Answer to Question #133220 in Geometry for Nade

Question

∠A and ∠B are complementary. Find m∠A and m∠B.
M∠A= (3x+2)
M∠B= (x-4’)
M∠A=
M∠B=

∠A and ∠B are supplementary. Find m ∠A and m∠B. M∠A=(8x+100) m∠B(2x+50)

M ∠ A =
M ∠B=

Answer 1

Given

$m∠A=(3x+2)$

$m∠B=(x-4)$

And

∠A and ∠B are complementary angel.

So, we know that

$m∠A+m∠B=90°$

After the plug in the value

m∠A and m∠B.

$(3x+2)+(x-4)=90°$

Combining the like tems

$4x-2=90°$

$4x=90°$

$x=23°$

Now we substitute the value of x in the ∠A and ∠B.

$m∠A=(3×23+2)=71°$

$m∠B=23-4=19°$

Given

$m∠A=(8x+100)$

$m∠B=(2x+50)$

And

∠A and ∠B are supplementary angel.

$∠A+∠B=180°$

After plug in the value of m∠A and m∠B.

$(8x+100)+(2x+50)=180°$

Combining the like tems

$10x+150=180°$

$10x=30°$

$x=3°$

Now we substitute the value of x in the ∠A and ∠B.

$m∠A=8×3+100=124°$

$m∠B=2×3+50=56°$