLEM AND GIVE ANSWER OF ALL FOUR QUESTIONS

TOMORROW MY MATHS EXAM​

Answer 1

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Answer 2

Step-by-step explanation:

• complementary agles

sum of two angles = 90°

• supplementary angles

sum of two angles = 180°

1) complement of 2/3 rd of right angle

$\frac{2}{3} ( {90)}^{o} + x = {90}^{o} \\ (2 \times 30) + x = 90 \\ 60 + x = 90 \\ x = 90 - 60 \\ x = {30}^{o}$

2) supplement of 3/5th of right angle

$\frac{3}{5} ( {90)}^{o} + x = {180}^{o} \\ (3 \times 18) + x = 180 \\ 54 + x = 180 \\ x = 180 - 54 \\ x = {126}^{o}$

3) measure of the angle which 30° less than it's supplement

• let the angle be x

$x + (x - 30) = 180 \\ 2x - 30 = 180 \\ 2x = 180 + 30 \\ 2x = 210 \\ x = \frac{210}{2} \\ x = {105}^{o}$

x - 30 = 105 - 30 = 75°

4) angle is 4 times it's complement

• let the angle be x

$x + 4x = 90 \\ 5x = 90 \\ x = \frac{90}{5} \\ x = {18}^{o}$

4 times the complement = 4x = 4 × 18

= 72°