Please help with these questions! I don't understand them!
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In ∆s HIJ and HKJ,
Angle I = Angle K (Each 90°)
IJ = KJ (Given)
HJ = HJ (Common)
By RHS, ∆HIJ congruent to ∆HKJ
Angle IHJ = Angle KHJ (CPCT)
3a + 6 = 5a - 18
=> 2a = 24
=> a = 12
Angle IHJ = 3×12+6
= 42°
Angle I = Angle K (Each 90°)
IJ = KJ (Given)
HJ = HJ (Common)
By RHS, ∆HIJ congruent to ∆HKJ
Angle IHJ = Angle KHJ (CPCT)
3a + 6 = 5a - 18
=> 2a = 24
=> a = 12
Angle IHJ = 3×12+6
= 42°
UmangTarang:
The second question is also similar.
No, the second answer is 112°
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