Question

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Answer 1

From ΔBHD,

m∠DBH + m∠BDH + m∠BHD = 180°

47° + 31° + m∠BHD = 180°

m∠BHD = 180° - 78°

m∠BHD = 102°

Since, ∠BHD and ∠BHC are the linear pair of angles,

m∠BHD + m∠BHC = 180°

102° + m∠BHC = 180°

m∠BHC = 180° - 102°

x° = 78°

Answer 2

Answer:

x° is 66°  

Step-by-step explanation:

From the given diagram, we have;

∠JIH = 105° Given

∠IDJ = 39° Given

Therefore, we have;

∠JID and ∠JIH are supplementary angles, by the sum of angles on a straight line

∴ ∠JID + ∠JIH = 180° by definition of supplementary angles

∠JID + 105° = 180° by substitution property

∠JID = 180° - 105° = 75° by angle subtraction postulate

∠JID = 75°

∠IDJ + ∠JID + ∠IJD = 180° by the sum of interior angles of a triangle

∠IJD = 180° - (∠IDJ + ∠JID)  = 180° - (39° + 75°) = 66° angle subtraction postulate

∠IJD = 66°

∠x° ≅ ∠IJD, by vertically opposite angles

∴ ∠x° = ∠IJD = 66° by the definition of congruency

∠x° = 66°