MATH SOLVE

4 months ago

Q:
# Angelique draws triangle GHK. If A.1.2 or 5.7B.2.2 or 4.7C.4.7D.5.7

Accepted Solution

A:

the correct question in the attached figure

we know that

applying the law of sines

g/sinG=k/sinK

g=3 units

G=30°

k=4 units

K=?

so

g/sinG=k/sinK------> g*sinK=k*sinG-----> sinK=k*sinG/g

sinK=4*sin 30°/3-----> 2/3

K=arc sin(2/3)-------> K=41.81°

case 1)

for angle K=41.81°

the sum of the angles (H +G+K)=180°

H=180-(G+K)------> 180-(30+41.81)--------> H=108.19°

h/sinH=g/sinG-------> h=g*sinH/sinG------> h=3*sin 108.19°/sin 30°

h=5.70 units

case 2)

for angle K=180°-41.81°---------> K=138.19°

the sum of the angles (H +G+K)=180°

H=180-(G+K)------> 180-(30+138.19)--------> H=11.81°

h/sinH=g/sinG-------> h=g*sinH/sinG------> h=3*sin 11.81°/sin 30°

h=1.2 units

therefore

the answer is the option

A. 1.2 or 5.7

we know that

applying the law of sines

g/sinG=k/sinK

g=3 units

G=30°

k=4 units

K=?

so

g/sinG=k/sinK------> g*sinK=k*sinG-----> sinK=k*sinG/g

sinK=4*sin 30°/3-----> 2/3

K=arc sin(2/3)-------> K=41.81°

case 1)

for angle K=41.81°

the sum of the angles (H +G+K)=180°

H=180-(G+K)------> 180-(30+41.81)--------> H=108.19°

h/sinH=g/sinG-------> h=g*sinH/sinG------> h=3*sin 108.19°/sin 30°

h=5.70 units

case 2)

for angle K=180°-41.81°---------> K=138.19°

the sum of the angles (H +G+K)=180°

H=180-(G+K)------> 180-(30+138.19)--------> H=11.81°

h/sinH=g/sinG-------> h=g*sinH/sinG------> h=3*sin 11.81°/sin 30°

h=1.2 units

therefore

the answer is the option

A. 1.2 or 5.7