# Why are B-O-H and B-O-C angles in B(OH)3 and B(OCH3)3, respectively are higher than normal?

In Concise Inorganic Chemistry by JD Lee (4th edition; adapted by Sudarshan Guha), page 77 under the topic "Back Bonding" it is given:

The $$\mathrm{2p}\pi$$-$$\mathrm{2p}\pi$$ back bonding in $$\ce{B(OH)3}$$ and $$\ce{B(OCH3)3}$$ explains why $$\ce{B-O-H}$$ and $$\ce{B-O-C}$$ angles are $$112°$$ and $$113°$$, respectively.

Based on the above statement I arrived at the following:

Due to back bonding in $$\ce{B(OH)3}$$ and $$\ce{B(OCH3)3}$$ there is a partial double bond character in the bond between boron and oxygen in both the compounds. The oxygen in both compounds is $$\mathrm{sp}^3$$ hybridised and ideal bond angle is $$109°28'$$. But due to the partial double bond character and as we know the repulsion between a double bond and single bond is greater than between two single bonds the bond angles $$\ce{B-O-H}$$ and $$\ce{B-O-C}$$ are larger than the ideal $$109°28'$$.

Kindly tell whether my understanding is correct or not. If not, kindly give an explanation for the observed bond angles and their deviation from ideality.