QUESTION IMAGE
Question
tuesday 8/26
each figure shows a triangle with one of its angle bisectors.
- (mangle2 = 27x - 2) and (mangle1 = 25x + 2). find (x).
- (mangle2 = 6x - 5) and (mangle1 = 5x + 1). find (x).
- given (angle r) and (angle t) are complimentary. if (mangle r) is 135 what is the (mangle t)?
- suppose (angle a) and (angle b) are supplementary angles, (mangle a=(3x + 5)^{circ}), and (mangle b=(2x - 10)^{circ}). solve for (x) and then find (mangle a) and (mangle b).
1.
Step1: Use angle - bisector property
Since an angle - bisector divides an angle into two equal angles, $m\angle1 = m\angle2$. So, $27x−2=25x + 2$.
Step2: Solve the equation for $x$
Subtract $25x$ from both sides: $27x-25x−2=25x-25x + 2$, which simplifies to $2x−2 = 2$. Then add 2 to both sides: $2x-2 + 2=2 + 2$, giving $2x=4$. Divide both sides by 2: $x = 2$.
Step1: Use angle - bisector property
Because an angle - bisector makes $m\angle1 = m\angle2$, we have the equation $6x−5=5x + 1$.
Step2: Solve the equation for $x$
Subtract $5x$ from both sides: $6x-5x−5=5x-5x + 1$, which simplifies to $x−5 = 1$. Then add 5 to both sides: $x-5 + 5=1 + 5$, so $x = 6$.
Step1: Recall the definition of complementary angles
Complementary angles add up to $90^{\circ}$. But $m\angle R=135^{\circ}$, and since $135^{\circ}>90^{\circ}$, there is an error in the problem statement as an angle in a complementary - angle pair cannot be greater than $90^{\circ}$. If we assume it was a mis - type and they are supplementary (add up to $180^{\circ}$), then $m\angle T=180^{\circ}-m\angle R$.
Step2: Calculate $m\angle T$
$m\angle T=180 - 135=45^{\circ}$
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$x = 2$