In triangle PQR, <p = 30 degrees
<Q = 80 degrees
<R = 70 degreees
QH is an altitude and RM is a median and H is on the line PR,
the measure of <MHP is:
a) 15 degrees or 20 degrees or 30 degrees or 40 degrees or 45 degreess ?
diagram is attatched
In triangle PQR, <p = 30 degrees
<Q = 80 degrees
<R = 70 degreees
QH is an altitude and RM is a median and H is on the line PR,
the measure of <MHP is:
a) 15 degrees or 20 degrees or 30 degrees or 40 degrees or 45 degreess ?
diagram is attatched
Multiple hellos. It is 30 degrees.
Solution: (All angular measurements in degrees)
$\displaystyle \angle P = 30$ degrees
$\displaystyle \angle Q = 80 $degrees
$\displaystyle \angle R = 70$ degrees
$\displaystyle QH $is an altitude and $\displaystyle RM $is a median and $\displaystyle H$ is on the line $\displaystyle PR$
If $\displaystyle QH $is an altitude, then $\displaystyle \angle QHR$ and $\displaystyle \angle QHP$ are $\displaystyle 90 $.
Filling in relevant angles angles: $\displaystyle \angle PHQ=60$,
Area of $\displaystyle \triangle PQR=\frac{PRQH}{2}$
A median splits a triangle into two triangles equal in area.
Construct altitude $\displaystyle MO$. (dotted line in diagram, forgot to label O)
Two similar triangles are discovered, $\displaystyle \triangle MOP$ and $\displaystyle \triangle PHQ
$ (as $\displaystyle MO$and $\displaystyle QH$are parallel)
Calculate area of $\displaystyle \triangle PMR=\frac{MOPR}{2}=\frac{PRQH}{4}$
(Because a median splits a triangle into two triangles equal in area.)
Hence,$\displaystyle MO=\frac{QH}{2}$
Which means that $\displaystyle PO=\frac{PH}{2}$ and $\displaystyle PM=\frac{PQ}{2}$
Thus $\displaystyle OH=PH-\frac{PH}{2}=\frac{PH}{2}$
Hence $\displaystyle \triangle MPO$ and$\displaystyle \triangle MOH$ are identical
Hence $\displaystyle \angle MHP=30$